Chem 1721
Brief Notes: Chapters 16 and 17
Chapter 16: Acid-Base Equilibria; Chapter 17: Solubility and Complex-Ion Equilibria

still largely focussed on acid/base chemistry; shift focus to solutions that contain both an acid and a base

Common Ion Effect and Buffer Solutions

Titrations, pH Curves and Indicators

Solubility Equilibria
Common Ion Solutions

solutions that contain a conjugate acid/base pair
presence of an ion common to the equilibrium from a secondary source
in sol’ns of acids & bases, the presence of a common ion results in a ΔpH (increase for acids, decrease for bases)
this is called the common ion effect (will see this again at the end of the chapter with solubility equilibria)

What we want to know: What happens to a solution of an acid or base if a salt containing its conjugate is added?
How is the dissociation equilibrium affected? Does the pH change? % dissociation?
ex. calculate the pH of 0.25 M HC2H3O2 (aq) AND 0.25 M HC2H3O2 (aq) + 0.15 M NaC2H3O2 (aq)
HC2H3O2 is a weak acid (Ka = 1.8 x 10 5); C2H3O2 is the conjugate base
⎯
note:
initial [ ]
HC2H3O2
0.25 M
equil [ ]
(0.25 – x) M
initial [ ]
HC2H3O2
0.25 M
equil [ ]
(0.25 – x) M
⎯
+
H2O
↔
C 2 H3 O2
0M
H3O+
0M
+
⎯
xM
+
H2O
↔
C 2 H3 O2
0.15 M
xM
H3O+
0M
+
⎯
(0.15 + x) M
xM
the big difference between these 2 calculations is that the initial [C2H3O2 ] in the common ion solution
IS NOT 0 M
⎯
some numbers:
0.25 M HC2H3O2
[H3O+] = 0.0021 M
pH = 2.68
% diss = 0.84%
0.25 M HC2H3O2 + 0.15 M NaC2H3O2
[H3O+] = 3.0 x 10 5 M
pH = 4.52
% diss = 0.012%
⎯
demonstration of LeChatelier’s principle; presence of common ion C2H3O2 causes equilibrium to shift to
left; lower [H3O+]; less acidic; higher pH
⎯
ex. calculate the pH of 0.66 M NH3 (aq) AND 0.66 M NH3 + 0.55 M NH4Cl (aq)
NH3 is a weak base (Kb = 1.8 x 10 5); NH4+ is the conjugate acid
⎯
note:
initial [ ]
NH3
0.66 M
equil [ ]
(0.66 – x) M
initial [ ]
NH3
0.66 M
equil [ ]
(0.66 – x) M
+
↔
H2O
NH4+
0M
+
OH
0M
⎯
xM
+
H2O
↔
NH4+
0.55 M
(0.55 + x) M
xM
+
OH
0M
⎯
xM
the big difference between these 2 calculations is that the initial [NH4+] in the common ion solution
IS NOT 0 M
some numbers:
0.66 M NH3
[OH ] = 0.0034 M
pH = 11.53
% diss = 0.52%
0.66 M NH3 + 0.55 M NH4Cl
[OH ] = 2.2 x 10 5 M
pH = 9.34
% diss = 0.0033%
⎯
⎯
⎯
demonstration of LeChatelier’s principle; presence of common ion NH4+ causes equilibrium to shift to left;
lower [OH ]; less basic; lower pH
⎯
Buffer Solutions

solutions that contain both an acid and a base; usually a conjugate acid/base pair

common ion solutions

the job of a buffer solution is to maintain an approximately constant pH when small amounts of strong acid
(H+) or strong base (OH ) are added
⎯

pH of a buffer solution is dependent on: Ka of the acid component, relative [acid] and [base]
dependence given by Henderson-Hasselbalch equation:
[A-]
[base]
pH = pKa + log
pH = pKa + log
OR
[acid]
[HA]

buffer solutions resist a change in pH when H+ or OH are added

H+ or OH consumed in a neutralization reaction; strong acid or base replaced by weak acid or base (HA or A )
⎯
⎯
⎯
added H+ reacts with the base component of the buffer, A : H+ + A  HA
added OH reacts with the acid component of the buffer, HA: OH + HA  A + H2O
⎯
⎯
⎯

