ionic equilbrium(class notes) - the...
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Ionic Equilibrium
Dr. S. S. Tripathy
IONIC EQUILIBRIUM
Bronsted-Lowry Theory(BL Theory) of Acids and Bases:Acids: Proton donors Bases: Proton acceptors
Examples: Acids: (i) All known protonic acids eg. HCl, HNO3, H
2SO
4, CH
3COOH, HCN etc..
(ii) H3O+ and NH
4+ ions
Bases: (a) All –ve ions e.g Cl–, CN–, HSO4– etc
(b) Neutral molecules having a lone pair in central atom: eg H2O, NH
3, ROH etc.
Conjugate Adi/Base pair:A pair of species differing from each other by a single H+ ion.eg. HCl/Cl–; CH
3COOH/CH
3COO–; NH
3/NH
4+; H
2O/OH–, NH
3/NH
2–; H
2O/H
3O+ etc.
Relative Strength: If a certain acid is stronger than the other acid, then the conjugate base of the stronger acidis weaker than the conjuagate base of the weaker acid.
HCl > CH3COOH (acid strength) Cl– < CH
3COO– (base strength)
Acid-Base Equilibra:
HCl + H2O Cl– + H
3O+
A1
B1
B2
A2
Here H2O acts as the base in the forward reaction. Since HCl is a strong acid, the equilibrium lies far to the
right, as at moderate concentrations(say 0.1M), it is assumed to be nearly 100% dissociated. However thereversible nature of the reaction should always be borne in mind. By some process, say increasing [Cl–], wecan shift the equilibrium towards left according to LCP. A
1/B
2 are conjugates while B
1/A
2 are conjugates.
CH3COOH + H
2O CH
3COO– + H
3O+
A1
B1
B2
A2
In this case, the equilibrium lies far to the left, as CH3COOH is a weak acid and the degree of dissociation is
small at moderate concentrations. Here also, water acts as the base. This equilibrium can be made to shift tofurther right or left with the help of some manipulation(LCP), which we shall study later in the title “CommonIon Effect”.We find that acids cannot show its acidic property in the absence of a base(acceptor). In this case the solventwater acts as base.
NH3 + H
2O NH
4+ + OH–
B1
A1
A2
B2
Here NH3 is the BL base which accepts proton from H
2O, which acts as acid. We often write NH
4OH in stead
of NH3, which is not correct. NH
3 is weak base and so the equilibrium lies far to the left. The degree of
dissociation is small. Most of NH3 that exists in the dissolved state are merely hydrated NH
3, not in the form of
NH4+ and OH– ions. So without an acid(here H
2O is the acid), a base cannot show basic property. Note that
in some reactions, NH3 can act as acid, donating H+ to a base, and converting to NH
2– which is its conjugate
base.Hence H
2O, NH
3 and many others(see the SAQ below) are amphoteric.
SAQ: Write the formula of the conjugates of the following. If amphoteric, give both the conjugates.H
3PO
4, H
3O+, HSO
4–, HCO
3–, SO
42–, OH–, H
2PO
4–, H
2CO
3, H
2PO
2–, H
2SO
4
Solution: H2PO
4–(b); H
2O(b); SO
42–(b) and H
2SO
4(a); CO
32–(b) and H
2CO
3(a); HSO
4–(a); H
2O(a);
HPO42–(b) and H
3PO
4(a); HCO
3–(b); H
3PO
2(a)(note that H
3PO
2 is a monoprotic acid); HSO
4–(b)
Ionic Equilibrium
Dr. S. S. Tripathy
Relative Strengths of Acids and their conjuagate Bases.If an acid is stronger than another acid, the conjugate base of the first acid weaker than that of the second acidand vice versa. This has already been told before. Just a revision.
HCl > HCN (acid) Cl– < CN– (base)
Let me give a long series of acids with decreasing strength.
Acids: HCl > RCOOH(organic carboxylic acid) > OH (phenol) > H2O > R-OH(alcohol) >
CH≡CH(alkyne) > NH3 > CH
2=CH
2(alkene) > CH
4(alkane) > H
2
Conjugate bases:
Cl– < RCOO– < O-
< OH– < RO– < CH≡C– < NH2– < CH
2=CH– < CH
3– < H–
Arrehenius Thoery: (outdated):Acids: produce H+ in aq. solution (but actually free H+ ion does not exist and H
2O is a player in the acid-base
reaction, not merely a solvent)Base: produce OH– in aq. solution (a base may not produce OH–. In fact OH– itself is a BL base).Merit: It was the starting point for pH calculation.Demerits: (i) H+ actually exists as H
3O+/ H
5O
2+/H
7O
3+ etc.) (ii) NH
3(g) is a base even in the absence of
H2O. It can react with HCl(g) to form dense white fumes(NH
4+Cl–).
(ii) Salt solutions can be acidic or basic too not always be neutral.
Strong Acids: HCl, HBr, HI, HNO3, H
2SO
4, HClO
3, HClO
4 . These 7 acids are strong from among mineral
acids. Their degree of dissociation is nearly 1 even at 1M solution. The order among them is as follows.HI > HBr > HCl ≈ HClO
4 > H
2SO
4 > HNO
3 > HClO
3
Strongest Acid: Super acid: HSbF6 etc.
Weak Acids: Other mineral acids are weak. Eg. H3PO
4, HCN, HF, HNO
2, H
2CO
3, H
2SO
3, H
2S, organic
carboxylic acids( organic sulfonic acids are strong), Phenol etc.
Strong Bases: alkali metal hydroxides/oxides eg NaOH, KOH etc.(more soluble in water)Moderately strong bases: alkaline earth metal oxides/hydroxides: Ca(OH)
2, Sr(OH)
2, Ba(OH)
2 etc.
Weak bases: oxides and hydroxides of other metals eg. Fe(OH)3, Al
2O
3, other BL weak bases like NH
3,
alphatic organic amines, anilines and other organic nitrogen bases.
Basicity of Acids: (a) Monoprotic/monobasic: HCl, HNO3, H
3PO
2 etc.
(b) diprotic/dibasic: H2SO
4, H
2CO
3, H
2S, H
2SO
3, H
3PO
3 etc
(b) Triprotic/tribasic: H3PO
4, H
3AsO
4 etc. (d) Tetraprotic: H
4P
2O
7
Acidity of Bases: (a) Monoacidic: NaOH, KOH etc (b) diacidic: Ca(OH)2, Mg(OH)
2, Fe(OH)
2 etc.
Triacidic: Al(OH)3, Cr(OH)
3 etc.
Salts: (a) Normal: Na3PO
4, K
2SO
4, NaCl, NaH
2PO
2(b) acidic: NaH
2PO
4, Na
2HPO
4, KSO
4 etc
(c) basic salt: Mg(OH)Cl, Fe(OH)2Cl etc.
Ionic Equilibrium
Dr. S. S. Tripathy
LEWIS THEORYAcids: Electron acceptors Bases: Electron donors.Lewis Acids: (i) All +ve ions: Na+, Mg2+, Al3+ etc.(ii) Electron deficient molecules having incomplete octet: BF
3, BCl
3, AlCl
3, BeCl
2 etc.
(iii) Molecules having incomplete d-orbitals: SiCl4, PCl
5, SF
6, IF
7
(iv) Acidic oxides: CO2, SO
3, P
2O
5 etc. (they can accept negative charge: SO
3 + OH– →HSO
4–)
Lewis Bases: (a) All –ve ions (b) neutral molecules having a lone pair in central atom eg. NH3, H
2O, ROH etc.
Hence all BL bases are also Lewis bases.Lewis acid-Lewis base reaction:
NH3 BF3+ H3N BF3 (adduct)
Lewis acids are called Electrophiles and Lewis bases are called Nucleophiles which are mostly used in organicchemistry.SAQ: Indicate which one of the following is Bronstend-Lowery Acid(BLA), Bronsted-Lowry Base(BLB),Lewis Acid(LA), Lewis Base(LB). Indicate if one can be designated with more than one of these.HCO
3–, NH
3, H
2SO
4, AlCl
3, NH
4+, SO
2, Na+, SF
6, CO
32–
Solution: HCO3–: BLB, LB, BLA; NH
3: BLB, LB, BLA; H
2SO
4: BLB; AlCl
3 : LA;
NH4+ : BLA SO
2 : LA Na+ : LA SF
6 : LA; CO
32– : BLB, LB
IONIC PRODUCT OF WATER(KW
) :In water (pure water), there is some dissociation taking place within two H
2O molecules like this
H2O + H
2O H
3O+ + OH–
For simplicity we write dissociation of H2O as follows.
H2O H+ + OH–
][
][][
2OH
OHHKD
Where; KD = Dissociation Constant of water
][][][ 2 OHHOHKD
But [H2O] of pure water = 55.55M
)25(10][][ 014 COHHK W
pKW
= – log10
(10–14) = 14 (250C)K
W = Ionic product of Water
KW
increases with increase in temperature as the degree of ionisation increases with it.
Temperature KW
pKW
pH for pure water100C 0.293 × 10–14 14.53 7.27250C 1.0 × 10–14 14.00 7.00500C 5.47 × 10–14 13.26 6.631000C 51.3 × 10–14 12.23 6.14
Ionic Equilibrium
Dr. S. S. Tripathy
Acid/Base Dissociation:Monoprotic Acid:
(i) HA type:
HA + H2O A– + H
3O+; For simplicity we write
HA H+ + A–
HA
AHK A
(Where KA = Acid Dissociation Constant)
For strong acids: KA = 10 – 1010 and for weak acids; K
A<1.
(ii) Xn+ type:Positive ions like NH
4+, Zn2+ etc. are also acids, while NH
4+ is a BL acid, Zn2+ is a LA.
Both react with water to form H+ ion. This aspect is often called hydrolysis of salt which has been discussed indetail later.
NH4+ + H
2O NH
4OH + H+
(Truly the above equation should have been: NH4+ + H
2O NH
3 + H
3O+)
][
][][
4
4
NH
HOHNHKK HA { molarity of H
2O is disregarded}
KH = Hydrolysis constant of the salt (eg. NH
4+ salt)
Note that acid dissociation of such BL acids which are positive ions are commonly called hydrolysis of the saltcarrying such ions as a part of respective salts eg. NH
4Cl, ZnCl
2 etc.
BASES:(i) Neutral type: (eg. NH
3, R-NH
2, Ph-NH
2, pyridine and other organic nitrogen bases)
NH3 + H
2O NH
4+ + OH–
For simplicity we often write this as : NH4OH NH
4+ + OH–
][
][][
3
4
NH
OHNHKB
; Often we write as : ][
][][
4
4
OHNH
OHNHKB
KB = Base dissociation Constant
(ii ) Ym– type: ( CN–, F– etc.)Negative ions from salts which are conjugates of weak acids like HCN, HF, H
2CO
3 etc. react
with water to form OH–. This is often called hydrolysis of salt carrying such ions eg. KCN, NaF etc..This willbe discussed in details later.
CN– + H2O HCN + OH–
][
][][
CN
OHHCNKK HB { molarity of H
2O is disregarded}
Where; KH = hydrolysis constant of the salt
N.B: OH–, NH2–, RO–, CH≡C–, R–, H– are strong bases as their K
H/K
B values are large(>1) . Except OH–,
others react with H2O to form OH–. The production of OH– is the cause of basicity of the solution, though
according to BL theory, definition of base is : any species which can accept a H+ ion. Note that ions likeCN–, F–, CH
3COO–, though conjugates of weak acids, their basic strength in water is not much. These no
doubt produce basic aq. solution, but their KH/K
B values are low(<1). More to come on this later.
Ionic Equilibrium
Dr. S. S. Tripathy
pH/pOH:In 1909, Sorensen(Danish Biochemist) introduced pH scale for easy expression of [H+], i.e hydrogen ionpotenz for dilute solutions (≤ 10–3).
pH = –log10
[H+]He used the scale range : pH = 0 –14.
Note that truly pH is the measure of activity of hydrogen ion i.e Ha .
pH = –log10 H
a
Where Ha = activity of hydrogen ion in solution
HaH
Where = activity coefficient
For dilute solution, =1, so HaH
, and we can find pH = –log10
[H+]; But for concentrated solutions
and in persence of added salts, pH cannot be determined from the above relation. In stead,we have to useactivity of hydrogen ion, as activity coefficient ( )is often less than 1.
Activity(a) of a speices is the effective concentration of that species in the solution. More about ‘activity’ isgiven at the end of this chapter. After readin that, you will have a changed idea about concentration. But for thischapter we shall use molar concentration i.e [H+] for our pH calculation.
