additional aspects of aqueous equilibria. aspects of aqueous equilibria: the common ion effect salts...
TRANSCRIPT
Additional Aspects of Aqueous Equilibria
Aspects of Aqueous Equilibria: The Common Ion Effect
Salts like sodium acetate are strong electrolytes
NaC2H3O2(aq) Na+(aq) + C2H3O2
-(aq)
The C2H3O2- ion is a conjugate base of a weak acid
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-(aq)
Ka = [H3O+] [C2H3O2
-]
[HC2H3O2]
The Common Ion Effect
Ka = [H3O+] [C2H3O2-]
[HC2H3O2]
Now, lets think about the problem from the perspective of LeChatelier’s Principle
What would happen if the concentration of the acetate ion were increased?
Q > K and the reaction favors reactant
Addition of C2H3O2- shifts equilibrium, reducing H+
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-(aq)
Since the equilibrium has shifted to favor the reactant, it would appear as if the dissociation of the weak acid(weak electrolyte) had decreased.
The Common Ion Effect Ka = [H3O+] [C2H3O2
-]
[HC2H3O2]
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-(aq)
So where might the additional C2H3O2-(aq) come
from? Remember we are not adding H+. So it’s not like we can add more acetic acid.
Aspects of Aqueous Equilibria: The Common Ion Effect
How about from the sodium acetate?
NaC2H3O2(aq) Na+(aq) + C2H3O2
-(aq)
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-(aq)
In general, the dissociation of a weak electrolyte (acetic acid) is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte
The shift in equilibrium which occurs is called theCOMMON ION EFFECTCOMMON ION EFFECT
Let’s explore the COMMON ION EFFECTCOMMON ION EFFECT in a little more detail
Suppose that we add 8.20 g or 0.100 mol sodium acetate, NaC2H3O2, to 1 L of a 0.100 M solution of acetic acetic acid, HC2H3O2. What is the pH of the resultant solution?
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-(aq)
NaC2H3O2(aq) Na+(aq) + C2H3O2
-(aq)
Now you try it!
Calculate the pH of a solution containing 0.06 M formic acid (HCH2O, Ka = 1.8 x 10-4)and 0.03 M potassium formate, KCH2O.
Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of HClHCl and 0.20 mol HF in 1.0 L
HF (aq) + H2O H3O+ (aq) + F-(aq)
HCl (aq) + H2O H3O+(aq) + Cl-
(aq)
Kb = [NH4+] [OH-]
[NH3]
Now, lets think about the problem from the perspective of LeChatelier’s PrincipleLeChatelier’s Principle
But this time lets deal with a weak base and a salt containing its conjugate acid.
Q > K and the reaction favors reactant
Addition of NHAddition of NH44++ shifts equilibrium, reducing OH shifts equilibrium, reducing OH--
NH3(aq) + H2O NH4+
(aq) + OH- (aq)
Calculate the pH of a solution produced by mixing 0.10 mol NH4Cl with 0.40 L of 0.10 M NH3(aq), pKb = 4.74?
NH3(aq) + H2O NH4+(aq) + OH-
NH4Cl(aq) NH4+
(aq) + Cl-(aq)
Common Ions Generated by Acid-Base ReactionsThe common ion that affects a weak-acid or weak-base
equilibrium may be present because it is added as a salt, or the common ion can be generated by reacting an acid and base directly (no salt would be necessary….which is kind
of convenient if you think about it)
HC2H3O2 (aq) + OH-(aq) H2O + C2H3O2
-(aq)
Suppose we react 0.20 mol of acetic acid (weak) with 0.10 mol of sodium hydroxide strong)
0.20 mol 0.10 mol
-0.10 mol +0.10 mol-0.10 mol
0
00.10 mol 0.10 mol
HC2H3O2(aq) + OH- (aq) H2O + C2H3O2-aq)
Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxide
0.20 mol 0.10 mol-0.10 mol 0.10 mol-0.10 mol
0
00.10 mol 0.10 mol
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
0.10 M0.10 M
Let’s suppose that all this is occurring in 1.0 L of solution
0
Sample problem: Calculate the pH of a solution produced by mixing 0.60 L of 0.10 M NH4Cl with
0.40 L of 0.10 M NaOHNH4Cl(aq) NH4
+(aq) + Cl- (aq) NH4
+ + OH- NH3 + H2O
0.06 mol 0.04 mol
-0.04 mol 0.04 mol
0
0.04 mol0-0.04 mol 0.02 mol
Don’t forget
to convert to
MOLARITIES
0.02 M NH4
+ + H2O H3O+ + NH3 0.04 M0
Now you try it!