⎯
result of neutralization rxn:
⎯
[H+] or [OH ] = 0 after reaction
[HA] and [A ] change (increase or decrease depending on reactant or product)
but [A ]/[HA] remains relatively constant
⎯
⎯
⎯

some calculations
ex. A buffer sol’n composed of 0.25 M HNO2 (pKa = 3.40) and 0.35 M KNO2 has a pH of 3.54. Calculate the
change in pH that occurs when 0.010 mol of strong acid or 0.050 mol strong base are added to 1.00 L of the
buffer solution.
 consider the addition of strong acid first:
strong acid (H+) will react with the base component of this buffer, NO2
⎯
neutralization reaction:
note:

initial mol
H+
0.010 mol
Δ mol
- 0.010 mol
- 0.010 mol
+ 0.010 mol
final mol
0 mol
0.34 mol
0.26 mol
+
NO2
0.35 mol
⎯
HNO2
0.25 mol
the neutralization reaction happens FIRST; it goes very fast and effectively to completion
after reaction: H+ completely consumed (limiting reactant);
[NO2 ] = 0.34 mol/1.00 L = 0.34 M;
⎯
note:
[HNO2] = 0.26 mol/1.00 L = 0.26 M
after the added strong acid is consumed, the acid ionization equilibrium of HNO2 is re-established
solve for pH using and equilibrium table OR using the Henderson-Hasselbalch equation
equilibrium table:
initial [ ]
HNO2 (aq)
0.26 M
equil [ ]
(0.26 ⎯ x) M
+
H2O (l)
----
↔
----
NO2 (aq)
0.34 M
+
⎯
H3O+ (aq)
0
(0.34 + x) M
xM
Ka = 4.0 x 10 4; solve for x
⎯
x = [H3O+] = 3.06 x 10 4
⎯
Henderson-Hasselbalch equation:
pH = pKa + log([NO2 ]/[HNO2]);
pH = 3.40 + log (.34/.26)
using either approach:
ΔpH = 0.02
⎯
pH = 3.52
 consider the addition of strong base:
strong base (OH ) will react with the acid component of this buffer, HNO
⎯
neutralization reaction:
initial mol
Δ mol
- 0.050 mol
- 0.050 mol
+ 0.050 mol
final mol
0 mol
0.20 mol
0.40 mol
⎯
note:
+

OH
0.050 mol
HNO2
0.25 mol
NO2
0.35 mol
+
⎯
H2O
the neutralization reaction happens FIRST; it goes very fast and effectively to completion
after reaction: OH completely consumed (limiting reactant);
⎯
[HNO2] = 0.20 mol/1.00 L = 0.20 M;
note:
[NO2 ] = 0.40 mol/1.00 L = 0.40 M
⎯
after the added strong base is consumed, the acid ionization equilibrium of HNO2 is re-established
solve for pH using and equilibrium table OR using the Henderson-Hasselbalch equation
equilibrium table:
initial [ ]
HNO2 (aq)
0.20 M
equil [ ]
(0.20 ⎯ x) M
+
H2O (l)
-------
↔
NO2 (aq)
0.40 M
⎯
(0.40 + x) M
+
H3O+ (aq)
0
xM
Ka = 4.0 x 10 4; solve for x
⎯
x = [H3O+] = 2.0 x 10
4
⎯
Henderson-Hasselbalch equation:
pH = pKa + log([NO2 ]/[HNO2]);
pH = 3.40 + log (.40/.20)
using either approach:
ΔpH = 0.30
⎯
pH = 3.70
buffer capacity and buffer failure:
 buffer capacity refers to how much H+ or OH can be added before a buffer fails; buffer capacity can also be
interpreted in terms of how large or small the ΔpH is upon addition of H+ or OH
 a buffer fails when ΔpH > 1 pH unit
⎯
⎯

consider 2 HF + NaF buffer solutions
0.25 M HF + 0.50 M NaF
0.050 M HF + 0.100 M NaF
pH = 3.76
pH = 3.76
[F ]/[HF] = 2
[F ]/[HF] = 2
after the addition of some strong acid (specifically 0.0020 mol to 0.100 L of solution)
pH = 3.71
pH = 3.51
[F ]/[HF] = 1.78
[F ]/[HF] = 1.14
⎯
⎯
⎯
⎯
⇑⇑⇑
this buffer is said to have a better buffer capacity – smaller ΔpH when H+ is added