KW
= [H+]×[OH–] ⇒ –log(KW
) = –log[H+] – log[OH–] = –log(10–14) (250C)
⇒ pH + pOH = 14 (250C) where pOH = –log10
[OH–]
Acidic Solution: [H+] > 10–7 and [OH–] < 10–7; so pH <7 and pOH > 7Basic Solution: [H+] < 10–7 and [OH–] > 10–7; so pH > 7 and pOH < 7Pure Water or Neutral Solution:[H+] = [OH–] = 10–7; So pH = pOH = 7
KA and K
B for a Conjugate Acid-Base Pair:
(a) Acids: HA H+ + A– HA
AHK A
A– + H2O HA + OH–
][
]][[
A
OHHAKK HB
KA × K
B = K
W⇒
A
WB K
KK
pKA + pK
B = pK
W =14
(b) Bases:
BOH B+ + OH–;][
]][[
BOH
OHBKB
B+ + H2O BOH + H+
][
]][[
B
HBOHKK HA
Ionic Equilibrium
Dr. S. S. Tripathy
KB × K
A = K
WB
WA K
KK
pKA + pK
B = pK
W=14
N.B: This aspect will be discussed once again in the chapter “hydrolysis of salts”.
SAQ: (a) If KA of HCN is 10–10 at certain temperature, then what isK
B(K
H) of CN– at that temperature ?
(b) KB of CH
3NH
2 is 10–4 at certain temperature, then what is the K
A of CH
3NH
3+ at the same
temerature ?Solution: (a) 10–4; (b) 10–10.
pH of monoprotic acids:
(a) Strong acids:You know that a strong acid is believed to be nearly 100% dissociated at moderate concentration i.e 1M orbelow. So [Acid] = [H+]
SAQ: Find the pH of 0.01M HCl.Solution: [HCl] = [H+] = 10–2 M. pH = – log(10–2) = 2.
IMPORTANT : For concentration upto 10–7 (> 10–7), the H+ ions coming from the dissociation of H2O is not
included for calculation. Only acid concentration is used. But when [acid] ≤10–7 , the H+ from dissociationfrom H
2O(i.e 10–7) has to be included to get the pH in the acidic range, otherwise pH will be greater than 7.
SAQ: Find the pH of 10–8M HCl.Solution: Check: pH = –log(10–8) =8, but 8 indicates basic nature, which is impossible.
[H+]T = 10–8(from acid) + 10–7 (from water) = 1.1 ×10–7.
pH = –log(1.1 ×10–7) = – log 1.1 + 7 = 6.958 (1)More Refined(Accurate) Calculation by taking K
W:
H2O H+ + OH–
HCl → H+ + Cl–
Let the degree of dissociation of H2O is ‘x’ in presence of HCl
[H+]T = (x + 10–8); [OH–] = x
KW
= [H+] [OH–] = x(x+10–8) = 10–14; x = 0.9512 ×10–7
[H+]T = (x + 10–8) = 1.0512 ×10–7; pH = 6.9783 (2)
Although the 2nd method is more accurate, we can use the first method for our calculation.
(b) Weak Acids: (Ostwald’s Dilution Law)HA H+ + A–
Initial 1 0 0Eqm c(1-α) cα cα where α = degree of dissociation
c = initial concentration of HA
)1()1(
2
c
c
ccK A (1) (Ostwald’s Dilution Law)
When α is small, it can be neglected in the denominator.
Ionic Equilibrium
Dr. S. S. Tripathy
CK A2
c
KA (2)
cKcH A ][ (3)
N.B: In stead of using the above Ostawald’s equation, one can solve numericals from KA expression as
follows.
][
][ 2
HA
HK A
(4)
Case-I : α small: While using equations 2, 3 and 4, we have made the approximation that α is small andtherefore it is neglected in the denominator. In eqn. (4), therefore, the initial concentration of HA is used for itsequilibrium concentration.But one must not rely the results of the above equations without checking the approximation. Intutitively onecan have an idea about it from both on K
A and [HA]. It is directly proportional to K
A and inversely to [H+].
First use eqn. (2) to find α, if it is less than 0.1, then the approxiation is justified. Then use either
[H+]=αc or cK A .
Case-II: α large: When comes large (> 0.1), then this result of α is discarded. Using original Ostwald dilutionlaw(quadratic equation), α is calculated again. Then [H+] is determined from [H+]=αc. Note that α becomeslarge on two grounds i.e (a) low concentration or/and (b) high K
A value. For weak acid like acetic acid [K
A =
1.8 × 10–5], α can be neglected upto [HA] = 10–3M. Below this concentration, it cannot be neglected. See thisexample.
SAQ: Find the degree of disscociation and pH of 1.8 × 10–3 M CH3COOH. (K
A = 1.8 × 10–5).
Approximation:
1.0108.1
108.13
5
43 108.1108.11.0][ cH M
pH = 3.74 (a)Since α is 0.1, it should not be neglected. Hence let us solve again for α using quadratic equation.
0108.1108.1108.1 5523 0951.033 101712.0108.10951.0][ cH M;
pH = 3.769 (b)Note that solution (b) is more accurate than (a). But one can avoid solving quadratic equation in this case as αis not very high. If α is high,then approximation method would give unacceptable result.
N.B: Note that in stead of using Ostwald dilution law given in eqn (1), one can solve as follows.
HA H+ + A–
Initial c 0 0Eqm c–x x x
xc
xK A
2
Solving this quadratic eqn. one can get ‘x’.
( Here x = number of moles/L dissociated from ‘c’ mole/L. In this case, α will be calculated on dividing
the c by x. c
x )
Ionic Equilibrium
Dr. S. S. Tripathy
Case-III: [H+] ≤ 10–7:In this case, first [H+] is determined by using original Ostwald’s dilution law and then H+ from water(10–7) to beadded to get the total [H+].
SAQ: Find the pH of 10–7M CH3COOH.
Solution: If you find on approximation method; α will come >1 which is unacceptable.
0108.1108.110 5527 9945.087 10945.9109945.0][ cH
(if you find pH from this value, it will come 7.00239 which is also unacceptable)
[H+]T = 778 109945.11010945.9
pH = 6.7(N.B: More accurate result of 6.79 is obtained by using dissociation of H
2O as explained before)
Important: For weaker acid like HCN (KA ≈ 10–10), pH cannot be obtained correctly even at concentrations
≤10–5M without including dissociation of water, as the [H+] will come ≤10–7M. In this case, H+ from water hasto be included to get correct pH.SAQ: Find the pH of 10–5M HCN. (K
A = 10–10)
Solution: 0101010 101025 31016.3 81016.3][ cH pH = 7.51 (wrong)
[H+]T = 778 10316.1101016.3
pH = 6.88 (acceptable) More accurate result is obtained by using KW
.___________________________________________________________________________________________________________________________________________________________________Charge and Mass Balance Concept(Alternative Approach): (to be studied only for academic interest andnot for solving numericals for examination purpose)Monoprotic Weak Acid:
][][][ AOHH (1) (Charge Balance)
][][ AHACA(2) (Mass Balance)
Where CA = Initial concentration of the acid; [HA] = equilibrium concentration of acid
Note that the ‘Mass Balance’ equation actually gives the mole or concentration balance. From this the massbalance is understood. For finding mass, we have be take the same number of moles of H+.
]][[ OHHKW (3)
][
]][[
HA
AHK A
(4)
From eqn. (1) we get
0][][
][
AH
KH W
(5)
Using eqn. (2) in eqn.(4), we get
][
]][[
AC
AHK
AA (6)
Ionic Equilibrium
Dr. S. S. Tripathy
From eqn. (6) we get;
A
AA
KH
CKA
][][ (7)
Substituting eqn. (7) in eqn. (5) we get
0][][
][
A
AAW
KH
CK
H
KH (8)
0][][][ 23 WAWAAA KKHKKCHKH (9)
This is cubic equation which is applicable for getting the most accurate result at all concentrations. This includesK
W also. Mostly this equation is used for academic interest, never by a student for examination point of view.
The approximation and short-cut method explained before are to be used by a student. The charge and massbalance concepts are given here only for academic interest.1st approxiation: When concentration [H+] ion remains greater than 10–7, we can neglect the dissociation ofwater. We approximate the charge balance equation as follows by neglecting the [OH–] coming from water.
][][ AH (10) (approximate charge balance equation)
Substituting the value of eqn. (10) in eqn (6) we getc.
][
][ 2
HC
HK
AA (11)
0][][ 2 AAA CKHKH (12)
Note that this is the same as using the original Ostwald’s dilution law at lower concentrations when α cannot beneglected.2nd approximation: When α is small (at higher concentration/lower K
A), further approximation can be done.
CA = [HA] (13) (Approximate mass balance eqn)
Using eqn. (13), in eqn. (6), that means neglecting [A–] in eqn. 6, we get;
AACKH ][ (14)
Note that this is the same as neglecting α in the Ostwald’s dilution law done before.
Charge and Mass Balance Concept for Strong monoprotic Acids:HA → H+ + A– (almost completely dissociated)
Mass balance equation can be approximated as follows.C
A = [A–],
Using the above in the charge balance equation we have
AW C
H
KAOHH
][][][][ (15)
0][][ 2 WA KHCH (16)
This gives the accurate [H+] at concentration ≤10–7M.Approximation: At higher concentration (> 10–7M), further approximation can be done as follows.
[H+] = [A–] = CA
This is what we find the pH for monoprotic strong acids at higher concentrations.__________________________________________________________________________________
Ionic Equilibrium
Dr. S. S. Tripathy
BASES:(a) Weak Bases: Neutral molecules like NH
3 and amines: (Ostwald Dilution Law)
We shall be proceeding in the same way as we did for monoprotic acids(HA) to get the Ostwald’s dilution law.
BOH B+ + OH–
Initial 1 0 0Eqm c(1-α) cα cα where α = degree of dissociation
c = initial concentration of BOH
)1(][
]][[
cc
BOH
OHBKB ⇒
)1(
2
c
KB (1)
At higher concentrations, α may be small(check approximation) and can be neglected in the denominator.
cKB2 ⇒
c
KB
⇒ cKcOH B ][
At lower concentrations, when α is large, original Ostwald’s dilution law has to be used to find a and then OH–
concentration is determined from cOH ][ .
When [OH–] ≤ 10–7: then OH– from H2O is included to get correct pH. The most accurate result is obtained
by using the data in KW
.In stead of using Ostwald dilution law, one can solve as follows.
BOH B+ + OH–
Initial c 0 0Eqm c–x x x
xc
xKB
2
( x is obtained by solving the quadratic equation. α is obtained from c
x )
SAQ: Find the pH of 0.00018M NH3 solution (NH
4OH). K
B[NH
4OH] = 1.8 × 10–5 .
(N.B: Coincidentally KA of CH
3COOH and K
B of NH
4OH are identical).
Solution:
316.0108.1
108.14
5
(quite high) So, this has to be discarded.
0108.1108.1108.1)1(
55242
cKB ⇒ 27.0
(Compare this value of α with that obtained for the approximated one)54 1086.4108.127.0][ cOH
pOH = 4.313 pH = 14– 4.313 = 9.687
Ionic Equilibrium
Dr. S. S. Tripathy
(b) Strong Base: (eg NaOH/KOH)Like strong acids, when c
B > 10–7, we take c
B = [OH–], but when c
B ≤ 10–7, [OH–] from H
2O is
to be included.SAQ: Find the pH of (a) 10–5M NaOH (b) 10–8M KOHSolution: (a) [NaOH] = [OH–] = 10–5M ⇒ pOH = –log
10[OH–] = –log 10–5 = 5
⇒ pH = 9(b) [OH–]
T = 10–8 + 10–7 = 1.1 × 10–7;
pOH = 6.95 ⇒ pH = 7.05Taking K
W: (More accurate result)
H2O H+ + OH–
KOH → K+ + OH–
Let the degree of dissociation of H2O is ‘x’ in presence of KOH
[OH–]T = (x + 10–8); [H+] = x
KW
= [H+] [OH–] = x(x+10–8) = 10–14; x = 0.9512 ×10–7
[OH–]T = (x + 10–8) = 1.0512 ×10–7; pOH = 6.9783 pH = 7.0217
Note that this is not to be used as the difference in the results is small. Learn this primarily for academic interest._____________________________________________________________________________________________________________________________________________________________________Charge and Mass Balance Concept in monoacidic Base(BOH): (to be studied only for academic interest)Weak Base(BOH):
[OH–] = [H+] + [B+] (Charge Balance)C
B = [BOH] + [B+] (Mass Balance)
Proceeding in the similar way as we did for monoprotic acid, we shall get the following cubic equation.
