acid-base equilibria and solubility equilibria chapter 16 dr. ali bumajdad

Acid-Base Equilibria andSolubility Equilibria

Chapter 16

Dr. Ali Bumajdad

Chapter 16 Topics

• Homogeneous Versus Heterogeneous Solution Equilibria• The Common Ion Effect• Buffer Solutions• Acid-Base Titrations and Indicators• Solubility Equilibria• Relation between solubility and Ksp

• The Common Ion Effect and Solubility• pH and Solubility

Acid –Base Equilibria

Dr. Ali Bumajdad

H2O (l) H+ (aq) + OH- (aq)

AgNO3(aq) +NaCl(aq) AgCl(s) + NaNO3 (aq)

Homogeneous Equilibria

Heterogeneous Equilibria

•Common ion effect is the shift in equilibrium to the left (reactant) caused by the addition of a compound (strong electrolyte) having an ion in common with the dissolved substance (weak electrolyte). This is because of the Le Chatelier Principle.

The presence of a common ion decreases the ionization of a weak acid or a weak base.

Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).

CH3COONa (s) Na+ (aq) + CH3COO- (aq)

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

common ion

The Common Ion Effect

pH will increase

Consider mixture of salt NaA and weak acid HA.

HA (aq) H+ (aq) + A- (aq)

NaA (s) Na+ (aq) + A- (aq)Ka =

[H+][A-][HA]

-log [H+] = -log Ka - log[HA]

[A-]

-log [H+] = -log Ka + log [A-][HA]

pH = pKa + log [A-][HA]

Henderson-Hasselbalch equationpH = pKa + log

[conjugate base][acid]

(2)

pKa = -log Kawhere

[H+] =Ka [HA]

[A-]

(1)

Q) What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Common ion effect

0.30 – x 0.30

0.52 + x 0.52

pH = pKa + log [HCOO-][HCOOH]

HCOOH pKa = 3.77

pH = 3.77 + log[0.52][0.30]

= 4.01

Mixture of weak acid and conjugate base!

(Ka of HCOOH = 1.710-4 )

- Na is spectator ion and do not affect pH

Another solution:

- Find [H+] then find pH

[H+] =Ka [HCOOH]

[HCOO-]

(1)

pH=-log [H+]

• Buffer solution is a solution that resists change in pH because it contain both acids to neutralize OH- and base to neutralize H+ ions

• Buffer composition:

1. A weak acid + its salt (that is salt of weak conjugate base)

2. A weak base + its salt (that is salt of weak conjugate acid)

Add strong acid

H+ (aq) + CH3COO- (aq) CH3COOH (aq)

Add strong base

OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

Consider an equal molar mixture of CH3COOH and CH3COONa

Buffer solution

HCl H+ + Cl-

HCl + CH3COO- CH3COOH + Cl-

Q) Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) HF is a weak acid and F- is its conjugate basebuffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is its conjugate acidbuffer solution

The greater the amount of the conjugate acid-base pair the greater the buffer capacity (that is the grater the resistance of pH change). Some buffer solution have the same pH but their capacity are different

Buffer capacity:

Q) Which of the following buffer solutions has greater pH? And which has greater buffer capacity?

1) 1M HC2H3O2 + 1 M NaC2H3O2

2) 0.1M HC2H3O2 + 0.1M NaC2H3O2

pH = pKa + log[conjugate base]

[acid]

(2)

= 9.20

Q) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? (Ka NH4

+ =5.610-10; Kb NH3 = 1.810-5)

NH4+ (aq) H+ (aq) + NH3 (aq)

pKa = 9.25 pH = 9.25 + log[0.30][0.36]

= 9.17

NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)

start (moles)

end (moles)

0.029 0.001 0.024

0.028 0.0 0.025

pH = 9.25 + log[0.25][0.28]

[NH4+] =

0.0280.10

final volume = 80.0 mL + 20.0 mL = 100 mL

[NH3] = 0.0250.10

pH = pKa + log[NH3][NH4

+]

(2)

(better to use mole then convert to M)

Sa. Ex. How many mole of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form Buffer solution with pH =9.00? (assume addingNH4Cl do not affect the total volume (Kb = 1.810-5)

pH = pKa + log[conjugate base]

[acid]

(2)

orpOH = 14 - pH

[OH-]= 10-pOH

Kb= [NH4+ ][OH-] / [NH3]

pKa+pKb = pKw=14 For base or acid and its salt

pH = pKa + log[conjugate base]

[acid]

(2)

The added strong acid (HCl) will be completely consumed by the reaction with the buffer and that’s way it will not affect the pH much

(a)

(b)

(Ka=1.810-5)

Start molarity:Change :End molarity:

CH3COO- + H+ CH3COOH1.0 0.1 1.0-0.1 -0.1 +0.10.9 0 1.1

Use equation 2

Sa. Ex. Buffer made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L solution. The pH of the buffer is 4.74.(a) Calculate the pH of this solution after 0.20 mol of NaOH is added (neglect any volume change) (b) calculate the pH when 0.020 mol of NaOH is added to 1.00L of water

(a) OH- will react with the acid and the no. of mole of the acid and the conjugate base will change

HC2H3O2 + OH- H2O + C2H3O2-

Before reactionChangeAfter reaction

HC2H3O2 H+ + C2H3O2-

pH = pKa + log[conjugate base]

[acid]

(2)

Titrations

In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

• Equivalence point : the point at which the reaction is complete

• Indicator : substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

Strong Acid-Strong Base Titrations

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

OH- (aq) + H+ (aq) H2O (l)