buffers are most efficient when [base]/[acid] ≈ 1
the acid component of the buffer solution should have pKa close to the desired pH (pKa = desired pH + 1)
ex.
Consider these acids:
acid, HA
pKa
base, A
H2PO4
NH4+
HF
HC7H5O2
7.21
9.25
3.14
4.19
HPO42
NH3
F
C7H5O2
⎯
ex.
⎯
⎯
⎯
⎯
pH range of best buffering;
pH = pKa + 1
6.21 – 8.21
8.25 – 10.25
2.14 – 4.14
3.19 – 5.19
Calculate the pH of a buffer prepared by mixing 1.04 M NaF and 0.86 M HF. (for HF pKa = 3.14)
answer: pH = 3.22
ex.
Calculate the [NaC7H5O2] required to prepare a buffer of pH 4.10 with 0.249 M HC7H5O2 (pKa = 4.19).
answer: [C7H5O2 ] = 0.202 M
⎯
Titrations
see Titration Calculation Strategies handout on Chem 173 web page:
http://www2.onu.edu/~sbates/chem173/handouts.html

an analytical technique; volulmetric analysis

determination of composition i.e. molar concentration, mass %, etc.

reaction between a titrant (known composition) and analyte (unknown composition)

one reactant added to the other until the reaction is just complete
stoichiometrically correct mole ratio of reactants
no limiting reactant, no excess reactant
reactants completely consumed, only products present
this is the stoichiometric point (or end point) of the titration
ex. for the reaction 3 A + 2 B  C + 2 D2
at the stoichiometric point the mol ratio of A:B is 3:2
A and B completely consumed; only C and D2 present

titrations require an indicator – some indication (usually visual) of when you have reached the stoichiometric point
color change
specific pH change

different types of titrations
acid/base
oxidation/reduction (i.e. your Solubility of Ca(IO3)2 lab)
complexiometric (i.e. determination of Ni concentration in a plating bath)

strong acid + strong base titrations
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
Cl and Na+ are spectator ions
net ionic equation: H+ + OH  H2O
⎯
⎯

ex. consider the titration of 40.0 mL of 0.110 M HCl titrated with 0.095 M NaOH
mol H+ = (0,0400 L)(0.110 mol/L) = 0.00440 mol H+
1:1 stoichiometry, so at stoichiometric point mol H+ = mol OH = 0.00440 mol
volume NaOH required to reach stoichiometric point? vol NaOH = (0.00440 mol)÷(0.095 mol/L) = 0.0463 L
⎯
before the stoichiometric point: after the addition of 10.00 mL NaOH (aq)
mol OH = (0.0100 L)(0.095 mol/L) = 9.5 x 10 4 mol OH added
⎯
⎯
⎯
mol before:
H+
0.00440 mol
Δ mol:
- 0.00095 mol
- 0.00095 mol
mol after:
0.0035 mol
0
+
OH
0.00095 mol
⎯

after neutralization reaction H+ remains in excess (OH is limiting reactant)
[H+] = 0.0035 mol/(0.0400 + 0.0100)L = 0.070 M
pH = 1.15
⎯
H2O
AT the stoichiometric point: after the addition of 46.30 mL NaOH (aq) (see calculation above)
mol OH = (0.04630 L)(0.095 mol/L) = 0.00440 mol OH added
⎯
⎯

mol before:
H+
0.00440 mol
Δ mol:
- 0.00440 mol
- 0.00440 mol
mol after:
0
0
+
OH
0.00440 mol
⎯
H2O
after neutralization reaction [H+] and [OH ] = 0, only H2O present
in pure water at 25°C, [H+] = 1.0 x 10 7 M
pH = 7.00
⎯
⎯
beyond the stoichiometric point: after the addition of 60.00 mL NaOH (aq)
mol OH = (0.0600 L)(0.095 mol/L) = 0.0057 mol OH added
⎯
⎯
mol before:
H+
0.00440 mol
Δ mol:
- 0.00440 mol
mol after:
0
+