0][][][ 23 WBWBBB KKOHKKCOHKOH
Solving we get the accurate value of [OH–] even at very low concentration below 10–7M.At concentrations > 10–7; approximation in the charge balance equation becomes [OH–] = [B+], and theabove cubic equation is reduced to a quadratic one(similar to Ostwald’s dilution law)
0][][ 2 BBB CKOHKOH
For higher concentration, when α is small, we can further approximate; CB = [BOH] i.e neglecting [B+] in the
denominator of KB expresssion, we get
BBCKOH ][
Strong base(NaOH typeFor C
B > 10–7; [OH–] = C
B;
But for CB ≤10–7 ; We can approximate C
B = [B+] in the mass balance equation and substututing this in
the charge balance equation, in the same way as we did for strong acid before, we shall get
0][][ 2 WB KOHCOH
__________________________________________________________________________________________________________________________________________________
Ionic Equilibrium
Dr. S. S. Tripathy
POLYPROTIC WEAK ACIDS:
Case-I: 21 AA KK
H2S H+ + HS– (1) 710
1
AK
HS– H+ + S2– (2) 14102
AK
____________________________________
H2S 2H+ + S2– (3) K = 10–21
In this case, [H+] is determined from 1st dissociation of the acid (1AK ). For [H
2X] > 10–7, C
A = [H
2X].
From eqn. (1), we get
][
][
][
]][[
2
2
21 SH
H
SH
HSHK A
[H+] = [HS–]
From eqn. (2) i.e by using 2AK , we can find [S2–]. In fact,
2AK = [S2–].
We can also use Ostawald dilution law(approximate) for the purpose. cKH A1][
Overall equation H2S 2H+ + S2– ( K = 10–21) can also be used to determne [S2–].
If [H+] from acid is ≤ 10–7 , H+ from from H2O is included to find pH.
SAQ: Find the [H+], [HS–] and [S2–] in a saturated solution of H2S(0.1M). The two acid constants are given
above.Solution:
][
][
2
2
1 SH
HK A
872 101.010][ H 410][ H M (pH =4)
[HS–] = [H+] = 10–4 M;Using the 2nd dissociation, we have
142
10][
]][[2
HS
SHK A
142 10][ S (same as 2AK )
SAQ: Find pH of a 10–7M H2S(aq.) (K values given before)
Solution:
Neglecting α in the denominator, we get; 777 101010][1
cKH A M; Hence, pH = 7(wrong)
Using original Ostwald’s dilution law we get;
772
10)1(
10
618.0
87 1018.610618.0][ cH pH = 7.2 (wrong)
We have to include H2O dissociation.
[H+]T = 6.18 × 10–8 + 10–7 = 1.618 × 10–7; pH = 6.79 (Acceptable)
(More accurate result will be obtained by usng KW
)
Ionic Equilibrium
Dr. S. S. Tripathy
POLYPROTIC STRONG ACIDS:pH of strong dibasic acid eg. H
2SO
4:
H2SO
4 H+ + HSO
4– (1)
1AK = 103
HSO4
– H+ + SO42– (2)
2AK = 10–2
Note that, H2SO
4 is a strong acid on the basis of its 1st dissociation constant. Its 2nd dissociation is weak.
Hence, it is wrong to always write [H+] = 2[H2SO
4]. Only at very dilute solutions i.e [H|SO
4] < 10–4M, we can
approximate [H+] = 2[H2SO
4]. At higher concentrations, it is not the case. The trick to solve problems at these
concentrations is as follows.We have to presume that the 1st dissociation is complete and find out [H+] from that(same as C
A) and from 2nd
dissociation to find out the the additional [H+] from the dissociation of HSO4– and then the total [H+]. Note that
in the case of H2S, the 2nd dissociation was completely disregarded for finding [H+] as 2nd dissociation
constant is much less that 1st dissociation constant and moreover the 1st dissociation constant itself is verysmall.SAQ: Find the pH of 0.1M H
2SO
4. (K values are given before)
H2SO
4 H+ + HSO
4–
Initial 0.1 0 0Eqm 0 0.1 0.1
HSO4
– H+ + SO42– (2)
2AK = 10–2
Initial 0.1 0 0Eqm (0.1-x) (0.1+x) x
210)1.0(
)1.0(
x
xx31044.8 x
MH 10844.01.01044.8][ 3 pH = 0.96
(Note that if we would have taken [H+] = 2 × 0.1 = 0.2M; pH = 0.699; However actual pH = 0.96. In someof the text books, I find that the authors take [H+] = 2 × [H
2SO
4] as matter of rule, which is wrong as per this
author)
SAQ: Find the pH of 0.1M a dibasic acid, H2X, having
1AK = 10-1 and 2AK = 10–2 .
Solution: In this case, we cannot presume the 1st dissociation is complete like the case of H2SO
4. We have to
use first dissociation to find [H+] and [HX–] and use these in 2nd dissociation to find the total [H+]. Note in boththe cases, we have to struggle through quadratic equations. There is no way out for this, as α cannot beneglected. You can check this approximation.
H2X H+ + HX–
1AK = 10–1
Initial 0.1 0 0Eqm (0.1-x) x x
12
10)1.0(
x
xMx 0618.0
HX– H+ + X2–2AK = 10–2
Initial 0.0618 0.0618 0Eqm (0.0618-y) (0.0618+y) y
210)0618.0(
)0618.0(
y
yy31089.8 y
[H+] = 0.0618 + 0.00889 = 0.07069M pH =1.16
Ionic Equilibrium
Dr. S. S. Tripathy
N.B: Note that if you solve the above numerical like H2S by igonoring α in the 1st dissociation, you shall get
pH= 1; Again if you solve like H2SO
4 by treating the first dissociation to be complete, you would have also got
wrong pH, as in case case of H2SO
4, [H+] in 0.1M acid is (0.1 + fraction), but in this case [H+] = (0.0618 +
fraction). Note the difference clearly.
Tribasic Weak Acid H3PO
4:
Like the case for H2S, [H+] is determined from 1st dissociation constant upto 10–3M by neglecting α and
below this concentration by using quadratic form Ostwald’s dilution law or by the alternative method. If [H+] ≤10–7, then H
2O dissociation is included.
SAQ: Calculate [H+], [H2PO
4–], [HPO
42–] and [PO
43–] in 0.1M H
3PO
4(aq.).
1AK =7.1 ×10–3, 2AK =6.3 ×
10–8;3AK = 4.5 ×10–13
Solution:
H3PO
4 H+ + H
2PO
4– (1)
1AK =7.1 ×10–3
32
101.71.0
][
H
MH 21066.2][ = [H2PO
4–]
H2PO
4– H+ + HPO
42– (2)
2AK =6.3 × 10–8
8
42
24 103.6][
]][[
POH
HPOHMHPO 82
4 103.6][
HPO42– H+ + PO
43– (3)
3AK = 4.5 ×10–13
1324
34 105.4][
]][[
HPO
POH13
8
34
2
105.4103.6
][1066.2
PO
1834 100657.1][ PO M
ACID & BASE Dissociation Constant Data Table for some Common Acids/Bases:
Ionic Equilibrium
Dr. S. S. Tripathy
Ionic Equilibrium
Dr. S. S. Tripathy
HYDROLYSIS OF SALTSAlthough some authors like to treat this aspect as hydrolysis of salt, i would prefer to call it as base and aciddissociation of BL base and acid respectively. However for salt of type-C(from weak acid and weak base) iwill have no choice than to call it as hydrolysis of salt, as both the cation and anion undergo dissociation withwater simultaneously. But for the sake of the convenience of students who will be perhaps more familiar inhydrolysis term, i woul be using this too.
(A) Salt of Weak acid and Strong base:Such salts have cation usually from Na+/K+(i.e alkali metal ion) and any one other than Cl–, Br–, I–, NO
3–,
SO42, ClO
3– and ClO
4–. (conjugate bases of the seven strong acids already mentioned)
Examples: KCN, NaF, Na2CO
3, K
3PO
4, HCOONa, CH
3COOK, K
2SO
3, NaClO etc.
Na+A– + H2O HA + Na+ OH–
A– + H2O HA + OH–
Initial 1 0 0Eqm (1-h) h hEqm Molarit c(1-h) ch ch
)1(][
]][[ 2
h
ch
A
OHHAKK HB
(similar to Ostwald Dilution law eqn)
(where h = degree of dissocation = degree of hydrolysisK
B =Base dissociation consant = K
H = Hydrolysis constant of the salt
This is to be called as the base dissociation of the BL base A–. Some authors take it as hydrolysis of such typesalt. The degree of dissociation(α) can also be called degree of hydrolysis and can be given a different symbol‘h’. Needless to say that such a solution is basic due to basic dissociation of A– or salt hydrolysis of A–.
Since ‘h’ is usually small, it can be neglected in the denominator.
2chKK HB c
Kh H
cKchOH H ][
While checking the above approxiation, if you find ‘h’ comes large, then the original quadratic equation needsto be solved. Note that in this case, very rarely you get such situations, as K
H is usually small. If [OH–] drops
to 10–7 or below, then dissociation of H2O is included like the normal acid/base dissociation.Lets see why K
H
is usually small ?In this case, alongwith the equilibria for base dissociation(hydrolysis) of A–, there is acid dissociation of
HA and dissociation of water(KW
).
HA H+ + A–
][
]][[
HA
AHK A
H2O H+ + OH– K
W = [H+] [OH–] =10–14.
Let us multiply KH with K
A.
WAH KOHHHA
AH
A
OHHAKK
][][][
]][[
][
]][[
Ionic Equilibrium
Dr. S. S. Tripathy
A
WBH K
KKK (Since K
W is very small, K
H is also small and so ‘h’)
cK
Kh
A
WA
W
K
cKOH ][
cKKOH AW logloglog2
1]log[ cpKpKpOH AW log
2
1
cpKpKpHpK AWW log2
1 cpKpKpH AW log
2
1
cpKpKpH AW log2
1
The above equation is handy to be directly used to find pH of such a salt solution having molarity of ‘c’.
(B) Salt of Strong Acid and Weak base:Such a salt should contain any cation other than Na+/K+(alkali metal ion) and anion one from Cl–, Br–, I–,NO
3–, SO
42-, ClO
3– and ClO
4–.
Example: ZnCl2, CuSO
4, NH
4Cl, Mg(NO
3)
2, Mn(ClO
4)
2 etc.
B+X– + H2O BOH + H+X–
B+ + H2O BOH + H+
Initial 1 0 0Eqm (1-h) h hEqm. molarity c(1-h) ch ch
)1(][
]][[ 2
h
ch
B
HBOHKK HA
We call this acid dissociation of B+ or hydrolysis of the salt of type (B).h = degree of acid dissociation = degree of hydrolysis
)1(][
]][[ 2
h
ch
B
HBOHKK HA
Neglecting ‘h’ in the denominator we get,
2chKK HA c
Kh H cKchH H ][
Note that you have to deal with it in the same way you did with other weak acids and bases. Here, like the saltsof type (A), ‘h’ is small. So at moderate concentrations, the above equations will give acceptable results.However, when ‘h’ is large or [H+] drops to 10–7 or below, you know how you will tackle.
Needless to say that such solution is acidic in nature due to acid dissociation of B+ or hydrolysis of salt.In addition to equilibrium of acid dissociation(salt hydrolysis), there is also base dissociation of BOH anddissociation of water(K
W).
Ionic Equilibrium
Dr. S. S. Tripathy
BOH B+ + OH–
][
]][[
BOH
OHBKB
WBH KBOH
OHB
B
HBOHKK
][
]][[
][
]][[
B
WH K
KK
Thus cK
Kh
B
WB
W
K
cKH ][
cKKH BW logloglog2
1]log[ cpKpKpH BW log
2
1
cpKpKpH BW log2
1
The above equation can be directly used to get the pH of such solutions.
SAQ: Calcualte pH of (i) 0.1M CH3COONa (K
A = 1.8 ×10–5) and (ii) 0.02M NH
4Cl (K
B = 1.8 ×10–5)
Solution: (i) You can check approximation for ‘h’. It will come very low. So directly we can use the following.
65
14
1045.7108.1
1.010][
A
W
K
cKOH ⇒ pOH = 5.127
⇒ pH = 14 – 5.126 = 8.874
OR: )1.0log()108.1log(142
1log
2
1 5 cpKpKpH AW
87.8174.4142
1 pH
(ii)
65
214
1033.3108.1
10210][
B
W
K
cKH
477.5)1033.3log( 6 pH
OR: )102log()108.1log(142
1log
2
1 25 cpKpKpH BW
479.5699.174.4142
1 pH
(C) Salt of Weak acid and Weak base:Such a salt should contain any cation other than Na+/K+(alkali metal ion) and any another other than l–, Br–,I–, NO
3–, SO
42-, ClO
3– and ClO
4–.
Example: Zn(CN)2, MgF
2, NH
4CN, CH
3COONH
4, NH
4F etc.