Calculate the pH of a solution formed by mixing 0.50 L of 0.015 M NaOH with
0.50 L of 0.30 M benzoic acid (HC7H5O2, Ka = 6.5 x 10-5)
adding acid or baseadding acid or base
calculate the pH of a solution that has .2 mol of NaOH added to a solution that is .25 M HC2H3O2 and .32M NaC2H3O2 HC2H3O2(aq) + OH- (aq) H2O + C2H3O2
-aq
..25 .20 .32
-.20 -.20 +.20
.05 0.52
HC2H3O2(aq) H+ + C2H3O2-aq
.05 0 .52
-X X X
X (.52+X) =1.8 x 10-5
.05-X
[H +] = [HX]
[X-]Ka
BUFFERED SOLUTIONSA buffered solution is a solution that resists change in
pH upon addition of small amounts of acid or base.
Suppose we have a salt: MX M+(aq) + X-
(aq) And we’ve added the salt to a weak acid containing the same conjugate base as the salt, HX:
HX +H2O H3O+ + X-
And the equilibrium expression for this reaction is
Ka = [H+ ] [ X-]
[HX]Note that the concentration of the H+ is dependent upon the Ka and the ratio between the HX and X- (the conjugateacid-base pair)
Two important characteristics of a buffer are buffering capacity and pH. Buffering capacity is the amount of acid
or base the buffer can neutralize before the pH beginsto change to an appreciable degree.
•This capacity depends on the amount of acid and base from which the buffer is made
•The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and
hence the pH, to change
•The pH of the buffer depends upon the Ka
[H +] = [HX]
[X-]Ka
-log[H +] = [HX]
[X-] -log Ka
Henderson-Hasselbalch EquationHenderson-Hasselbalch Equation
pH = pKa +log [X-][HX]
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-(aq)
NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)
0.1 M 0 0.1
0.1 M 0.1 M 0.1 M
-x x x
x0.1 - x 0.1 + x
1.8 x 10-5 = x(0.1 + x )
0.1 - x x = 1.8 x 10-5 pH = 4.74
Using the Henderson-Hasselbalch Equation
pH = 4.74 + log [.1][.1]
Note that these are initialconcentrations
A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH after 0.020 mol NaOH
is addedis added, (b) after adding 0.020 mol HCladding 0.020 mol HCl is added.
HC2H3O2(aq) + OH- H2O + C2H3O2-
(aq)
0.1 M 0.02 M0.02 M
-0.02 M-0.02 M-0.02 M
0.1 M
NaC2H3O2(aq) Na+(aq) + C2H3O2
-(aq)
0.1 M 0.1 M 0.1 M
0.02 M0.02 M
0.12 M0.00 M0.08 MHenderson-Hasselbalch Equation
pH = 4.74 + log [.12][.08]
Note that these are initial concentrations
pH = 4.92
Step 1
Step 2
A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH0.020 mol NaOH is added, (b) after adding 0.020 mol HCl is addedafter adding 0.020 mol HCl is added.
C2H3O2-(aq) + H+ HC2H3O2
Step 10.10 M 0.10 M0.02 M
-0.02 M
0.00 M
-0.02 M 0.02 M0.12 M0.08M
Henderson-Hasselbalch Equation
pH = 4.74 + log [.08][.12]
Note that these are initial concentrations
pH = 4.56
Step 2
Now consider, for a moment, what would have happened if I had added 0.020 mol of NaOH or
0.02 mol HCl to .1 M HC2H3O2.