OH
0.0057 mol
⎯
H2O
- 0.00440 mol
0.0013 mol
+
after neutralization reaction OH remains in excess (H is limiting reactant)
[OH ] = 0.0013 mol/(0.0400 + 0.0600)L = 0.13 M
pOH = 1.89
pH = 12.11
⎯
⎯

pH curves for acid/base titrations (see Titrations handout)
strong acid + strong base: H+ + OH  H2O
start at low pH (excess H+)
stoichometric point at pH = 7
end at high pH (excess OH )
⎯
⎯

ex. weak acid + strong base: HA + OH  A + H2O
start at acidic, but slightly higher pH (excess HA, but weak acid)
stoichiometric point at pH > 7; HA and OH consumed but A (a weak base) in solution
end at high pH (excess OH )
⎯
⎯
⎯
⎯
⎯
before the stoichometric point – a buffer solution exists (HA and A present)
⎯
special point in a weak/strong titration:
at the half-way point (i.e. half way to the stoichometric point): pH = pKa
+
OH
0.025 mol

mol before:
HA
0.050 mol
Δ mol:
- 0.025 mol
- 0.025 mol
+ 0.025 mol
mol after:
0.025 mol
0
0.025 mol
⎯
A
0
⎯
+
H2O
mol HA = mol A ∴ [HA] = [A ]
pH = pKa + log([A ]/[HA])
pH = pKa + log(1); pH = pKa
⎯
⎯
⎯
Consider specifically the titration of 30.0 mL of 0.125 M HC2H3O2 (aq) with 0.100 M NaOH (aq)
note:
see Titrations handout for details of calculations, reaction tables, pH calculations, and pH curve
pH after the addition of 20.0 mL NaOH (aq)
pH = 4.80
pH at the half-way point
pH = 4.74
pH at the stoichiometric point
pH = 8.75
pH after the addition of 60.0 mL NaOH (aq)

pH = 12.40
ex. weak base + strong acid: H+ + B  BH+
start at basic pH (excess B, but weak base)
stoichiometric point at pH < 7; H+ and B consumed but BH+ (a weak acid) in solution
end at low pH (excess H+)
before the stoichometric point – a buffer solution exists (B and BH+ present)
special point in a weak/strong titration: see discsusion above too
at the half-way point (i.e. half way to the stoichometric point): pH = pKa; here pKa of BH+ ( or 14 ⎯ pKb)
Consider specifically the titration of 20.0 mL of 1.20 M CH3NH2 (aq) with 0.75 M HNO3 (aq)
note:
see Titrations handout for details of calculations, reaction tables, pH calculations, and pH curve
pH at the half-way point
pH = 10.57
pH after the addition of 20.0 mL HNO3 (aq)
pH = 10.35
pH at the stoichiometric point
pH = 5.46
pH after the addition of 60.0 mL HNO3 (aq)
pH = 1.00

miscellaneous acid + base and titration calculations
ex. What volume of 0.368 M Sr(OH)2 is required to reach the stoichiometric point in the titration of 75.0 mL of
0.44 M HBr (aq)?
HBr (aq) + Sr(OH)2 (aq)  SrBr2 (aq) + 2 H2O (l)
at stoichiometric point: mol Sr(OH)2 = 1/2 mol HBr
logical pathway: vol HBr ⇒ mol HBr ⇒ mol Sr(OH)2 ⇒ vol Sr(OH)2
ex. Determine the molar concentration of HNO3 (aq) if 33.3 mL of 1.04 M LiOH (aq) is required to reach the
stoichiometric point in the titration of 25.5 mL of the acid.
HNO3 (aq) + LiOH (aq)  LiNO3 (aq) + H2O (l)
at stoichiometric point: mol HNO3 = mol LiOH
logical pathway: vol LiOH ⇒ mol LiOH ⇒ mol HNO3; [HNO3] = mol HNO3 ÷ 0.02525 L
ex. 25.00 mL of 0.750 M HCl (aq) and 45.00 mL of 0.840 M KOH (aq) are mixed. Is the resulting solution
acidic, basic, or neutral? What is the pH?
HCl
+
mol before:
Δ mol:
mol after:
identify limiting reactant and excess reactant
determine concentration of the excess reactant
pH = ????
KOH