B+ + A– + H2O BOH + HA
Initial 1 1 0 0Eqm (1–h) (1–h) h hEqm molarity c(1-h) c(1-h) ch chIn this case both cation and anion of the salt will undergo undergo hydrolysis as they have come from their
Ionic Equilibrium
Dr. S. S. Tripathy
weak conjugates. Here the terms ‘acid dissociation’ and ‘base dissociation’ cannot be used. Here I have nooption than to use the term ‘hydrolysis of salt’. Since both the products of hydrolysis BOH and HA are weak,the ultimate nature of solution is dependent on the relative strength of the two. In any case the solution will benearly neutral and pH can vary within 6 – 8, in the vicinity of mid point 7. If K
A of HA and K
B of BOH are
coincidentally same, then the salt solution will be perfectly netural (pH = 7). The case of CH3COONH
4 is an
example of this type.
2
2
)1()1()1(]][[
]][[
h
h
hchc
chch
AB
HABOHKH
Where KH = hydrolysis constant and h= degree of hydrolysis
As ‘h’ is usually small, it can be neglected in the denominator.2hKH HKh
In this case, in addition to equilibrium for hydrolysis of salt, there is acid dissociation of HA and base dissociationof BOH and K
W as well.
WBAH KBOH
OHB
HA
AH
AB
HABOHKKK
][
]][[
][
]][[
]][[
]][[
BA
WH KK
KK
BA
WH KK
KKh
Let us look to dissociation of HA to find [H+].
ch
hcH
HA
AHK A
)1(][
][
]][[
Neglecting ‘h’ inside parenthesis, we have
h
H
HA
AHK A
][
][
]][[
BA
WAA KK
KKhKH ][
So,B
AW
K
KKH ][
BAW KKKH logloglog2
1]log[
So, BAW pKpKpKpH 2
1
Note that pH is independent of concentration of the reactant.Here we presume that ‘h’ values for B+ and A– are the same. But if K
A and K
B are widely different, then
‘h’values ought to be different . But here the student should keep in mind that if [Salt] > 10–3M, pH isdetermined from the above equation accurately even if K
A and K
B values are widely different. But when the
concentration is lower than this, then other method is to be adopted to get the accurate pH(See later for moreon this).
SAQ: Calculate the pH and degree of hydrolysis of (a) 0.01M CH3COONH
4 (K
A =K
B = 1.8 ×10–5)
(b) 0.01M NH4F [K
A (HF) ] =6.7 ×10–4; K
B = see above)
Ionic Equilibrium
Dr. S. S. Tripathy
Solution:
(a) 00555.01055.5108.1108.1
10 355
14
BA
W
KK
Kh (very small indeed)
774.474.4142
1pH The solution is perfectly neutral as K
A = K
B
(b)4
54
14
101.9108.1107.6
10
BA
W
KK
Kh
215.674.417.3142
174.4)107.6log(14
2
1 4 pH
__________________________________________________________________________________________________________________________________________________________________Charge and Mass Balance Concept for Hydrolysis of a salt type B+A– : (only for academic purpose)
B+ + A– + H2O BOH + HA
][
]][[
HA
AHK A
(1)][
]][[
BOH
OHBKB
(2)
Mass Balance: CA = [A–] + [HA] (3) C
B = [B+] + [BOH] (4)
Charge Balance: [A–] + [OH–] = [B+] + [H+] (5)In a salt solution; C
A = C
B
From eqn. (3), we have [HA] = CA – [A–], Putting this in equn (1), we have
A
AA
KH
CKA
][][ (6)
From eqn. (4) we have, [BOH] = CB – [B+], Putting this in eqn. (2), we have
][
][
][][
][
HKK
HCK
KH
K
CK
KOH
CKB
BW
BB
BW
BB
B
BB
(7)
Using [A–] from eqn. (6) and [B+] from equn. (7) in eqn. (5), we have
][][
][
][][
HHKK
HCK
H
K
KH
CK
BW
BBW
A
AA(7A)
This is a 4th degree eqn. which can be used to find [H+] for any mixture of HA and BOH (CA ≠ C
B ) or salt
solution BA at any concentration(CA = C
B ). The solution of this needs a mathematical software and hence
serves only academic interest.Simplified Approach for BA type of salt solution:For salt of BA type; we have C
A = C
B (same salt). If solution is not too dilute (>10–3M), then we can have
another assumption [A–] = [B+].From eqn. (1), and eqn. (2), we get
][
][][
A
HAKH A
(8) ][
][][
B
BOHKOH B
(9)
Ionic Equilibrium
Dr. S. S. Tripathy
][
][
][
][][][
B
BOHK
A
HAKOHHK BA
W
BA
W
KK
K
B
BOH
A
HA
][
][
][
][(10)
In the charge balance equation (5), since most of H+ and OH– ions react withe each other to form undissociatedH
2O and the small excess of one of these will not affect the charge balance significantly, we can make further
assumption that [A–] = [B+]Since, C
A = C
B = C(say) for salt BA type, we get the mass balance equations as follows.
][][][][ BOHBHAAC ; Since [A–] = [B+], it implies that [HA] = [BOH]So, we can use this in eqn. 10 to get
BA
W
KK
K
A
HA
2
2
][
][
BA
W
KK
K
A
HA
][
][
Substituting in this in eqn.(8) we get
B
AW
BA
WA
A
K
KK
KK
KK
A
HAKH
][
][][
BAW pKpKpKpH 2
1
This we had aready seen before.
For Salt of Strong Base and Strong Acid type:(NaCl type)In the case K
A and K
B are so large that [H+] is neglected in the denominator of the 1st term of LHS in the 4th
degree equn(7A) and KW
term in the denominator of the 1st term in RHS is also neglected. So the 4th degreeeqn is reduced to
][][
HC
H
KC
BW
A Since ABCC for any binary salt
][][
H
H
KW7][
WKH
Thus such salts produce perfectly neutral solutions.Salt of Strong Base and Weak Acid: (CH
3COONa/KCN type)
In such case the cation(B+) does not undergo hydrolysis. So presume 0BC , the 4th degree eqn(8) is
reduced to
0][][][
HH
K
KH
KCW
A
AA
(this can be used for KCN type of salt at any concentrations)Salt of Strong Acid and Weak Base: (NH
4Cl/CuSO
4 type)
In this case, the anion(A–) does not undergo hydrolysis and hence we presume; 0AC , the 4th degree
eqn(8) is reduced to
0][
][][
][
H
KH
HKK
HKCW
BW
BB
(this can be used for CuSO4 type of salt at any concentration)
_______________________________________________________________________________________________________________________________________________________________________
Ionic Equilibrium
Dr. S. S. Tripathy
(D) Salt of Strong Acid and Srong Base:
Such salt will contain cation onen from Na+/K+(alkali metal ion) and anion one from Cl–, Br–, I–, NO3–, SO
42 ,
ClO3– and ClO
4–.
Example: NaNO3, K
2SO
4, NaCl, KClO
4 etc.
Since both cation and anion have come from strong base and acid respectively, they do not undergo anyhydrolysis with water. Hence normal H
2O dissociation occurs there and its a neutral solution (pH =7)
H2O H+ + OH–
pH of Amphiprotic(Amphoteric) Salts:(1) NaHCO
3 i.e HCO
3– type:
Let us find the pH of 0.1M NaHCO3 solution. Given two disociation constants of H
2CO
3 :
1AK =4.5 × 10–7.
2AK = 4.7 × 10–11
HCO3– H+ + CO
32– (1)
2AK =4.7 × 10–11
HCO3– + H
2O H
2CO
3 + OH– (2)
8102.21
A
WHB K
KKK
Since KB > K
A, the solution will be basic. But we cannot find the pH from eqn. (2), because the two are not
widely different. Most of OH– produced from equation (2) will be eaten away by H+ ions produced fromeqn.1. But there is will be preponderance of OH– after the neutralisation.Here we shall have to make an approximation in the charge balance concept. That after self neutralisationbetween H+ and OH–, no appreciable effect would be there on charge balance from the excess of OH–. Since[Na+] = [HCO
3–], for every HCO
3– ion removed to form CO
32–, another HCO
3– must disappear to form
H2CO
3 to make charge balance. So the presumption;
[CO32– ] = [H
2CO
3] = x (say)
So [HCO3
–] = (0.1 – 2x) ≈ 0.1(2x is neglected. We can find the value of ‘x’ to check approximation)
][
]][[
3
23
2
HCO
COHK A
1.0
][107.4 11 xH
(3)
][
]][[102.2
3
328
HCO
OHCOHKK HB
1.0
][102.2 8 xOH
(4)
Dividing eqn (3) by eqn (4) we get
38
11
1012.2102.2
107.4
][
][
OH
H 31012.2
][
][
HK
H
W
3142 1012.210][ H 91061.4][ H M
pH = 8.3Note that the pH of salt is independent of the salt concentration as that term vanished during division.
Ionic Equilibrium
Dr. S. S. Tripathy
General Formula:
Dividing the expression for 2AK and K
B as given above we get,
B
A
K
K
OH
H2
][
][
W
AA
W K
KK
K
H21
2][
21][ AA KKH
212
1AA pKpKpH
Note this relation is valid when [Salt] is high ( >> 1AK ) and moreover
2AK and KB are not widely different.
(2) NaH2PO
4 (H
2PO
4– ion) :
H2PO
4– H+ + HPO
42– (1)
2AK = 6.3 × 10–8
H2PO
4– +H
2O H
3PO
4 + OH– (2)
123
14
10408.1101.7
10
1
A
WHB K
KKK
Since 2AK is greater than K
B, the solution is acidic.
Like the previous case, we can presume; [HPO42–] = [H
3PO
4]; Proceeding in the same manner as we did for
HCO3–, we get the pH of the solution, which is independent of salt concentrations, but with restrictions.
212
1AA pKpKpH
(3)Na2HPO
4 (HPO
42– ion) :
HPO42– H+ + PO
43– (1)
3AK = 4.5 × 10–13
HPO42– +H
2O H
2PO
4– + OH– (2)
78
14
10587.1103.6
10
2
A
WHB K
KKK
Since KB is greater than
3AK , solution will be basic. Like the previous cases, wepresume [PO43–]=[H
2PO
4–]
322
1AA pKpKpH
(4) Salt containing protic cation and protic anion : NH4HCO
3 or NH
4HS etc.
There are three equilibria here.
NH4+ + H
2O NH
4OH + H+ K
A= K
H = K
W/K
B
HCO3– H+ + CO
32– K
A2
HCO3– +H
2O H
2CO
3 + OH–; K
B = K
H = K
W/K
A1
Taking all the three, we shall get [H+] as follows: (you can try derivation yourself)
21 AB
WA K
K
KKH
Ionic Equilibrium
Dr. S. S. Tripathy
__________________________________________________________________________________________________________________________________________________________________Charge and Mass Balance Concept for Amphiprotic salt: (HA– type)Dissociations for H
2A:
H2A H+ + HA– (1) ][
]][[
21 AH
HAHK A
HA– H+ + A2– (2) ][
]][[ 2
2
HA
AHK A
Balance for salt hydrolysis:
][][][ 22
HAAAHCHA
(mass balance)
][][][ 22 AHAH (3)
The above relation is due to the fact that in H+ is formed in 2AK of the HS– and disappears in K
B(K
H)
of the HS–.
HA– H+ + A2– HA– + H2O H
2A + OH–
Using eqn. (1) and (2) in eqn. (3), we have
1
2]][[
][
][][
A
A
K
HAH
H
HAKH
1
21
][
][][ 2
A
AA
KHA
HAKKH
Case-I : ][ HACHA
, which is the case with all concentrations, the above equation is reduced to
1
21][AHA
HAAA
KC
CKKH
This relation is is to be used to find [H+] at all concentrations.
Case-II: If 1AHA
KC ; 1AK can be neglected in the denominator, i.e at higher concentration of the salt.
21][ AA KKH
___________________________________________________________________________________________________________________________________________________________________
COMMON ION EFFECT:The degree of dissociation of a weak acid or base is lowered in the presence of a strong electrolyte furnishinga common ion with respect to the weak acid or weak base. LCP can be used to explain the phenomenon.
CH3COOH CH
3COO– + H+
CH3COONa CH
3COO– + Na+
In the presence of CH3COONa which is a strong electrolyte and ionic, the dissociation of CH
3COOH is
further lowered. The common ion here is CH3COO–, which is present in abundant quantity from the strong
electrolyte. Thus according to LCP, the equilibrium (1) shifts to further left. Thus degree of dissociation islowered. Same thing would have happened if we would take HCl in stead of CH
3COONa. In case of HCl the
common ion is H+. While solving numericals on common ion effect, we have to make approximation that theconcentration of common ion is equal to concentration of the strong electrolyte. The contribution from weakacid or base is ignored. Firstly it is a weak electrolyte and moreover its dissociation has been futher supressed.
