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Ch. 14/15: Acid-Base Equilibria
Sections 14.6, 14.7, 15.1, 15.2
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The mineral fluorite (CaF2) is deposited through a precipitation process. Note
that pure fluorite is colorless, and that the color in this sample is due to the
presence of other metals in the crystal.
Neutralization Reactions
Buffers and the Common Ion Effect
!  Write the molecular, ionic, and net ionic equations for the
following reactions:
!  1) HCl(aq) + NaOH(aq) !
H+(aq) + OH-(aq) ! H2O(l)
!  What happens to HF in solution?
HF(aq) +
+
!
HF(aq) + OH-(aq) ! F-(aq) + H2O(l)
+
F-(aq)
+ H2O(l)
Calculate the pH of a 0.50 M CH3COOH solution (Ka = 1.8x10-5).
pH = 2.52
What will happen to pH after we add 0.10 M NaCH3COO?
14.6 Buffer Solutions (Common Ions)
A solution that contains a weak acid and its conjugate base (or
a weak base and its conjugate acid) is a buffer.
acid
conjugate base
Buffer Solutions
The pH of a buffer solution can be calculated using two initial
concentrations in an ICE table, or we can use the HendersonHasselbalch equation (derived from the Ka equation).
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq)
reacts with
added base
H3O+(aq) + CH3COO–(aq)
H 2O
Na+(aq) + CH3COO-(aq) addition
Equilibrium is driven toward reactants
!  2) HF(aq) + NaOH(aq) !
Na+(aq)
CH3COOH(aq) + H2O(l)
NaCH3COO(aq)
!  or H3O+(aq) + OH-(aq) ! 2H2O(l)
OH-(aq)
What will happen when we add sodium acetate to a solution of
acetic acid?
A system at equilibrium will shift in response to being stressed.
The addition of a reactant or a product can be an applied stress.
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) ! Na+(aq) + Cl-(aq) + H2O(l)
Na+(aq)
2
1
Chemistry: OpenStax
reacts with
added acid
•  Buffer solutions resist changes in pH when small amounts of
strong acid or base are added. They do this because they already
contain both a weak acid and a weak base (that don’t neutralize
each other).
•  When a small amount of strong acid is added, the buffer’s base
will neutralize the added acid.
•  When a small amount of strong base is added, the buffer’s acid
will neutralize the base.
pH = pK a + log (
base
)
acid
What will be the pH of 0.50 M CH3COOH (Ka = 1.8x10-5) after 0.10
M NaCH3COO is added?
pKa = -log Ka = -log (1.8x10-5) = 4.7447
pH = pKa + log (base/acid) = 4.7447 + log (0.10 / 0.50)
pH = 4.05
Example 14.20
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Group Work
!  Identify the solutions below that would make good
buffer solutions (what criteria need to be met?):
! 
! 
! 
! 
! 
! 
! 
! 
HF and NaF
Ammonia and ammonium bromide
KOH and KF
CH3COOH and LiCH3COO
Nitric acid and sodium nitrate
Chlorous acid and potassium chlorite
NaOH and NaCl
HCl and NaCH3COO – think about what these would do
in solution!
15.3 Buffer Solutions (Common Ions)
1) Add strong acid (e.g., HCl)
HCl (aq) + NaCH3COO(aq) ! CH3COOH (aq) + NaCl(aq)
Draw the product beaker of:
Reactant beaker 1 = 2 moles of buffer solution (can use HA/NaA)
Reactant beaker 2 = 1 mole of HCl
2) Add strong base (e.g., NaOH):
NaOH(aq) + CH3COOH(aq) ! NaCH3COO(aq) + H2O(l)
Draw the product beaker of:
Reactant beaker 1 = 2 moles of buffer solution (can use HA/NaA)
Reactant beaker 2 = 1 mole of NaOH
Buffer Solution plus Strong Acid
Buffer Solution plus Strong Base
Buffer Solutions
Buffer Solutions – Figure 14.17
(a) The buffered solution on the left and the unbuffered solution on the right have the
same pH (pH 8); they are basic, showing the yellow color of the indicator methyl
orange at this pH.
(b) After the addition of 1 mL of a 0.01-M HCl solution, the buffered solution has
Figure 14.18: This diagram shows the buffer action of these reactions.
not detectably changed its pH but the unbuffered solution has become acidic, as
indicated by the change in color of the methyl orange, which turns red at a pH of
about 4. (credit: modification of work by Mark Ott)
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Buffer Solutions
Buffer Solutions with Strong Acid
Calculate the pH of 75.0 mL of a 0.10 M acetic acid (Ka =
1.8x10-5) and 0.10 M sodium acetate solution.