KCl
+
H2O
Acid Base Indicators

used to visually indicate the stoichiometric point of a titration with a color change

important to choose an indicator with a color change that occurs very close to the stoichiometric pt of the titration;
minimizes experimental error
pH of indicator color change should be close to pH of stoichiometric point

acid/base indicators are typically large organic weak acids; can exist as acid (HIn) OR conjugate base (In )
acid form (HIn) and base form (In ) are 2 different colors
⎯
⎯
HIn (a q) + H2O (l) ↔ In (aq) + H3O+ (aq)
color 1
color 2
⎯
Ka OR KIn =
[In-][H3O+]
KIn
[HIn]
[H3O+]

color is dependent on pH and [In ]/[HIn]

consider 2 extremes:
=
[In-]
[HIn]
⎯
very low pH; acidic solution
[H3O+] is high; common ion pushes equilibrium to left
[HIn] is large, [In ] is small
⎯
∴color of solution is the color of HIn
very high pH; basic solution
[OH ] is high; reaction occurs between OH and H3O+ pushing equilibrium to right: OH + H3O+  2 H2O
[In ] is large, [HIn] is small
⎯
⎯
⎯
⎯
∴ color of solution is the color of In
⎯
in the middle, range of colors depends of pH and [In ]/[H3O+]
⎯

pH of visible color change?
for the human eye to detect the color change 1 part in 10 of the indicator must be converted to the other form

ex. bromothymol blue; KIn = 1.0 x 10 7; HIn is yellow; In is blue
⎯
⎯
If a few drops of bromothymol blue indicator is added to a solution of HCl (aq), what color is the solution
initially? If this HCl (aq) is then titrated with NaOH (aq), at what pH will the color change be visible?
HCl (aq) is an acidic solution ∴ the indicator is in its HIn form; the solution is yellow
for the color change to be visible, 1 part in 10 must change from HIn  In
OR [In ]/[HIn] = 1/10
⎯
⎯
pH = ? Can solve this 2 ways:
[In-][H3O+]
KIn =
[HIn]
OR
[H3O+] = KIn
pH = pKIn + log
[HIn]
[In-]
[In-]
[HIn]
answer: pH = 6.00
note: this same indicator in a basic solution would start blue (indicator in In form); the color change would
occur at [In ]/[HIn] of 10/1; pH = 8.00
⎯
⎯

the general rule is: an indicator will change color at pH = pKIn + 1; for a titration choose an indicator with a pKIn
value close (+ 1) to the pH of the stoichiometric point of the titration,
Solubility Equilibria

equilibrium established between an ionic solid and its ions in solution: (s) ↔ (aq);
a saturated solution is one in which solid and ions exist in dynamic equilibrium

heterogeneous equilibrium

equlibrium constant, Ksp, solubility product constant

ex. write the chemical equation for the solubility equilibrium and Ksp expression for:
magnesium fluoride:
MgF2 (s) ↔ Mg2+ (aq) + 2 F (aq);
⎯
Ksp = [Mg2+][F ]2
⎯
calcium phosphate:
Ca3(PO4)2 (s) ↔ 3 Ca2+ (aq) + 2 PO43 (aq);
⎯
Ksp = [Ca2+]3[PO43 ]2
⎯
Calculating Ksp for Saturated Solutions

consider 2 different ways to prepare a saturated solution of calcium fluoride:
1.
solid CaF2 added to water and let stand; over time the solution will become saturated
CaF2 (s) ↔ Ca2+ (aq) + 2 F (aq)
measure experimentally [Ca2+] and [F ]; calculate Ksp
⎯
⎯
so . . . if [Ca2+] = 3.3 x 10 4 M and [F ] = 6.7 x 10 4 M
⎯
⎯
⎯
Ksp = [Ca2+][F ]2 = (3.3 x 10 4)(6.7 x 10 4)2 = 1.5 x 10 10
⎯
2.
⎯
⎯
note: here [F ] = 2•[Ca2+]
⎯
⎯
mix 2 solutions together – one with Ca2+ (aq) and the other with F (aq); CaF2 (s) will precipitate from solution,
but some Ca2+ and F will remain in solution and equilibrium will be established
⎯
⎯
CaF2 (s) ↔ Ca2+ (aq) + 2 F (aq)
measure experimentally [Ca2+] and [F ]; calculate Ksp
⎯
⎯
so . . . if [Ca2+] = 1.5 x 10 4 M and [F ] = 1.0 x 10 3 M
⎯
⎯
⎯
Ksp = [Ca2+][F ]2 = (1.5 x 10 4)(1.0 x 10 3)2 = 1.5 x 10 10
⎯