Ionic Equilibrium
Dr. S. S. Tripathy
(2) NH4OH NH
4+ + OH–
NH4Cl NH
4+ + Cl–
The degree of dissociation is further lowered in presence of a strong electrolyte like NH4Cl due to common ion
effect.N.B: Note that in the next section, we shall know that a mixture of CH
3COOH/CH
3COONa or NH
4OH/
NH4Cl are named as buffer solutions, which have special capacity to resist the change of pH.
In heterogeneous ionic equilbrium, we shall see that the solubility of solid salts in water is lowered due tocommon ion effect.SAQ: Calculate the [H+] of a solution which is 0.3M in CH
3COOH and 0.2M in CH
3COONa. (Use K
A of
CH3COOH)
Solution:
CH3COOH CH
3COO– + H+
CH3COONa CH
3COO– + Na+
][
]][[
3
3
COOHCH
HCOOCHK A
55
107.22.0
3.0108.1][
H M (1)\
In this case, we used the molarity of salt(strong electrolyte) i.e CH3COONa for [CH
3COO–] as a strong
electrolyte is believed to be nearly completely dissociated. Moreover, the contribution from the weak acidCH
3COOH is negligible.
More refinement in calculation can be done by solving a quadratic equation as follows.
CH3COOH CH
3COO– + H+
Initial 0.3 0.2 0(0.3–x) (0.2+x) x
We can get the expression for KA and solve the quadratic equation. But you are also sure that such solution
would not lead to any appreciable change in the [H+] value. Because, we can check our approximation that xis too small against 0.2 and 0.3. Hence to be neglected inside the two parenthesis and calculated as shownbefore to get [H+]. Of course, when the acid is too dilute, then we have to go for the quadratic eqution.
In the abosence of CH3COONa i.e for only acid solution of 0.3M strength, [H+] is as follows.
][
][
3
2
COOHCH
HK A
MH 35 103.23.0108.1][ (2)
Due to common ion effect, the degree of dissociation has bee decreased and so [H+] is nearly 100 times lowerin the presence of common ion effect(1).Without solving more numericals on common ion effect, let us study buffer solutions, in which the guidingprinciple is common ion effect.SAQ: Calculate [S2–] in a saturated solution of H
2S(0.1M) in presence of 0.1M HCl. Compare this with [S2]
in saturated solution(0.1M) solution of H2S. Also calculate the [HS–](Import two K
A values of H
2S from
previous SAQs)Solution:
H2S H+ + HS– (1) 710
1
AK
Initial 0.1 0 0Eqm (0.1–x) (0.1+x) xSince x will be small, it can be deleted from parentheis. Virtually [H+] = 0.1M from the added HCl
Ionic Equilibrium
Dr. S. S. Tripathy
1.0
][1.010 7
HS710][ HS M (see how small it is =x)
Then you can find [S2–] by using 2nd dissociation.
HS– H+ + S2– (2) 14102
AK
7
214
10
1.0][10
S
202 10][ S M
Alternatively for only [S2–], we can use
H2S 2H+ + S2– K
A = 10–21.
][
][][10
2
2221
SH
SH
202
212 10
)1.0(
1.010][
S M
In saturated solution of 0.1M H2S, [S2–] =
2AK = 10–14. In presence of 0.1M HCl, it is 10–20M. Hence it has
been decreased due to decrease in the degree of dissociation of H2S(common ion effect)
BUFFER SOLUTIONSSolutions which have the capacity to resist the change of pH by adding a small quantity of strong acid/base ordiluting with small quantity of water.Types: (a) Acidic (b) BasicAcidic Buffer: It consists of a weak acid and a salt containing the conjugate base of the weak acid.Examples: CH
3COOH + CH
3COONa; HF+KF; HCN/KCN etc.
HA H+ + A– Presume CHA
= [HA] i.e α is very small.
NaA Na+ + A– [A–] = [NaA}; salt is completely dissociated
][
]][[
HA
AHK A
][
][][
A
HAKH A
][
][loglog]log[
HA
AKH A
Acid
SaltpKpH A [
][log (Henderson-Hasselbalch Equation)
Limitation of Herderson-Hasselbalch(HH) equation:* Note that H.H equation holds good when acid and salt concentrations are higher( i.e >10–3M). In caseconcentrations are lower than 10–3, fresh calculations are made to find [HA] and [A–] from the equilibrium ofHA. If concentration drops to 10–7 or lower, then dissociation of H
2O is taken into consideration. You will get
feel of all these in the succeeding SAQs.* HH eqauation is also not applicable when K
A of the weak acid is relatively higher or pK
A lower(< 2.5).
In this case also fresh calculations as stated above have to be made.Preparation of Buffer solution:Buffer solution can be prepared by directly mixing the weak acid(HA) with salt(NaA) in aqueous solution orby carrying partial neutralisation of HA(weak acid) with NaOH(strong base), such that NaOH is taken aslimting reagent. This will produce NaA(salt) alongwith excess HA forming buffer solution.SAQ: Calculate pH of a buffer which is 0.05M in CH
3COOH and 0.05M in CH
3COONa. (use K
A)
Ionic Equilibrium
Dr. S. S. Tripathy
Solution: Since the concentrations are not low and KA is not high, we can safely use HH equation.
74.4074.405.0
05.0log)108.1log( 5 pH
SAQ: Calculate the pH of 10–5M CH3COOH + 10–5M CH
3COONa buffer.
Solution: Since the concentration is low, α cannot be neglected. Fresh calculation is necessary.
HA A– + H+
10–5 10–5 0(10–5–x) (10–5+x) x
55
5
108.1)10(
)10(
x
xx61039.5 x pH = 5.3
Note that if you would have solved by HH equation, then pH would have been 4.74 like the previous example.This would be wrong.N.B: If the molarites are 10–7 or below, then first you find [H+] by the above method by solving the quadraticequtaion. Then add 10–7 from H
2O to find acceptable pH.
SAQ: Calculate pH of a solution which is 0.1M in dichloroacetic acid and 0.05M in NaOH. (pKA=1.5).
Solution: Instantaneous molarities are given. There will be chemical reaction to give salt and excess weakacid(buffer solution). NaOH is the limting reactant.
[HA]= 0.1–0.05 =0.05M [A–] = 0.05MSince the pK
A of acid is low, we have to calculate fresh and cannot use HH equation even though the
concentrations are higher.pK
A = 1.5; K
A = 0.0316.
Since the limting reactant is NaOH, a buffer solution will form.
HA A– + H+
0.05 0.05 0(0.05–x) (0.05+x) x
0316.0)05.0(
)05.0(
x
xxMx 016.0 ; So pH = 1.7
Note that if you use HH equation, then pH would be 1.5, which is wrong.
Basic Buffer:It consists of a weak base and salt carrying the conjugate acid of the weak base.Examples: NH
4OH+NH
4Cl; C
5H
5N(pyridine)+ C
5H
5NH+Cl–; C
6H
5NH
2(aniline) + C
6H
5NH
3+Cl–;
Many other organic nitrogen bases alongwith salts carrying their conjugates also make basic buffer.
BOH B+ + OH– Presume CBOH
= [BOH] i.e α is very small.
BX B+ + X– [B+] = [BX]; salt is completely dissociated
][
]][[
BOH
OHBKB
][
][][
B
BOHKOH B
][
][loglog]log[
BOH
BKOH B
][
][log
Base
SaltpKpOH B (Henderson-Hasselbalch Equation)
Ionic Equilibrium
Dr. S. S. Tripathy
Limitation: This is applicable at higher concentrations and higher pKB values. At lower concentrations and
lower pKB values, fresh calcuations are to be made just like acidic buffer done before.
SAQ: Find the pH of a buffer which is 0.2M in NH4OH and 0.1M in NH
4Cl. (pK
B = 4.74). If 0.01 mole of
HCl is added per litre of solution, what is the new pH. Comment on the change of pH.Solution:
439.4301.074.42.0
1.0log74.4 pOH ; pH = 14 – 4.439 = 9.561
0.01 mole of H+, reduce [OH–] by that amount and hence there will be increase in [NH4+] by that
amount and decrease in [NH4OH] by that amount too. For details, read the next item i.e buffer action.
503.4237.074.401.02.0
01.01.0log74.4
pOH ; pH = 9.497
Comment: The pH changes to a small extent by adding 0.01 mole of strong acid per litre. This change is notappreciably large. If the amount added is less than this, the change will be still less. Hence every buffer has acapacity within which it resists the change of pH effectively. We shall know about this just after some time.
N.B: Note that if you add 0.01 mole of HCl to a 0.2M NH4OH, then partial neutralisation would occur to
result a buffer solution with 0.01M NH4+ and 0.19M NH
4OH . Then you can calculate the pH as usual.
In stead, if you add 0.01 mole of NaOH with 0.2M NH4OH, then the pH change will appreciably large, as the
resulting solution is not a buffer.
Buffer Action:Acidic Buffer:(a) Addition of strong acid:
HA H+ + A– (1)
NaA Na+ + A– (2)When a small quantity of strong acid (HX) is added to an acidic buffer, the eqm. (1) shifts a bit to left. There isa slight decrease in [A–] and slight increase in [HA]. But they are already present in large excess for which theration [Salt]/[Acid] in HH equation will not change significantly. Hence pH remains nearly constant.(b) Similarly if a small quantity of base is added to an acidic buffer, the OH– will remove H+ forming H
2O. The
equilibrium will shift to bit right. Thus there is a slight increase in [A–] and decrease in [HA]. But they are alradypresent in excess. So their ratio will remain nearly constant and hence pH.Basic Buffer:(a)
BOH B+ + OH– (1)
BX B+ + X– (2)
(a) When we add a small quantity of strong base(NaOH), eqm. (1) shifts to a bit left. Thus there is a decreasein [B+] and bit increase in [BOH] concentration. But they are already present in large excess for which the ratio[salt]/[base] remains nearly constant. Thus pOH/pH does not vary much. This is similar to addition of strongacid to an acidic buffer.(b) When we add a small quantity of acid to a basic buffer, eqm (1) shift to right as OH– is removed by addedH+. Thus the slight increase in [B+] and decrease in [BOH] cannot change their ratio to any appreciable extent.So the pH remains nearly constant. This is similar to addition of base to an acidic buffer.
Ionic Equilibrium
Dr. S. S. Tripathy
The resistance to change of pH will occur only when you add strong acid or base to a buffer within certain limit.Hence every buffer has a capacity. Greater is its capacity, greater is its resistance to change in pH.Buffer Capacity(β)
dpH
dn
pH
n
Where Δn = no. of moles of strong acid/base added to 1.0 L of buffer solution.ΔpH = Change of pH.
So buffer capacity of a buffer solution is the number of moles of strong acid/base added to one litre of solutionfor unit change of pH.Greater the β value, greater is its resistance to change of pH and hence more efficient is the buffer. Theinstantaneous values of β varies with pH between pure acid(HA) and pure pure base(A–). The maximumappears close to pK
A.
For acidic buffer, the following equation can be used to find β.
2][
][][
][303.2
HK
HKCH
H
K
A
AbuffW (derivation not given here)
Where Cbuff
= [HA] + [A–]
Theory of Indicators(Titration Curves):Indicators are organic dyes which are either weak acids or weak bases which are used to get equivalence pointof a titration by its sharp colour change(end point). Let us represent an acidic indicator as follows. In– representsthe conjugate base of the indicator.
HIn H+ + In–
][
]][[
HIn
InHKIn
][
][][
In
HInKH In
][
][loglog]log[
HIn
InKH In
][
][log
HIn
InpKpH In
(where KIn = Indicator Constant)
According to Ostwald’s theory of indicators, HIn has a distinct colour while In– has another distinct colour.
When 100][
][In
HIn, then HIn colour dominates while that of In– is masked.
Ionic Equilibrium
Dr. S. S. Tripathy
When 10][
][
HIn
In, then In– colour dominates while that of HIn is masked. Note that In– has greater colour
intensity, that is why this dissimilarity.Hence for every indicator, there is a pH range, below which HIn colour prevails and above which In– colourprevails.In presence of an external acid, eqm (1) shifts to further left and HIn colour prevails, while in presence of anexternal base, eqm (1) shifts to further right and In– colour prevails. But in between this pH range, the colour ismixed does not carry any diagnostic importance.Litmus is red for HIn colour and blue for In– colour.SAQ: Calculate the pH range of litmus(K
In = 10–7).
Solution:
5100
1log7 pH 8
1
10log7 pH
Hence the pH range of litmus is 5 – 8. Below pH =5, the colour is red(HIn colour) and above pH=8, the colouris blue(In– colour). In between pH = 5 – 8, the colour is mixed. This range will be byepassed in titrationdrasticallay during titration, so no need to bother about it. Read next.
