Solubility Equilibria (Sec 6-4)
Ksp = solubility product
AgCl(s) = Ag+(aq) + Cl-(aq)
Ksp =
CaF2(s) = Ca2+(aq) + 2F-(aq)
Ksp =
in general AmBn = mAn+ + nBmKsp = [An+]m[Bm-]n
We use Ksp to calculate the equilibrium
solubility of a compound.
Calculating the Solubility of an Ionic Compound (p.131)
PbI2 = Pb2+ + 2I-
in general for AmBn = mAn+ + nBm-
The Common Ion Effect (p. 132)
What happens to the solubility of PbI2 if we add a
second source of I- (e.g. the PbI2 is being
dissolved in a solution of 0.030 M NaI)?
The common ion =
PbI2 = Pb2+ + 2I-
Ch 12: A Deeper Look at Chemical
Equilibrium
Up to now we've ignored two points1.
2.
PbI2(s) = Pb2+(aq) + 2I-(aq)
Ksp = 7.9 x 10-9 (ignoring PbOH+, PbI3-, etc)
K'sp =
Activity Coefficients - concentrations
are replaced by "activities"
aA + bB = cC + dD
[C] c [D] d
K
[A]a [B]b
K
old definition
γ cC [C]c γ dD [D]d
γ [A] γ B [B]
a
A
a
b
b
new definition using activity
We can calculate the activity coefficients if we know what
the ionic strength of the solution is.
Charge Effects - an ion with a +2 charge
affects activity of a given electrolyte more
than an ion with a +1 charge
 = ionic strength, a measure of the magnitude of the
electrostatic environment
1
μ
2

Ci Zi2
Ci = concentration
Zi = charge
e.g. calculate the ionic strength of an aqueous soln of
0.50M NaCl and 0.75M MgCl2
The Extended Huckel-Debye Equation
log A 
 0.51z 2 
1    / 305
A = activity coefficient
Z = ion charge
 = ionic strength (M)
 = hydrated radius (pm)
works well for   0.10M
Example (p. 262) - Find the activity
coefficient in a solution of 3.3 mM Mg(NO3)2
Data from Table 12-1:
Example (p. 264) –
A Better Estimate of the Solubility of PbI2
PbI2 = Pb2+ + 2I-
The Real Definition of pH
pH  logΑ H  logγ H [H ]
What is the concentration of H+ in (a) pure H2O and (b)
0.10M NaCl?
Systematic Treatment of Equilibria
(Sec 12-3 and 12-4)
A procedure for solving any equilibrium problem no matter
how complicated.
Charge Balance - the sum of the positive charges in
solution must equal the sum of negative charges.
e.g. sulfate ion CSO42- = 0.0167 M
Charge balance equation, p. 266
Solution containing H+, OH-, K+, H2PO4-,
HPO42-, and PO43-
General charge balance equation n1[C1] + n2[C2] +…. = m1[A1] + m2[A2] + …
where
C = cation concentration
n = cation charge
A = anion concentration
m = anion charge
e.g. write the charge balance equation for a soln of
Na2SO4 and NaCl in water.
Mass Balance
The sum of all substances in solution containing a
particular atom (or group of atoms) must equal the
quantity added to solution.
e.g. solution of 0.050 M HAc
HAc
HAc
HAc
HAc
HAc
HAc
HAc
HAc
HAc
HAc
Mass Balance
e.g. solution of 0.025 M H3PO4
H3PO4
H3PO4
H3PO4
H3PO4
H3PO4
H3PO4
H3PO4
H3PO4
H3PO4
H3PO4
Mass balance for a sparingly soluble
salt is different:
e.g. CaF2
General Procedure
1.
2.
3.
4.
5.
6.
Write down all the relevant chemical equations
Write the charge blance
Write the mass balance
Write down the equilibrium constant expressions
(only step where activities may be used)
Make sure that the number of unknowns equals
the number of equations
Solve the system of equations
-make approximations
-use a computer
Coupled Equilibria: Solubility of CaF2
1. Relevant equations
Ca2+ + 2 F-
CaF2(s)
F- + H2O
H2O
HF + OH-
H+ + OH-
2. Charge Balance
Coupled Equilibria: Solubility of CaF2
3. Mass Balance
4. Equilibrium Expressions
Coupled Equilibria: Solubility of CaF2
5. Number of equations = number of unknowns
[H+], [OH-], [Ca2+], [F-], [HF] = unknowns
6. Simplifying Assumptions and Solution
• fix the pH using a buffer { [H+] = CH+ } , removes one
unknown
• adding a buffer and associated ions nullifies the
charge balance equation
• so now we have 4 equations and 4 unknowns
After buffering to pH = 3.0, [H+] = 1.0 x 10-3 M
[OH-] = Kw/[H+] = 1.0 x 10-11 M and now subst into Kb
Kb
[HF]
1.5 x 10 11


 1.5


11
[F ] [OH ] 1.0 x 10
[HF] = 1.5[F-] and now subst 1.5[F-] for [HF] in the mass
balance equation
[F-] + [HF] = 2[Ca2+]
[F-] + 1.5[F-] = 2[Ca2+]
[F-] = 0.80[Ca2+] and finally subst 0.80[Ca2+] for [F-] into Ksp
[Ca2+][F-]2 = Ksp
[Ca2+](0.80[Ca2+])2 = Ksp
[Ca2+] = (Ksp/0.802)1/3 = 3.9 x 10-4 M