Weak Acid-Strong Base Titrations

CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)

CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)

At equivalence point (pH > 7) because of the hydrolysis:

Strong Acid-Weak Base Titrations

HCl (aq) + NH3 (aq) NH4Cl (aq)

NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

At equivalence point (pH < 7) because of the hydrolysis:

H+ (aq) + NH3 (aq) NH4Cl (aq)

Q) Exactly 100 mL of 0.10 M HNO2 are titrated with 100 ml of 0.10 M NaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal volume = 200 mL

Kb =[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05 – x 0.05 x 1.05 x 10-6 = [OH-]

pOH = 5.98

pH = 14 – pOH = 8.02

Acid-Base Indicators

HIn (aq) H+ (aq) + In- (aq)

10[HIn]

[In-]Color of acid (HIn) predominates

10[HIn]

[In-]Color of conjugate base (In-) predominates

• Indiators are weak organic acid or base •End point: the point when the indicator change color

pH

The titration curve of a strong acid with a strong base.

Methyl red and ph.ph are good indicators but Thymol blue is not

Q) Which indicator(s) would you use for a titration of HNO2 with KOH ?

Weak acid titrated with strong base.

At equivalence point, will have conjugate base of weak acid.

At equivalence point, pH > 7

Use cresol red or phenolphthalein

Solubility Equilibria

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-] Ksp is the solubility product constant

MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO4

3-]2

Heterogeneous equilibrium

Ksp is the product of the concentrations of the ions involve in the equilibrium all rise to a power = the coefficient

•Molar solubility, s, (mol/L) is the number of moles of solute dissolved in 1 L to form a saturated solution.

•Solubility (g/L) is the number of grams of solute dissolved in 1 L to form a saturated solution.

Relation between Molar solubility and Ksp

Molar solubility = the highest concentration of solution

Molar solubility change upon adding another solutes

Molar solubility change upon changing temperature

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp= [Ag+] [Cl-] But [Ag+] = solubility of AgCland [Cl-] = solubility of AgCl

Molar solubility M.m. = solubility (3)

Ksp= s s = s2 (4) s = (Ksp)1/2 (5)

Solubility Product versus Molar Solubility

Salt type KSP S AB

(e.g. NaCl , KI O3) KSP =S2 S = SPK

A2B or AB2 (e.g. CaF2 , Ag2CrO4)

KSP =4 S3

S = 3

1

4

SPK

A3B or AB3 (e.g. Ce(NO3)3 , AlCl3)

KSP = 27 S4

S =4

1

27

SPK

A2B3 or A3B2

(e.g. Bi2S3)

KSP = 108 S5

S =5

1

108

SPK

/ M.m.

M.m.

Q) What is the solubility of silver chloride in g/L ?

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp = s2

s = Ksps = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl = 1.3 x 10-5 mol AgCl

1 L soln143.35 g AgCl

1 mol AgClx = 1.9 x 10-3 g/L

Ksp = 1.6 x 10-10

Ksp = 1.6 x 10-10

Ksp of Cu(OH)2 = 2.210-20

Dissolution of an ionic solid in aqueous solution:

Q = Ksp Saturated solution

Q < Ksp Unsaturated solution No precipitate

Q > Ksp Supersaturated solution Precipitate will form

Ion concentration not at equilibriumIon concentration at equilibrium

Q) If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form? Ksp of Ca(OH)2 = 8.0 x 10-6

The ions present in solution are Na+, OH-, Ca2+, Cl-.

Only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

Ksp of BaSO4 = 1.110-10

Q) What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M?

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]

Ksp = 1.6 x 10-10

AgBr (s) Ag+ (aq) + Br- (aq) Ksp = 7.7 x 10-13

Ksp = [Ag+][Br-]

[Ag+] = Ksp

[Br-]7.7 x 10-13

0.020= = 3.9 x 10-11 M

[Ag+] = Ksp

[Cl-]1.6 x 10-10

0.020= = 8.0 x 10-9 M

3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M

The Common Ion Effect and Solubility

The presence of a common ion decreases the solubility of the salt. This due to LeChatelier’s principle

Q) What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr?

AgBr (s) Ag+ (aq) + Br- (aq)

Ksp = 7.7 x 10-13

s2 = Ksp

s = 8.8 x 10-7

NaBr (s) Na+ (aq) + Br- (aq)

[Br-] = 0.0010 M

AgBr (s) Ag+ (aq) + Br- (aq)

[Ag+] = s

[Br-] = 0.0010 + s 0.0010

Ksp = 0.0010 x s

s = 7.7 x 10-10

Solubility not molar solubility

pH and Solubility• The presence of a common ion decreases the solubility.• Insoluble bases dissolve in acidic solutions• Insoluble acids dissolve in basic solutions

Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)

Ksp = [Mg2+][OH-]2 = 1.2 x 10-11

Ksp = (s)(2s)2 = 4s3

4s3 = 1.2 x 10-11

s = 1.4 x 10-4 M

[OH-] = 2s = 2.8 x 10-4 M

pOH = 3.55 pH = 10.45

At pH less than 10.45

Lower [OH-]

OH- (aq) + H+ (aq) H2O (l)

remove

Increase solubility of Mg(OH)2

At pH greater than 10.45

Raise [OH-]

add

Decrease solubility of Mg(OH)2

S-2 is a conjugate base of a weak acid Weak baseCl- is a conjugate base of strong acid Extremely weak baseSO4

2- is a conjugate base of a weak acid Weak base