* In basic indicator : InOH In+ + OH–;][
]][[
InOH
OHInKIn
Like acidic indicator, InOH has a certain distinct colour and In+ has different distinct colour. We can find the pHrange of such indicator and its direction of equilibrium in presence of external acid or base can be known. Inpresence of external acid, equilibrium shifts to right and In+ colour prevails and in presence of external base, theequilibrium shifts to left and InOH colour prevails.Note that most indicators belong to acidic category.
Some Common Acidic Indicators:
Indicator Colour in colour in pKIn
pH rangeacid base
Thymol blue red yellow 1.5 1.2 – 2.8(1st change)Methyl Orange red yellow 3.7 3.2 - 4.4Bromocresol green yellow blue 4.7 3.8 – 5.4Methyl Red yellow red 5.1 4.8 – 6.0Litmus red blue 7 5.0 – 8.0Bromothymol blue yellow blue 7 6.0 – 7.6Phenol Red yellow red 7.9 6.8 – 8.4Thymol blue yellow blue 8.9 8.0 – 9.6(2nd change)Phenolphthalein colourless pink 9.4 8.2 - 10.0Thymolphthalein yellow red 10.0 9.3 – 10.5
In acid-base titration the indicator is so chosen that the drastic colour change of indicator(end point) andequivalence point(exact end of reaction) occur simultaneously. In other words, the pH range of the indicatorfall within the equivalence point range(vertical portion of the graph) in the titration curve(see below).
Ionic Equilibrium
Dr. S. S. Tripathy
Titration Curves:(A) Strong Acid with Strong Base: (monoprotic acid)When we titrate NaOH solution gradually from a burette to strong monoprotic acid solution, eg HCl taken inconical flask, pH increases as shown in the diagram. At equivalence point, there is a sharp change of slope. Theexact equivalence point cant be monitored through a pH meter interfaced with a computer. At equivalence
point dV
pHd= 0, where V= volume of alkali added.
The graph will have long vertical portion and the mid point of the vertical portion is the exact equivalence point.In this case it is pH = 7. Such salt solutions are exactly neutral as there is no hydrolysis.All types of indicators are suitable for this titration. The pH range of all indicators(expect thymol blue’s firstchange) fall within the long vertical portion of equivalence for such case (3 – 11).(B) Weak Acid with a Strong BAase (monoprotic acid) :When we gradually add NaOH solution from a burette to a weak acid solution(eg CH
3COOH) taken in a
conical flask, then buffer solution is formed with changing composition. Hence pH changes very gradually.The equivalence point appears in the alkaline region due to salt hydrolysis. The vertical portion near equivalencepoint is shorter in this case( pH = 8 –11).
At half neutralisation, pH = pKA as [Salt] = [Acid]. In case of CH
3COOH, pK
A = 4.74.
Phenolphthalein, Thymol blue(2nd change), thymolphthalein are suitable for such titration.
Ionic Equilibrium
Dr. S. S. Tripathy
(C) Polyprotic acid with Strong Base:Case-I: H
2SO
4(strong diprotic acid) with NaOH:
The pKA1
and pKA2
values are not widely different and moreover the pKA1
is –ve. The first half neutralisationto produce HSO
4– does not give a sharp equivalence point(no detectable vertical portion), which falls in the
acidic region(E1). But the 2nd equivalence point is sharp and occurs at pH =7 (E
2).
Case-II: Polyprotic weak acids like H2S, H
2CO
3(diprotic), H
3PO
4(triprotic):
In such acids, the successive pK values are widely different. Hence as many replaceable H atoms are there,same number of vertical portions(jumps) with detectable inflection points are observed. H
2S gives two sharp
jumps while H3PO
4 give three jumps. The mid point of each jump is the equivalence point of the respective
neutralisation. Before each jump respective pKA value will appear when pH = pK
A in the buffer solution.
(Diprotic weak acid like H2CO
3) (Triprotic weak acid like H
3PO
4)
(D) Strong Acid with Weak Base:Example: NH
4OH + HCl
When HCl is gradually added to a fixed volume of weak base(say NH4OH), buffer solution is formed for
which pH decreases very gradually from the highest value.The equivalence point appears in the acid region due to salt hydrolysis. The vertical portion lies between pH=3– 6.5. So methyl orange, methyl red and bromocresol green are suitable indicators for such titration.
Ionic Equilibrium
Dr. S. S. Tripathy
(Strong Acid HCl with weak base NH4OH) (Weak Acid with Weak Base)
(E) Weak Acid with Weak Base:Example: CH
3COOH + NH
4OH
The titration curve in such case has no sharp vertical portion(inflection point) i.e no sharp jump near equivalence
point. So indicator method is not suitable for such titration. The point of inflection, dV
pHd= 0 is obtained
accurately by carrying out the titration in a pH meter. See the graph given above.
(F) Titration Curve for Na2CO
3 with HCl:
This is opposite to the addition of NaOH to H2CO
3. In this case, first equivalence point comes at pH
=8.3(NaHCO3) and second at about pH ≈ 4(H
2CO
3)
(Titration of Na2CO
3 with HCl) (Titration of glycine with NaOH)
(F) Titration of Glycine(ααααα-amino acid) with NaOH :Starting from lower pH in which glycine remains as a cation, addition of NaOH gives a zwitterion at isoelectricpoint(pI) and before that pK
1. After pI, it gives pK
2 before 2nd inflection.
H3N+
CH2COOHpK1
H3N+
CH2COO-
pI
pK2H2N CH2COO
-
Ionic Equilibrium
Dr. S. S. Tripathy
Approximate pH of some common substances:Substance pH Substance pHHCl(1M) 0 Coffee 5Battery acid 0.5 Tea or healthy skin 5.5Gastric juice 1.5-2.0 Milk 6.5Lemon juice 2.4 Pure water 7.0Cola 2.5 Human saliva 6.5-7.4 (healthy >7)Vinegar 2.9 Blood 7.34-7.45Orange or apple juice 3.5 Seawater 7.7-8.3Tomato juice 4.1 Tap water depends on region(usually 7-8)Beer 4.5 Hand soap 9.0-10.0Acid Rain <5 Household ammonia 11.5Human urine 4.5-8 caustic soda(1M) 14(healthy person pH ≈ 7)Carrot juice 6.0
MORE ABOUT INDICATORSLitmus: it is a purple colour dye extracted from lichen(Roccella tinctoria). It consists a mixture of several dyesand the most important among them is 7-hydroxyphenoxazone(a chromophore). Red in acid and blue in base.
N
O
O
HO
(7-hydroxyphenoxazone)
Methyl Orange: Yellow in base and organe(pink) in acid. It is an azo dye whose structure is given below. Itis prepared by the reaction of N,N-dimethylaniline with diazotized sulfanilic acid.
Methyl Red: Yellow in basic medium and red in acid medium. It is prepared by the reaction of N,N-dimethylaniline with diazotized anthranilic acid.
Phenolphthalein: Pink(or magenta) in base and colourless in acid. It is prepared from 2 moles of phenol andone mole of phthalic anhydride in presence of ZnCl
2/AlCl
3 catalysts.
Ionic Equilibrium
Dr. S. S. Tripathy
Universal Indicator : Natural: contains the extracts from red cabbage, red onion, red hibiscus(Mandarflower): Show different colour at different pH.Synthetic: It consists of a mixture of indicators such as phenolphthalein sodium salt, methyl red, bromothymolblue sodium salt, thymol blue sodium salt dissolved in water containing propan-1-ol and NaOH.pH paper marketed is prepared from this synthetic universal indicator.The approximate colour range of a solution is known from the pH paper.pH paper.Colour Range : pH < 3 : strong acids : red ; pH = 3-6 : yellow : pH = 7 (light green), pH=8: greenpH : >9 Blue,
Olfactory Indicators:Become odourless in alkali(base) but restores the original odour in acidExamples: Onion, Vanila and Clove Oil
OH
OCH3
(eugenol : clove oil) 4-allyl-2-methoxyphenol
OH
OCH3
C
O
H(vanillin : vanilla essence : 4-hydroxy-3-methoxybenzaldehyde)
SH (thiophenol : from onion)OH
(Thymol)
Theories of Indicators:(A) Ostwald’s Thoery : Already discussed.(B) Benzenoid-Quinonoid Theory:This theory compliments Ostwald’s theory. Most dyes(indicators) have aromatic(benzene) structure. In caseof some indicators, it remains in benzenoid form in acidic medium and in quinonoid form in basic medium .Some other indicators the opposite is the case. That means it remains in benzenoid form in basic medium andin quinonoid form in acidic medium. These forms have different colours and hence the colour change.Methyl Orange:
(benzenoid form in basic medium: yellow)
Ionic Equilibrium
Dr. S. S. Tripathy
Methyl Red:
Phenolphthalein:
Heterogeneous Equilbrium:So far, we have been discussing homogenous ioinc equilibrium. Acids and bases dissolve completely in waterbut the degree of dissociation may be high or low. All the species in RHS and LHS in equilibrium were (aq) andhence homogneous.However the solubility of a solid in a solvent like water to form a saturated solution can give rise to heterogenousequilibrium. If some excess solid remains in equilibrium with the aqueous phase of the saturated solution, thenthat is the case,we are now interested in. In the soluble aqueous part, the solute predominantly remains in ionicform if its an ionic solid and may be some of them remain as undissociated molecules.
Solubility Product(KSP
)It is the equilbrium constant for the the solubility of a solid in water. The term is more significant for sparinglysoluble ionic substances like AgCl, BaSO
4 etc. though their solubilty and hence solubility product for a lay
person’s point of view is shockingly and negligibly small. But their study is of great importance in analyticalinorganic chemistry.
Defintion of KSP
:It is the product of molar concentrations of all the ions raised to the power their stoichiometric coefficientsformed by the dissociation of a solid to form a saturated solution at a fixed temperature.* Solubility: Every solid has a fixed solubility at a given temperature. Solubility usually means the number ofmoles of solute dissolved per litre to form a saturated solution at fixed temperature. Sparingly soluble solids likeAgCl, BaSO
4 have very low solubility while highly soluble solids like NaCl, Mg(NO
3)
2 etc. have high solubilities.
Ionic Equilibrium
Dr. S. S. Tripathy
Examples:
(1) AgCl(s) Ag+ + Cl–
SC AgCl
ClAgK
][
]][[
]][[][ ClAgAgClK SC
Since the molarity of a pure solid is fixed, it is deletd from the constant.]][[ ClAgKSP
KSP
is the solubility product which only is the product of ionic concentrations, of course, raised to thepower their coefficients. In this case coefficients are each 1. K
SP of a solid is fixed at a given temeperature.
(2) PbCl2(s) Pb2+ + 2Cl–
22 ][][ ClPbKSP
(3) Al(OH)3 Al3+ + 3OH–
33 ][][ OHAlKSP
Relationship between Solubility and Solubility Product:(1) AgCl type:
AgCl(s) Ag+ + Cl–
Let solubity of AgCl in water at 250C = x mol.L–1. So; [Ag+] = x and [Cl–] = x2xxxKSP SPKx
(2) PbCl2 type:
PbCl2(s) Pb2+ + 2Cl–
Let solubility of PbCl2 = x mol.L–1. So, [Pb2+] = x; [Cl–] = 2x
32 4)2( xxxKSP 3
4SPK
x
(3) Al(OH)3 type:
Al(OH)3 Al3+ + 3OH–
Let the solubility of Al(OH)3 = x mol.L–1. So, [Al3+] = x; [OH–] = 3x
43 27)3( xxxKSP From this x(solubility) can be determined.
SAQ: Find the relationship between solubility and solubility product of a salt having general formula XmY
n.
Solution:
XmY
n mXn+ + nYm–
Let solubility = S; [Xn+] = mS; [Ym–] = nSnmnmmn
SP nSmSYXK )()(][][ )( nmnmSP SnmK
N.B: Actually in the definition of KSP
, activity(a) of ions are to be used in place of molar concentrations. Butoften approximate activity = molar concentrations for simplicity in calculation. More on activity is discussed inlater in this chapter. Hang on.SAQ: Calculate the solubility of AgCl in water in g/L at 250C, if its K
SP = 10–10 at this temperature.
Solution: Refer the expression above.
510 1010 x mol. L–1. So solubility in g/L = 10–5 ×143.5 = 1.435 × 10–3 g/L.
Ionic Equilibrium
Dr. S. S. Tripathy
SAQ: If the solubility of PbCl2 is 0.278 g/L at certain temperature, then what is its K
SP at that temperature(Pb
= 207).