Calculate the pH of 75.0 mL of a 0.10 M acetic acid and
0.10 M sodium acetate solution after 1.0 mL of 1.0 M
HCl is added.
The volume doesn’t affect the calculation of the HendersonHasselbalch equation. However, we’ll need volumes when we
calculate pH after adding strong acid or base. Because volumes are
often measured in mL, it is helpful to use a new unit, mmol
(millimole).
The added acid (H3O+) reacts with the conjugate base (acetate ion).
Since a strong acid is added, the reaction goes to completion. Use
a change table to determine how many moles remain after
reaction. Then use Henderson-Hasselbalch eqn to calculate pH.
mol 1000 mmol mmol
M=
=
=
L
1000 mL
mL
M = mmol / mL
CH3COO-(aq)
H3O+(aq) +
! H2O(l)
CH3COOH(aq)
mmol = M x mL
pH = 4.7447 + log (6.5 / 8.5) = 4.63
Buffer Solutions with Strong Base
Calculate the pH of 75.0 mL of a 0.10 M acetic acid and
0.10 M sodium acetate solution after 1.0 mL of 1.0 M
NaOH is added.
The added base (OH-) reacts with the acid (acetic). Since a strong
base is added, the reaction goes to completion. Use a change
table to determine how many moles remain after reaction. Then
use Henderson-Hasselbalch eqn to calculate pH.
CH3COOH(aq)
OH-(aq) +
! H2O(l)
CH3COO-(aq)
Buffer Solutions – Practice Calculations
!  Calculate the pH of a 50.0 mL solution made by
combining 0.20 M HCOOH with 0.50 M NaHCOO.
Ka = 1.8x10-4
!  pH = 4.14
!  Practice: What is pH when 10.00 mL of 0.50 M HCl is
added? pH = 3.87
!  Work in class: What is pH when 30.00 mL of 2.0 M HCl is
added? pH = 0.36
!  Practice: What is pH when 10.00 mL of 0.50 M NaOH is
added? pH = 4.52
Problem 14.20
pH = 4.7447 + log (8.5 / 6.5) = 4.86
Buffer Solutions – Practice Calculations
Buffer Solutions
!  The pH of a 50.0 mL buffer solution made by combining 0.20
!  What is the pH of a 1.00 L solution that is 0.75 M NH3
M HCOOH and 0540 M NaHCOO is 4.14. Ka = 1.8x10-4
!  What is pH when 30.00 mL of 2.0 M HCl is added?
HCOO-(aq)
H3O+(aq) +
! H2O(l)
HCOOH(aq)
and 0.25 M NH4Cl? Kb = 1.8 x 10-5
!  Kb ! Ka ! pKa OR Kb ! pKb ! pKa
!  pH = pKa + log ([NH3]/[NH4+]) = 9.2552+log (0.75/0.25)
!  pH = 9.73
!  What is the pH of this solution after 0.01 mol HBr is
added to it?
!  [H3O+] = 35 mmol / (50 + 30) mL = 0.4375 M
!  pH = 0.36
!  [H3O+] from HCOOH = 0.00887 M (relatively negligible)
!  pH = 9.25 + log (0.74/0.26) =
!  pH = 9.70
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ICE vs Change Tables
Arrows
When to use
Units
Buffer Capacity
ICE table
Change Table
!
!
Weak acid or
base in water
molarity
At least one strong
reactant
moles
Solve for x at Find x by determining
How to find x
equilibrium
the L.R.
Initial, Change, Initial, Change, Final
Labels
Equilibrium
Buffer Solutions
Buffers have a certain range in which they can maintain a
constant pH. Outside of this range, we say the buffer’s
capacity has been exceeded and we see large changes in pH.
The best capacity for a buffer is when the conjugate acid and
base concentrations are within a factor of 10 (the best capacity
is when the acid and base concentrations are equal).
10 ≥
[base]
≥ 0.1
[acid]
The pH of a buffer cannot be more than one pH unit different than
the pKa of the weak acid it contains.
Buffer Solutions
To make a buffer with a specific pH:
1)  Pick a weak acid whose pKa is close to the desired pH
(usually within one unit of desired pH).