⎯
⎯
note: here [F ] ≠ 2•[Ca2+]
⎯
⎯
and knowing Ksp you can calculate the concentration of an ion in an saturated solution
Calculate [Ca2+] in a saturated solution of CaF2 with [F ] = 0.010 M.
⎯
answer: [Ca2+] = 1.5 x 10 6 M
⎯
Quantifying Solubility in mol/L and g/L

solubility is defined in terms of mol of solid that will dissolve per L of solution (molar solubility), OR g of solid
dissolved per L of solution

use an equilibrium table to determine solubility if you know Ksp, OR to determine Ksp if you know solubility

some examples
ex. Calculate the solubility of MgF2 in mol/L and g/L. For MgF2, Ksp = 7.4 x 10 11.
let x = mol of MgF2 (s) that will dissolve per L of sol’n; i.e. x = molar solubility of MgF2
⎯
↔
Mg2+ (aq)
0
initial [ ]
MgF2 (s)
----
+
2 F (aq)
0
Δ[]
----
+x
+ 2x
equil [ ]
-----
xM
2x M
⎯
Ksp = [Mg2+][F ]2;
7.4 x 10 11 = (x)(2x)2 = 4x3
⎯
⎯
x = molar solubility = 2.6 x 10 4 mol/L; i.e. 2.6 x 10 4 mol MgF2 will dissolve per L of solution at 25°C
⎯
⎯
g/L solubility? use molar mass of MgF2 (62.31 g/mol) to convert from mol/L to g/L
0.016 g MgF2/L
ex. A saturated solution of silver chromate is found to have [Ag+] = 1.3 x 10 4 M. Calculate Ksp for Ag2CrO4.
⎯
initial [ ]
Ag2CrO4 (s)
----
Δ[]
----
equil [ ]
2 Ag+ (aq)
0
↔
----
+ 2
2⎯
2
CrO42 (aq)
0
+
⎯
+ 2x
+x
2x M
xM
3
Ksp = [Ag ] [CrO4 ] = (2x) (x) = 4x
[Ag+] = 2x = 1.3 x 10 4 M ∴ x = 6.5 x 10 5
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Ksp = 4(6.5 x 10 5)3 = 1.1 x 10
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12
Relative Solubilities of Ionic Compounds

Which is more soluble: PbCl2 (Ksp = 1.6 x 10 5) OR PbF2 (Ksp = 4 x 10 8)?
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Which is more soluble: PbCl2 OR CaSO4 (Ksp = 6.1 x 10 5)
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When can you compare Ksp’s directly to determine relative solubility?