Solution: Solubility in mol.L–1 001.0278
278.0
93332 104)10(44)2( xxxKSP
IONIC PRODUCT(IP)/Reaction Quotient(Q):(Condition for precipitation during a reaction):Suppose you are adding an aq. solution of AgNO
3 with aq. solution of NaCl, whether AgCl(s) will be precipitated
or not. Suppose you are adding Pb(NO3)
2 solution with NaCl solution, whether PbCl
2 will be precipitated or
not - all these things can be predicted by comparing KSP
with IP(Q). See below.
Ionic product(IP) or Reaction Quotient(Q) is analgous to KSP
in expression but the concentrations are usedin non-equlibrium conditions. In this case, the initial concentrations of the ions at the time of mixing are used todetermined IP(Q). For example, when AgNO
3(aq.) is added to NaCl(aq.), the product of initial concetrations
of [Ag+] and [Cl–] gives IP(Q). Note that IP(Q) is variable i.e not fixed which depends on us. But KSP
of asubstance is fixed, which is the IP at the saturated conditions(equilibrium).Case-I: IP(Q) > K
SP :
This makes a supersaturated solution and their will be precipitation as the equilibrium will shift to left. Afterprecipitation, a saturated solution will be formed.Case-II: IP(Q) < K
SP :
This makes an unsaturated solution and there will not be any precipitation.Case-III: IP(Q) = K
SP :
This is just a saturated solution and there will be no precipitation.
SAQ: Equal volumes of 10–3M AgNO3 is mixed with 10–6 M NaCl solutions. Predict whether precipitation will
occur or not ? [KSP
(AgCl) =10–10]Solution:
2
10][
3 Ag M
2
10][
6 Cl M (As diluted to two times by mixing)
1063
105.22
10
2
10
IP ; Since IP > KSP
, precipitation will occur.
SAQ: Equal volumes of 10–3M MgCl2 and 10–4M NaOH solutions were mixed. Predict whether precipitation
will occur or not ? [KSP
(MgCl2) =10–11]
Solution:
2
10][
32
Mg M;
2
10][
4 OH M
12
243
1025.12
10
2
10
IP Since IP < K
SP, the precipitation will not occur.
COMMON ION EFFECT in Heterogenous Equilibrium:The solubility of a solid substance is lowered in presence of a strong electrolyte having a common ion withrespect to the substance. This is due to common ion effect which can be explained with the help of LCP.(1) Sparingly soluble substance:
AgCl(s) Ag+ + Cl–
HCl H+ + Cl–
In the presence of HCl, the eqm. for AgCl solubility shifts to further left. This solubility is lowered as comparedto that in pure water. Suppose we add, HCl after making a saturated solution of AgCl in water, then fresh
Ionic Equilibrium
Dr. S. S. Tripathy
AgCl(s) will be precipitated as its solubility will be lowered in presence of HCl. We can explain this by sayingthat IP of AgCl exceeds its K
SP and precipitation occurs.
SAQ: Caculate the solubilty of AgCl in 0.1M HCl. Compare this solubility with its solubility in pure water. (KSP
of AgCl = 10–10).Solution:
AgCl(s) Ag+ + Cl–
x (0.1+x) Since ‘x’ small, it is neglected inside parenthesisHCl H+ + Cl–
xxxClAg 1.0)1.0(]][[10 110 910 x M
In pure water, solubility = 510 1010 M. In presence of 0.1M HCl it is 10–9M. Hence solubility of AgCl is
10,000 times greater in pure water than in 0.1M HCl.(2) Strong Electrolyte:
NaCl(s) Ag+ + Cl–
HCl(g) H+ + Cl–
When HCl gas is bubbled through a saturated solution of NaCl, pure NaCl(s) is precipitated due to commonion effect. The IP of NaCl exceeds its K
SP, so precipitation occurs. We can explain the percipitation more
generally by applying LCP.SAQ: Calculate [H+] necessary to just prevent precipitation of ZnS from a solution of 0.2M ZnCl
2 saturated
with H2S(0.1M). [K
SP(ZnS) = 1.2 × 10–22]. Import the K
A values of H
2S from elsewhere.
Solution:
H2S 2H+ + S2– K
A = 10–21.
HCl H+ + Cl–
ZnS(s) Zn2+ + S2– KSP
= 1.2 × 10–22
Let us see, that is [S2–] in a saturated solution of ZnS with 0.2M [Zn2+].
]][[ 22 SZnKSP
2222
2 100.62.0
102.1][
S M
We have to reduce the [S2–] in H2S by using a common ion effect, i.e by mixing with a strong acid. You know
that in H2S alone [S2–] = 10–14. But we need now at least 6 × 10–22 to prevent precipitation. Let us use the
dissociation of H2S given below to find [H+] required to achieve that.
H2S 2H+ + S2– K
A = 10–21.
1.0
100.6][10
22221
H
MH 408.0][
So, the [H+] should be at least 0.408M. If it is more, no harm. It will form a unsaturated solution. Note that [H+]in only 0.1M H
2S is 10–4M which is negligible. So the requirement will be met from another strong acid like
HCl.SAQ: H
2S was continuously bubbled through 0.2M ZnCl
2 till no more precipitation occured. Calculate
[Zn2+] in the resulting solution. (Use data from the previous SAQ)Solution: After precipitation is over, a junior school student will say, [Zn2+] =0, but you won’t say that. Ofcourse, the value is close to zero(as we shall find now), but not zero. To a lay person, such calculation willappear a maddening exercise but to you it is not. It is of great importance in analytical chemistry, when you
Ionic Equilibrium
Dr. S. S. Tripathy
separate one group as precipitate and prevent precipitation of metal ions belonging to higher groups. Theapplication of solubility product principle in analytical chemistry will be dealth with in a separate chapter.After precipitation is over, [H
2S] = 0.1M(saturated solution) and [S2–] = 10–14(K
A2). From all these, we can
find [Zn2+] left out in solution.
]][[ 22 SZnKSPMZn 8
14
222 102.1
10
102.1][
How small is the [Zn2+] ? Is it not close to zero ? But we will have to give its value.SAQ: Calculate the solubility of AgCN in a buffer solution of pH = 3.00. Assume that no cyano complex isformed. K
SP (AgCN) = 2.2 × 10–16; K
A (HCN) = 6.2 × 10–10.
Solution:
AgCN(s) Ag+ + CN–
Eqm x y
CN– + H+ HCN
Eqm y 10–3 xLets try to understand the happenings in a sequential manner. Suppose we forget pH=3. In absence of acidicmedium, [Ag+] = [CN–]. But CN– have all then disappeared by reacting with H+ in pH=3. This reaction isalmost complete as it is the revese of the dissociation of HCN. Now after the final equilibrium is reached, if thesolubility of AgCN is ‘x’ then [Ag+] = x; and [CN–] + [HCN] = x, Here we can neglect [CN–] and write[HCN] = xLet [CN–] = y. This we can get in terms of ‘x’ by using K
A.
10102.6
1
]][[
][
CNH
HCN103 102.6
1
10
y
x
xy 7102.6 Now let us use K
SP of AgCN to get ‘x’.
]][[ CNAgKSP16102.2 xy
167 102.2102.6 xx 5109.1 x MNote that the solubility of AgCN has increased in pH=3 buffer as compared to in pure water in which the
solubility is 816 1048.1102.2 M. This is opposite to common ion effect in which solubility is decreased.
Likewise, if Ag+ formed is allowed to further react with CN– by using KCN, then complex formation occurs.This again drives the solubility forward. We shall solve this type of numercial later.SAQ: Calculate NH
4+ concentration needed to prevent the precipitation of Mg(OH)
2 in 1 L solution which
contains 0.01 mole of NH3 and 0.001 mole of Mg2+. K
SP(Mg(OH)
2 = 1.2 × 10–11, K
B(NH
3) = 1.8 ×10–5)
Solution:
Mg(OH)2(s) Mg2+ + 2OH–
NH4OH NH
4+ + OH–
NH4Cl NH
4+ + Cl–
Let us find out [OH–] required for making a saturated solution of Mg(OH)2, and thus there would not be any
precipitation.211 ][001.0102.1 OH 410095.1][ OH
Ionic Equilibrium
Dr. S. S. Tripathy
Now we shall use the dissociation of NH4OH, to get the [NH
4+] required for the purpose. Note that this
[NH4+] is to be provided by using a strong electrolyte like NH
4Cl.
01.0
10095.1][108.1
445
NHMNH 3
4 1064.1][
Note that if we use only NH4OH, then (not any common ion effect), then [OH–] would be 4.24 × 10–4 and
then IP would be 18 × 10–11 and hence precipitation would have definitely occured.SAQ: Calculate the solubility of AgBr in 0.4M NH
3 . Assume that [Ag(NH
3)
2]+ is the only complex formed.
KSP
(AgBr) = 5.0 × 10–13; KF [Ag(NH
3)
2]+ = 1.0 × 108.
Solution:
AgBr(s) Ag+ + Br–
Eqm y x
Ag+ + 2NH3
[Ag(NH3)
2]+
Eqm y (0.4–2x) xThis is similar to the numerical we have already solved i.e solubility of AgCN in buffer pH=3. Let the solubilityis ‘x’(final). [Br–] =x, So x = [Ag+] + [Ag(NH
3)
2]+. But looking to the formation constant of the complex, we
can presume that [Ag(NH3)
2]+ = x; and x can be neglected in the face of 0.4, so we have
28
4.0101
y
x81016.0
x
y
Using the value of y = [Ag+] in KSP
expression we have,
xx
BrAg
8
13
1016.0]][[105 Mx 3108.2
So the solubility of AgBr in 0.4 NH3 solution is 2.8 × 10–3M. Note that pure AgBr has a solubility of
713 107105 M which is much less that the solubility we got in this numerical.
Simultaneous solubility:SAQ: Calculate the simultaneous solubility of AgSCN and AgBr in water. K
SP(AgSCN)=1.1 ×10–12; K
SP(AgBr)
= 5.0 ×10–13
Solution:
AgSCN(s) Ag+ + SCN–
(x+y) x
AgBr (s)0 Ag+ + Br–
(y+x) y1.1 ×10–12 = [Ag+][SCN–] = (x+y) × x (1)5.0 ×10–13 = [Ag+][Br–] = (x+y) × y (2)
Solving these two equations, we get x = 8.689 ×10–7 M (solubility of AgSCN)y =3.949 ×10–7 M (solubility of AgBr)
Ionic Equilibrium
Dr. S. S. Tripathy
Numerical involving mixing of two solutions:Find the pH of the following mixtures.1. Strong acid + weak acid:SAQ: 200 mL of 0.1M HCl + 300 mL of 0.1M CH
3COOH
Hint: Only strong acid to be taken and weak acid to be ignored, except dilution effect.
Solution: MH 04.0500
2001.0][
2. Strong base + Weak baseHint: Ignore weak base except dilution effect.3. Strong acid + Weak base : Strong acid is in excess:Hint: Ignore the salt. Take only the excess strong acid.SAQ: 200 mL of 0.1M HCl + 100 mL of 0.1M NH
4OH
Solution: Excess HCl = 10 meq = 10 mmol; Total volume = 300 mL
MH30
1
300
10][
4. Weak acid + Strong base : Weak acid excessHint. Makes acidic buffer. Both salt and excess acid are used in Henderson equation.SAQ: 200 mL of 0.1 CH
3COOH + 150 mL of 0.1 NaOH (Use K
A value)
Solution: mmole of salt = 15; mmole of excess CH3COOH = 5
3log74.45
15log ApKpH
5. Strong Acid + Weak base: Weak base in excessHint: Makes basic buffer. Both salt and excess base are used in Henderson equation.Note that for buffer solution, molarity of the two are not required, as we use them in ratio form.SAQ: 200 mL of 0.1M HCl + 300 mL of 0.1M NH
4OH (K
B value is used)
Solution: mmole of salt = 20; mmole of excess weak base = 10
2log74.410
20log BpKpOH
6. Weak acid + Strong Base: Exact equivalence .Hint: Salt hydrolysis equations to be used.SAQ: 200 mL of 0.1M CH
3COOH + 100 mL of 0.2M NaOH (use K
A value)
Solution: mmole of salt = 20; [Salt] = M30
2
300
20
302log74.414
2
1log
2
1 cpKpKpH AW
7. Weak base + Strong acid : Exact equivalenceHint: Salt hydrolysis equation to be used.SAQ: 100 mL of 0.1M NH
4OH + 200 mL of 0.05M HCl (pK
B = 4.74).