2)  Substitute the pH and pKa into the HendersonHasselbalch equation to calculate the [base]/[acid] ratio.
pH = pK a + log (
base
)
acid
Figure 14.20: The graph, an illustration of buffering action, shows change of
pH as an increasing amount of a 0.10-M NaOH solution is added to 100 mL of
a buffer solution in which, initially, [CH3CO2H] = 0.10 M and [CH3CO2−] =
0.10 M.
Buffer Solutions
Example:
1)  If we want to make a buffer with a pH of 5.00, which
acid solution should we start with?
Ascorbic acid: Ka = 8.0x10-5
Acetic acid: Ka = 1.8x10-5
Hydrofluoric acid: Ka = 7.1x10-4
Citric acid: Ka = 3.5 x 10-4
2) Acetic acid gives a pKa of 4.7447.
pH – pKa = log (base/acid) = 0.2553
14.7 Strong Acid-Strong Base Titrations
The reaction between the strong acid HCl and the strong base NaOH
can be represented by:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Review terminology:
Titrant – solution in buret, being added
Analyte – unknown solution
Equivalence point – stoichiometric equivalents
End point – lightest color change
Calculate the volume of base needed to reach the equivalence point
when 25.00 mL of 0.10 M HCl is titrated with 0.10 M NaOH.
(base / acid) = 100.2553 = 1.80 : 1 (the base should be 1.8x more
concentrated than the acid)
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Titration Beaker Drawings
!  Draw 4 beakers with strong acid.
!  HCl + NaOH ! NaCl + H2O
!  1) 2 moles strong acid, 0 moles base
Acid-Base Titrations
!  4 key points on a titration curve
!  At the beginning (before titrant is added)
Figure 14.21
!  0.00 mL NaOH
!  Before the equivalence point
!  10.00 mL NaOH
!  At the equivalence point
!  25.00 mL NaOH
!  After the equivalence point
!  30.00 mL NaOH
!  2) 2 moles strong acid, 1 mole base
!  What is the limiting reagent?
!  3) 2 moles strong acid, 2 moles base
!  What is the limiting reagent?
!  4) 2 moles strong acid, 3 moles base
!  What is the limiting reagent?
Acid-Base Titrations: HCl + NaOH
1) 0.00 mL NaOH
HCl(aq) +
NaOH(aq)
Acid-Base Titrations: HCl + NaOH
2) 10.00 mL NaOH
! H2O(l)
NaCl(aq)
Test Question Example: What is the pH of 40.00 mL of 0.10 M HCl
titrated with 10.00 mL of 0.10 M NaOH?
HCl(aq) +
NaOH(aq)
! H2O(l)
NaCl(aq)
Only strong acid in solution
[HCl] = [H3O+]
NaCl is neutral – doesn’t affect pH
[HCl] = [H3O+] = 1.50 mmol / (25+10 mL) = 0.04286 M
pH = 1.37
pH = -log (0.10) = 1.00
Acid-Base Titrations: HCl + NaOH
3) 25.00 mL NaOH added
At the equivalence point, what is in solution?
Acid-Base Titrations: HCl + NaOH
4) 30.00 mL NaOH added
HCl(aq) +
HCl(aq) +
NaOH(aq)
! H2O(l)
NaOH(aq)
! H2O(l)
NaCl(aq)
NaCl(aq)
mmol acid = mmol base; only neutral salt in solution
pH = 7.00 (always 7 for strong acid-strong base!)
Base is in excess.
[NaOH] = [OH-] = 0.50 mmol / (25+30 mL) = 0.0090909 M
pH = 14 – (-log 0.0090909) = 11.96
Problem 14.21
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Acid-Base Titration Indicators
Weak Acid-Strong Base Titrations
!  Figure 14.22: Different indicators can be used depending on
where the pH at the equivalence point is expected to be.
Consider the neutralization between acetic acid and sodium
hydroxide:
CH3COOH(aq) + NaOH(aq) → NaCH3COO (aq) + H2O(l)
The acetate ion that results from this neutralization undergoes
hydrolysis:
CH3COO– (aq) + H2O(l)(aq) ⇌ CH3COOH(aq) + OH–(aq)
Calculate the volume of 0.10 M NaOH needed to reach the
equivalence point in a titration of 25.00 mL of 0.10 M acetic acid.
Ka = 1.8x10-5
Titration Beaker Drawings
!  Repeat except now use a WEAK acid instead of strong.