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-aq
0.1 M 0 0-x x x
x0.1 - x
x
1.8 x 10-5 = x2
0.1 - x x = 0.0013
pH = 2.9
HC2H3O2(aq) + OH- H2O + C2H3O2-(aq)
0.1 M 0.02 M0.02 M
-0.02 M-0.02 M-0.02 M-0.02 M
0.00
0.02 M0.02 M
0.02 M0.00 M0.00 M0.08 M
Henderson-Hasselbalch Equation
pH = 4.74 + log [.02][.08]
pH = 4.13
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2
-aq
0.10 00.02 M
pH = 1.7
H+ from HCl Completely dissociates therefore the pH is calculated without regard for the weak acid
pH = -log [0.02]
add 2 ml
10 M HCl
to 1.8 x 10-5M HCl
pH = 4.74
add 2 ml 10 M HClto .1M HC2H3O2
+ .1M NaC2H3O2 pH = 4.74
.02 M HClpH= 1.7a drop of 3.04
to .12M HC2H3O2 + .06M NaC2H3O2 pH = 4.56a drop of .18
Sample exercise: Consider a litter of buffered solution that is 0.110 M in formic acid (HC2H3O)
and 0.100 M in sodium formate (NaC2H3O ). Calculate the pH of the buffer
(a)before any acid or base are added, (b) after the addition of 0.015 mol HNO3, (c) after the addition of 0.015 mol KOH
Titration CurvesTitration Curves
HCl(aq) + NaOH(aq) H2O + NaCl
Stoichiometrically equivalent quantities of acid and base have reacted
End points
Titration of a weak acid and a strong base results in pH curves that look similar to those of a strong acid-strong base curve except that the curve (a) begins at a higher pH, (b) the pH rises more rapidly in the early part of the titration, but more slowly near the equivalence point, and (3) the equivalence point pH is not 7.0
Calculating pH’s from TitrationsCalculating pH’s from TitrationsCalculate the pH in the titration of acetic acid by NaOH after 30.0 ml of 0.100 M NaOH has
been added to 50 mL of 0.100 acetic acid
HC2H3O2(aq) + OH- H2O(l) + C2H3O2-
0.005 mol 0.003 mol 00.003 mol
0.003 mol
-0.003 mol-0.003 mol
-0.002 mol 0
pH = 4.74 + log [.0370]
[.0250] pH = 4.91
Determining the KDetermining the Kaa From the Titration Curve From the Titration Curve
pKa = pH = 4.74
Solubility Equilibria Ksp
The equilibrium expression for the following reaction is CaF2(s) Ca2+
(aq) + 2F-(aq)
[Ca2+] [F-]2
[CaF2]
Look at the table on page 759 or the appendices A26, where you will find the
value of the ksp to be 4.1 x 10-11
= Ksp
Calculating Ksp from solubility
The solubility of Bi2S3 is 1.0 x 10-15M what would be the Ksp?
Bi2S3(s) 2Bi3+(aq) + 3S2-
(aq)
1.0 x 10-15 2(1.0 x 10-15) 3(1.0x 10-15)
Ksp = [Bi3+]2 [S2-]3
Ksp = (2.0 x 10-15)2(3.0 x 10-15)3 = 1 x 10-73
Calculating solubility from Ksp
The Ksp of Cu(IO3)2 is 1.4 x 10-7 what would be the solubility?
Cu(IO3)2(s) Cu2+(aq) + 2IO3
1-(aq)
X X 2X
Ksp= 1.4 x 10-7 = (X)(2X)2 = 4X3
X = 3.3 x 10-3 mol/L
Common ion effectWhat would be the solubility of Ag2CrO4
in a solution that is .1M AgNO3,
the Ksp = 9.0 x 10-12
Ag2CrO4 2Ag1+ + CrO42-
X 2X + .1 X
9.0 x 10-12 = (2X+.1)2(X)
Drop the 2X as insignificant
X = 9.0 x 10-10 mol/L
pH and SolubilityMg(OH)2 Mg2+ + 2OH-
If the pH is raised (the OH- is raised) then we have the common ion effect
and the solubility is decreased.
If the pH is lowered (the H+ is raised) then OH- is reacted with H+ to make water and the solubility is increased.
pH and solubility of salts
Ag3PO4 3Ag1+ + PO43-
If the pH is lowered (the H+ is raised) the PO4
3- reacts with H+ to make HPO42- ,
which essentially removes PO43-
from the equation, shifting the reaction to the right and the solubility is
increased.
Will a precipitate form?
If 750 mL of 4.00E-3M Ce(NO3)3 is mixed with 300 mL of 2.00E-2M KIO3, will a precipitate form?
Ce(NO3)3(aq) + KIO3 (aq) → Ce(IO3)3(s) + KNO3(aq)
(750)(4 x 10-3) = (1050) X; X= 2.86 x 10-3M for Ce3+
(300)(2 x 10-2) = (1050) Y; Y= 5.71 x 10-3 M for IO31-
Ce(IO3)3(s) Ce3+ (aq)
+ 3IO31-
(aq)
Ksp = 1.9 x 10-10
Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x 10-10
Q is greater than K so a precipitate will form