When comparing 2 compounds that produce the same number of ions in solution, Ksp values can be compared
directly to determine relative solubility
larger Ksp  equilibrium position farther to right  higher [ion] in solution  greater solubility
i.e. PbCl2 vs PbF2
PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl (aq); Ksp = [Pb2+][Cl ]2 = 4x3
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PbF2 (s) ↔ Pb2+ (aq) + 2 F (aq); Ksp = [Pb2+][F ]2 = 4x3;
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
answer: PbCl2 is more soluble
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When comparing 2 compounds that do not produce the same number of ions in solution, Ksp values can NOT be
compared directly; you must calculate x (molar solubility) to determine which compound is more soluble
i.e. PbCl2 vs. CaSO4
PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl (aq); Ksp = [Pb2+][Cl ]2 = 4x3
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CaSO4 (s) ↔ Ca2+ (aq) + SO42 (aq); Ksp = [Ca2+][SO42 ] = x2
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answer: PbCl2 is more soluble
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Factors the Affect Solubility
1.
The Common Ion Effect
same idea as with acids and bases; the presence of an ion common to the solubility equilibrium will cause the
equilibrium position to shift and affect solubility
for ionic compounds, the presence of a common ion results in decreased solubility
ex. Calculate the solubility of MgF2 in 0.10 M NaF (aq).
Above we calculated the solubility of MgF2 in water at 25°C: 2.6 x 10 4 mol/L, 0.016 g/L.
⎯
↔
Mg2+ (aq)
0
initial [ ]
MgF2 (s)
----
+
2 F (aq)
0.10 M
Δ[]
----
+x
+ 2x
equil [ ]
-----
xM
(.10 + 2x) M
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Ksp = [Mg2+][F ]2;
⎯
7.4 x 10 11 = (x)(.10 + 2x)2 ≈ (x)(0.10)2
⎯
x = molar solubility = 7.4 x 10 9 mol/L OR 4.6 x 10 7 g MgF2/L at 25°C
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ex. Calculate the solubility of Pb(OH)2 in a solution with pH = 10.0. For Pb(OH)2 Ksp = 1.2 x 10 15
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in a solution with pH = 10.0, pOH = 4.0 and [OH ] = 10 4 M; this is a common ion solution
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↔
⎯
Pb2+ (aq)
0
initial [ ]
Pb(OH)2 (s)
----
+
2 OH (aq)
10 4 M
Δ[]
----
+x
+ 2x
equil [ ]
----
xM
(10 4 + 2x) M
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answer: solubility = 1.2 x 10 7 mol/L
vs. 6.7 x 10 6 mol/L in a neutral sol’n
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2.
pH of the solution
general rule: the solubility of an ionic compound that contains a basic anion will increase as the pH of the
solution decreases
also an application of Le Chatelier’s Principle
consider calcium carbonate: CaCO3 (s) ↔ Ca2+ (aq) + CO32 (aq)
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in an acidic solution [H3O+] is relatively high: H3O+ + CO32  HCO3 + H2O
[CO32 ] decreases as the ion is removed from solution by reaction with H3O+
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equilibrium position shift to the right (to replenish CO32 ), and solubility increases
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other common basic anions: OH , CN , PO43 , S2 , F , SO42 , C2H3O2 , NO2
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note: the solubility of salts with neutral anions (Cl , Br , I , NO3 and ClO4 ) will NOT be affected by pH
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3.
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formation of complex ions
the solubility of an ionic compound will increase dramatically if the solution contains a Lewis Base that can
form a Lewis acid-base adduct with a metal cation
again, Le Chatelier’s Principle in action:
consider AgCl; insoluble in water and acid, but will dissolve in solutions containing NH3 (aq)
AgCl (s) ↔ Ag+ (aq) + Cl (aq)
in solutions containing NH3 (aq): Ag+ (aq) + 2 NH3 (aq) ↔ Ag(NH3)2+ (aq)
⎯
as Ag(NH3)2+ forms, [Ag+] decreases; equilibrium shifts to the right (to replenish Ag+); solubility increases
the formation of Ag(NH3)2+ (aq) is an equilibrium; Kf = formation constant or stability constant
consider a series of equations:
step 1:
AgCl (s) ↔ Ag+ (aq) + Cl (aq)
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step 2:
Ag+ (aq) + NH3 (aq) ↔ AgNH3+ (aq)
step 3:
AgNH3+ (aq) + NH3 (aq) ↔ Ag(NH3)2+ (aq)
net:
AgCl (s) + 2 NH3 (aq) ↔ Ag(NH3)2+ (aq) + Cl (aq)
⎯
Ion Product (Q) and Predictions about Precipitation:
 compare Q and Ksp
 if Q = Ksp the solution is at equilibrium; the solution is saturated – no more solid will dissolve and no precipitation
will occur
 if Q < Ksp the solution is unsaturated – reaction will proceed in forward direction; more solid will dissolve, no
precipitation will occur
 if Q > Ksp the solution is supersaturated – reaction will proceed in the reverse direction; precipitation will occur
 remember to calculate the [ion] in the solution that results from combining 2 volumes;
i.e. [ion] = mol ion/total solution volume
 some examples
ex. 250.0 mL of 0.0062 M AgNO3 (aq) mixed with 250.0 mL of 0.00014 M Na2CO3 (aq). Will a precipitate of
Ag2CO3 (s) form? For Ag2CO3 Ksp = 8.1 x 10 12. Total solution volume after mixing = 500.0 mL.
⎯
answer: Q > Ksp ∴ Ag2CO3 precipitate will form
ex. Calculate the minimum [CO32 ] required to precipitate Ag2CO3 from 5.8 x 10 4 M AgNO3.
Precipitation occurs when Q > Ksp
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⎯
answer: [CO32 ] = 2.4 x 10 5 M
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