!  CH3COOH + NaOH ! NaCH3COO + H2O
!  1) 2 moles weak acid, 0 moles base
Acid-Base Titrations: CH3COOH + NaOH
!  4 regions in titration curve (same as SA-SB titration):
!  1) Before adding base
!  0.00 mL base
!  2) 2 moles weak acid, 1 mole base
!  2) Before equivalence point
!  What is the limiting reagent?
!  10.00 mL base
!  3) At equivalence point:
!  3) 2 moles weak acid, 2 moles base
!  What is the limiting reagent?
!  25.00 mL base
!  4) After equivalence point
!  4) 2 moles weak acid, 3 moles base
!  30.00 mL base
!  What is the limiting reagent?
Acid-Base Titrations – Comparing Curves
SA – SB Titration
Acid-Base Titrations – Comparing Curves
WA – SB Titration
Figure 14.23: The graph shows a titration curve for the titration of 25.00 mL of
0.100 M CH3CO2H (weak acid) with 0.100 M NaOH (strong base) and the titration
curve for the titration of HCl (strong acid) with NaOH (strong base). The pH ranges
for the color change of phenolphthalein, litmus, and methyl orange are indicated by
the shaded areas.
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Acid-Base Titrations – Comparing Curves
Acid-Base Titrations: CH3COOH + NaOH
1) 0 mL base.
HA(aq) +
NaOH(aq)
! H2O(l)
NaA(aq)
This is just weak acid. How do we find the pH of a weak acid?
Ka = x2 / (0.10 M – x); [H3O+] = 0.001342 M
pH = 2.87
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Acid-Base Titrations: CH3COOH + NaOH
Acid-Base Titrations: CH3COOH + NaOH
2) 10.00 mL base.
3) 25.00 mL base.
This is the buffer zone (both weak acid and its conjugate base are
present). Set up a change table to find mmoles excess acid and
mmoles cong. base produced. All strong base is converted to conjugate
acid (strong base is completely consumed).
HA(aq) +
NaOH(aq)
! H2O(l)
HA(aq) +
NaOH(aq)
! H2O(l)
NaA(aq)
NaA(aq)
pH = pKa + log ([A-] / [HA]) = 4.7447 + log (1.00 / 1.50)
pH = 4.57
Calculate pH after adding 12.50 mL of NaOH.
At equivalence point: only salt in solution
Basic salt determines pH
NaCH3COO ! Na+ (aq) + CH3COO- (aq)
Acetate ion hydrolyzes to shift pH
Acid-Base Titrations: CH3COOH + NaOH
Acid-Base Titrations: CH3COOH + NaOH
3) 25.00 mL base, continued.
At equivalence point: Find pH of salt. We’ve done this.
Need to find concentration of salt:
2.50 mmoles / (25+25 mL) = 0.05 M
4) 30.00 mL base.
CH3COO– (aq) + H2O(l)(aq) ⇌ CH3COOH(aq) + OH–(aq)
Need Kb because OH- is produced
Kb = Kw / Ka = 1.0x10-14 / 1.8x10-5 = 5.5556x10-10
Kb = x2 / (0.05 – x); x = [OH-] = 5.2705x10-6 M
pOH = 5.2782
pH = 8.72
HA(aq) +
NaOH(aq)
! H2O(l)
NaA(aq)
After the equivalence point:
[NaOH] = [OH-] = 0.50 mmol / (25+30 mL) = 0.0090909 M
[A-] is a minor contributor and doesn’t need to be calculated
[OH-] = 5.2705x10-6 M (doesn’t change [OH-]).
pH = 11.96
Problem 14.22
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Acid-Base Titrations: HCl + NaOH
10-5). pH = 2.37
Table 14.4 – Check pH values in titrations
Volume of 0.10 mmoles of
M NaOH
NaOH
Practice Calculations
!  1) Calculate the pH of a 1.0 M solution of acetic acid (Ka = 1.8 x
pH value
with 0.10 M
HCl
pH value with
0.10 M
CH3COOH
0.00
0.00
1.00
2.87
5.00
0.50
1.18
4.14
10.0
1.0
1.37
4.57
15.0
1.5
1.60
4.92
20.0
2.0
1.96
5.35
25.0
2.5
7.00
8.72
30.0
3.0
11.96
11.96
35.0
3.5
12.22
12.22
40.0
4.0
12.36
12.36
!  2) Calculate the pH of the above solution after 0.50 moles sodium
acetate is added to it. Volume = 1.00 L. pH = 4.44
!  3) Calculate the pH of the above solution (#2) after 0.10 mole of
NaOH is added. pH = 4.57
!  4) Calculate the pH of the solution in #2 after 1.2 moles of
NaOH are added. pH = 13.30
!  What is the pH of a solution made by combining 12.00 mL of
0.10 M HCl with 10.00 mL of 0.15 M NaOH? pH = 12.13
!  What is the pH of a solution made by combining 15.00 mL of
0.12 M ascorbic acid (Ka = 7.9 x 10-5) with 10.00 mL of 0.10 M
NaOH? pH = 4.20
SA-SB and SB-SA Titrations
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WA-SB and WB-SA Titrations
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Polyprotic Acid Titration: H3PO4 with NaOH
!  One eq. pt. for each proton.