Solution: [Salt] = M30
1
300
10
301log74.414
2
1log
2
1 cpKpKpH BW
Ionic Equilibrium
Dr. S. S. Tripathy
8. Salt of Weak acid-strong base + Strong acid: Salt in excess.Hint: Makes acidic buffer. Henderson equation to be used.SAQ: 200 mL of 0.1M CH
3COONa + 100 mL of 0.1M HCl (pK
A = 4.74)
Solution: CH3COO– + H+ CH
3COOH
mmole of weak acid= 10; mmole of excess salt = 10pH = 4.74 +0 = 4.74
9. Salt of strong acid-weak base + strong base : salt in excess.Hint: Makes basic buffer. Henderson equation to be used.SAQ: 200 mL of 0.1M NH
4Cl + 150 mL of 0.1M NaOH (pK
B = 4.74)
Solution: NH4+ + OH– NH
4OH
mmole of weak base = 15; mmole of excess salt = 5
3log74.415
5log74.4 pOH
10. Two weak acids:Hint: Use charge balance conceptSAQ: Calculate the [H+], [CH
3COO–] and [C
6H
5COO–] in a mixture which 0.02M in CH
3COOH and 0.01M
in C6H
5COOH, K
A(Acetic acid) = 1.8 × 10–5; K
A(Benzoic acid) = 6.8 × 10–5.
Solution:
][
][][ 3
3
H
COOHCHKCOOCH A
][
][][ 56
'
56
H
COOHHCKCOOHC A
Charge Balance: ][][][ 563 COOHCCOOCHH
][
][
][
][][ 56
'3
H
COOHHCK
H
COOHCHKH AA
][][][ 56'
32 COOHHCKCOOHCHKH AA
Substituting the values, we get[H+] = 1.02 × 10–3M; Then substsituting the values in K
A and K
A’ we get
[CH3COO–] = 3.53 × 10–4M [C
6H
5COO–] = 6.68 × 10–4M
Ionic Equilibrium
Dr. S. S. Tripathy
ACTIVITY/ACTIVIY COEFFICIENT
Dr. S. S. Tripathy
Chemical Potential, Gibbs Free Energy and Activity
CHEMICAL POTENTIAL(μμμμμ):It is the molar free energy of a substance (G/n). More correctly, during a chemical transformation, the chemical
potential is determined from TPn
G
,
form the grpah of G vs. n at constant temperature and pressure.
The standard chemical potential is μ0 at the standard conditions. Chemical potential(μ) at other non-standardconditions can be obtained from the following relationships.GAS:
(a) Single Ideal Gas:
00 ln
P
PRT (where P0 = 1 atm/ 1 bar : standard pressure)
μ0= standard chemical potential at P = P0.If P is taken in atm. or bar the chemical potential equation is reduced to
PRT ln0 ( P in atm. or bar)
(b) Mixture of Ideal Gases:
iii pRT ln0 (P in atm/bar)
where μi = chemical potential of the ith component; p
i = partial pressure of the ith component
(c) Single Non-ideal Gas:
fRT ln0 ( f = fugacity/activity of the gas/Effective Pressure)
Pf (φ = fugacity coefficient and P = Observed pressure)
(d) For a mixture of Non-ideal Gases:
iii fRT ln0 (fi = partical fugacity of ith component=φ
ip
i )
LIQUID MIXTURE:(a) Ideal: (Raoult’s law)
iii xRT ln0 (xi = mole fraction of ith component)
(b) Non-ideal:
iii aRT ln0 (ai = activity of of ith component)
ai = γ
i × x
i( γ
i = activity coefficient of the ith component)
Note that when γ =1, it is an ideal solution (ai = x
i). If γ >1, it is a case of positive deviation to Raoult’s law and
if γ < 1, it is negative deviation to Raoult’s law.
)(Pr
)(Pr
IdealessurVapour
ActualessureVapour
AQUEOUS SOLUTION(with non-volatile solute)
iii aRT ln0 (ai = activity of of ith component)
ai = γ
i × c
i(c
i = molar concentration of ith component)
1Activity(a) and Fugacity(f) :These two terms are synonymous. ‘a’ is used for liquid mixture or aqueous solution while ‘f’ is used for gases.Activity is the effective concentration of any species in non-ideal mixture. This determines the realchemical potential for a real solution(non-ideal mixture).
ACTIVITY/ACTIVIY COEFFICIENT
Dr. S. S. Tripathy
Chemical potential(μ) and activity(a) of a substance is dependent on several factors such as P, T, composition.But at constant temperature and pressure(for gas), they are dependent primarily on composition and chemicalenverionment.Fugacity is the effective pressure/partial pressure of a gas in non-ideal conditions.Activity/fugacity is to be used for getting more accurate results in concentrated solution/ under high pressure/non-ideal liquid mixture.Activity/fugacity has no unit as it is made unitless by dividing with c0(1M), p0(1 atm).Activity of ion also expressed in the same way as activity of any molecular species.
a(ion) = γ(ion) × c(ion)Activity of of an ion is greatly dependent on theh IONIC STRENGTH of the solution.IONIC STRENGTH(I):
The ionic strength of a solution is determined from the following
21
232
211 2
1.......
2
1zczczcI i
Where ci = molarity of ith ion, z
i = the magnitude of charge of ith ion.
SAQ: Find the ionic srength of a solution which is 0.0087M in KOH and 0.0002M in La(IO3)
3.
Solution:
01.00099.010002.0330002.010087.010087.02
1 2222 I
Effect of I on γγγγγ :At low ionic strength, γ =1, so that a = c/x. γ decreases with increase in I utpo 0.1M and at higher I, γincreases again(not shown in the graph).
Activity Coefficient(γ) can be determined by using the Debye-Huckel Extended equation.
3051
51.0log
2
Ir
Iz
h
where z= ionic charge; rh = radius of hydrated ion in pm.
ACTIVITY/ACTIVIY COEFFICIENT
Dr. S. S. Tripathy
We can approximate the above equation for less accurate determination as follows.
I
Iz
1
51.0log
2
γ → 1 when(a) I is low (b) lower concentration of ith component (c) lower charge of the ith component
and large effective radius(hydrodynamic radius).
SAQ: Find the γ for Hg22+ ion in 0.033M Hg
2(NO
3)
2 solution.
Solution:
MI 1.01033.022033.02
1 22
51.0149.01.01
1.0251.0
1
51.0log
22
I
Iz
⇒ γ = 0.3236SAQ: Find γ for H+ ion at I = 0.025M . [Given r
h(H+) = 900 pm]
Solution:
9451.010549.0
305025.0900
1
025.0151.0
3051
51.0log
22
Ir
Iz
h
⇒ γ = 0.881
_________________________________________________________________________________N.B: For neutral moleculess γ ≈ 1 when I ≤ 0.1M, so a
i = c
i. For gases φ =1, when pressre ≤ 1 atm, so f
i =
pi .
____________________________________________________________________________________MOLECULAR EQUILBRIUM
aA + bB cC + dD
At equilibrium: PR ⇒ 0G
BADC badcG
bBB
aAA
dDD
cCC aRTdaRTaaRTdaRTc lnlnlnln 0000
bB
aA
dD
cC
BADC aa
aaRTbadc
ln0000
QRTGG ln0 (Where Q = Reaction Quotient)
At equilibrium; 0G
PKRTG ln0 For gases, equilibrium constant, by default is K
P while for aq. and liquid systems it is K
C.
IMP: For molecular equilbrium without involving ions; γ ≈ 1 (as I < 0.1M),
bB
aA
dD
cC
bB
aA
dD
cC
bB
aA
dD
cC
bB
aA
dD
cC
C cc
cc
cc
cc
aa
aaK
ACTIVITY/ACTIVIY COEFFICIENT
Dr. S. S. Tripathy
We can use molar concentrations in the KC expression without encountering any appreciable error as the ratio
containing the activity coefficients approximately becomes equal to 1.Similarly for gases, ratio containing fugacity coefficients(φ) becomes approximately equal to 1 and hence wecan use partial pressures in stead of fugacities of gases, particularly when the pressure is not too high.
But for ionic equilibrium, however, molar concentration terms cannot be used. Always we have to use activityeven while calculating pH of solution. Only when ionic strength is low, we can safely say γ=1 and then we canuse molar concentration in stead of activity.
Examples showing the importance of activity in ionic equilibrium:(1) In a solution of KH(IO
3)
2 at 0.02M, ‘a’ is 40% of 0.02, for which the actual pH becomes greater than the
value we get by using 0.02M
][loglog HapHHH
(2) When 0.1M HCl solution containing methyl green indicator is added to 5M solution of MgCl
2, the colour
of the indicator changes from green to yellow(same acidic medium) indicating increase in [H+]. At higher ionicstrength, γ >1 and a > c for which effective hydrogen ion concentratin becomes greater.N.B : In ionc equilibrium, K(eg K
W, K
SP etc) values are usually obtained by conductance measurement which
is based on effective concentration(activity) rather than the actual molar concentration(c).SAQ: Find the pH of a solution containing 0.1M HCl and 0.04M KClO
4.
Solution: KClO4 enhances the ionic strength and hence activity becomes different from molar concentration.( ionic strength calculation not shown)
9069.010931.005.01
05.0151.0log
2
⇒ γ = 0.8054
)008054.0log(01.08054.0loglog HapH
⇒ 097.23903.038log pH(N.B: if you would not have included γ, then pH would be been exactly 2.0. But it is really 2.097 as determinedfrom a precision pH meter. This shows you the significance of activity)IMP: Ionic product of water is actually the product of activity of H+ and OH–.
1410]][[ OHHaaKOHHOHHW (at 250C)
SAQ: What is [Hg22+] in a saturaed solution of Hg
2Br
2 with 0.0010M KCl, where K
SP(Hg
2Br
2)=5.6 ×10–23. In
other words, find the solubility of mercurous bromide in presence of the added salt.Solution:
Hg2Br
2(s) Hg
22+ + 2Br–
KCl is an inert salt which increase the ionic strength only.
MI 001.01001.01001.02
1 22
9375.010625.0001.01
001.0251.0log
2
22
Hg
⇒ 865.022Hg
9844.010156.0001.01
001.0151.0log
2
Br
⇒ 993.0Br
ACTIVITY/ACTIVIY COEFFICIENT
Dr. S. S. Tripathy
222222
2 2993.0865.0][][( 22
22
xxBrHgaaKBrHgBrHgSP
⇒ Mxx 8323 10541.248529.0106.5 Solubilty obtained by using only molar concentration:
233 106.54 x Mx 81041.2 Note that the solubility in presence of added salt is convincingly more than in the absence of it. This is oppositeto common ion effect as the solubility here is increased in presence of an inert ionic salt.SAQ: What is the [Hg
22+] in a saturated soloution of Hg
2Br
2 with 0.001 M KBr. In other words, find the
solubility in presence of given KBr solution. (Use KSP
given before)Solution:
Hg2Br
2(s) Hg
22+ + 2Br–
KBr → K+ + Br–
This is a case of common ion effect in which we expect the solubility is bound to decrease. But lets see whatreally happens due to the effect of ionic strength in changing the effective concentration.We shall use the results of previous question to get the activity coefficients. As KBr is a uni-univalent type ofstrong electroyte and the concentration is the same(same ionic stregnth).
865.022Hg
993.0Br
[Hg22+] = x, [Br–]= (2x + 0.001)
222222
2 001.02993.0865.0][][( 22
22
xxBrHgaaK
BrHgBrHgSP
Neglecting ‘2x’ inside parenthesis as ‘x’ is incredibly small(check approximation).
Mxx 1723223 105658.610993.0865.0106.5 Using only molar concentration:
Mxx 172323 106.510106.5 Did you notice a remarkable thing here ? The solubility, no doubt, is decreased from the order of 10–8 in theabsence of common ion effect to the order of 10–17in presence of it. But due to effect of ionic strength, theactual solubility is greater than that expeced from ideal behaviour.N.B: Student to handle this carefully in examination, so as to read the mind of the question setter whether he/she is keen to get more accurate result or an approximate one in the face high ionic strength. For getting I, wedo not need any supporting data and we can find approximate value of γ from the approximate Debye-Huckelequation.SAQ: How can you justify the following ?
00 ln
P
PRT
Solution:Refer the chapter “Thermodynamics”. In the derivation of Maxwell relations, we had the following
relation.
VP
G
nT
,
(we have mentioned here, the number of moles ‘n’ to be constant as G is a extensive property. We did not writethere, as we were not needing its use in it)
ACTIVITY/ACTIVIY COEFFICIENT
Dr. S. S. Tripathy
dPP
nRTVdPdG nT ,
P
P
G
G P
dPnRTdG
0
1
0
001 ln
P
PnRTGGG
001 ln
P
PnRTnn
001 ln
P
PRT
where μ0= chemical potential at the standard state and μ
1 is the potential at any other non-standard state. Note
that similar type of relationship holds good for mole fraction in binarly liquid mixure or in concentrations insolutions containing non-volatile solutes.