!  3 eq. pts. and 3 half eq. pts.
15.1 Insoluble Salts (Ksp)
!  Some combinations of ions in solution form precipitates. Use
! 
! 
! 
! 
the solubility rules from Chapter 4 for the following:
!  NaNO3, PbCl2, NH4Cl, Ag2S
Even insoluble salts dissolve to a small extent. The solid
(undissolved) salt and its dissociated ions establish equilibrium
in solution.
Soluble salt: > 0.10 M aqueous solution dissolves at room
temperature (> 1 g dissolves in 100 mL H2O)
Insoluble salt: < 0.001 M aqueous solution at room temperature
(< 1 g dissolves in 100 mL H2O)
The value of the equilibrium constant, Ksp, (solubility product) is
determined at the saturation point.
http://www.youtube.com/watch?v=BLq5NibwV5g
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Solubility Equilibria
Calculations Using Ksp
Ksp is the solubility product constant. Its value indicates how soluble
or insoluble a salt will be in water. See Appendix J for Ksp values.
Ca2+(aq)
2F–(aq);
[Ca2+][F–]2 =
CaF2(s) ⇌
+
Ksp =
3.4 x
What does a large Ksp value indicate? A small value?
10-11
Write the dissolution equation and Ksp expression for silver sulfide.
Do the same for calcium phosphate.
Review nomenclature!!!!! (Chapter 2)
Figure 15.2: Silver chloride is a sparingly
soluble ionic solid. When it is added to water,
it dissolves slightly and produces a mixture
consisting of a very dilute solution of Ag+ and
Cl– ions in equilibrium with undissolved silver
chloride.
Example 15.1
Solubility Equilibria
We can solve for concentrations (or gram solubility) given Ksp values
or solve for Ksp given concentrations (or gram solubility).
Appendix J has a more complete list of Ksp values!
Substance
Ppt color
Ksp value
AgCl
AgBr
white
off-white
1.6x10-10
5.0x10-13
AgI
PbS
yellow
black
1.5x10-16
7x10-29
PbCl2
white
1.6x10-5
Fe(OH)3
CaCO3
rust red
white
4x10-38
8.7x10-9
BaSO4
white
2.3x10-8
Solubility/Ksp Practice
The Ksp of silver bromide is 5.0 x 10–13. Calculate the molar
solubility and solubility (or just gram solubility).
1) Lead (II) chloride has a Ksp value of 1.6x10-5. Calculate the
molar and gram solubility of lead (II) chloride.
AgBr(s) ⇌ Ag+(aq) + Br–(aq)
Initial concentration (M)
0
0
Change in concentration (M)
+x
+x
x
x
Equilibrium concentration (M)
Ksp =
[Ag+][Br–]
= 5.0 x
10–13
=
x2
x = 7.07107 x 10–7 M (molar sol.)
Solubility: use molar mass of salt;
7.07107x10-7 mol/L * 187.8 g/mol = 1.3x10-4 g/L (gram sol.)
Solubility/Ksp Practice Answers
1) PbCl2: x = 0.016 M; x = 4.4 g/L
2) If [Br-]eq = 0.0210 M, then [Pb2+] = 0.0105 M,
!  Ksp = [Pb2+][Br-]2 = (0.0105)(0.0210)2 = 4.6 x 10-6
3) CaCO3: x = 9.292 x
10-5
M; Ksp = 8.6 x
10-9
4) Fe(OH)3: x = 2.0 x 10-10 M; x = 2.1 x 10-8 g/L
2) If [Br -]eq = 0.0236 M, what is Ksp for PbBr2?
3) The solubility of calcium carbonate is found experimentally
to be 9.3x10-3 g/L. Calculate Ksp for calcium carbonate.
4) Ksp for iron (III) hydroxide is found to be 4.0 x 10-38.
Calculate the molar and gram solubility.
Examples 15.2 – 15.6
Predicting Precipitation of Ionic Compounds
To determine if a precipitate will form as a product of a double
replacement reaction or if all of a salt will dissolve in a given
amount of water, we calculate Qsp and compare to Ksp.
1)  Q < Ksp; no precipitate forms (unsaturated)
•  concentration of ions is not high enough to precipitate
2)  Q = Ksp; no precipitate forms (saturated)
•  solution has maximum conc. of ions to dissolve.
3)  Q > Ksp; precipitate forms (supersaturated; ppt forms)
•  excess ions in solution will cause precipitation
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Precipitation of Ionic Compounds
!  A solution contains 1.5 x 10-2 M hydrochloric acid and 1.2 x
10-2 M lead(II) nitrate. Will a precipitate form? Ksp (PbCl2) =
1.6 x 10-5
! 
! 
! 
! 
! 
! 
2 HCl (aq) + Pb(NO3)2 (aq) ! PbCl2 (s) + 2 HNO3 (aq)
PbCl2 (s) ! Pb2+ (aq) + 2Cl- (aq)
Qsp = [Pb2+][Cl-]2 = (1.2 x 10-2)(1.5 x 10-2)2
Qsp = 2.7 x 10-6
Dissolution of Ionic Compounds
!  245 mg of magnesium carbonate is placed in 1.00 L of water.
Will it all dissolve? Ksp = 4.0 x 10-5
!  Two methods to solve:
!  1) Find molarity of solid. Calculate Qsp. Compare to Ksp
to determine if precipitate will occur.
!  2) Calculate gram solubility from Ksp. Compare to given
mass of solid (same as what we’ve already worked).
Qsp < Ksp
No precipitate will form.
!  If Qsp > Ksp, supersaturated and salt won’t dissolve
!  If Qsp < Ksp, unsaturated and salt will dissolve
!  Examples 15.7 – 15.9
Dissolution of Ionic Compounds
!  245 mg of magnesium carbonate is placed in 1.00 L of water.
Will it all dissolve? Ksp = 4.0 x 10-5
Dissolution of Ionic Compounds
!  245 mg of magnesium carbonate is placed in 1.00 L of water.
Will it all dissolve? Ksp = 4.0 x 10-5
!  Method 1: Find molarity of solid. Calculate Qsp. Compare to
Ksp to determine if precipitate will occur.
!  Method 2: Calculate gram solubility from Ksp. Compare to
given mass of solid (same as what we’ve already worked).
!  245 mg = 0.0245 g/L x (1 mol/84.32 g) = 2.906 x 10-3 M
!  Ksp = [Mg2+][CO32-] = x2 = 4.0 x 10-5
!  Ksp =
!  x = √(4.0 x 10-5) = 6.325x10-3 M
[Mg2+][CO32-]
=
x2
= 4.0 x
10-5
!  Qsp = (2.906 x 10-3)2 = 8.442 x 10-6
!  x = 6.325x10-3 M x (84.32 g/1 mol) = 0.53 g/L
!  Qsp < Ksp, unsaturated and salt will dissolve
!  0.245 g/L < 0.53 g/L unsat., salt will dissolve
Selective Precipitation
Some compounds can be separated based on selective precipitation.
This is the separation of precipitates based on different Ksp values.
For example,
AgCl (Ksp = 1.6x10-10),
AgBr (Ksp = 7.7x10-13), and
AgI (Ksp = 8.3x10-17) all precipitate.
Because silver iodide’s Ksp value is so low (it is less soluble than the
other salts), it would precipitate before the other two. Almost all of
the AgI would precipitate before AgCl and AgBr would.
Qualitative analysis - Lab
involves the principle of
selective precipitation and
can be used to identify the
types of ions present in a
solution. This is the
qualitative analysis
lab you’ll be doing.
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12/20/16
Solubility and the Common Ion Effect
!  What will happen if we add solid KI to a solution of
saturated AgI?
!  AgI(s) ! Ag+(aq) + I-(aq)
!  Common ion effect: add more I- to precipitate more Ag+
from solution.
How does adding a common ion affect the solubility of the
solid?
A common ion will decrease the solubility of a solid
(increase the amount of solid/precipitate that forms).
Factors Affecting Solubility – Common
Ion Effect
!  Calculate the molar solubility of calcium fluoride (in
pure water) (Ksp = 3.5 x 10-11).
x = 2.1 x 10-4 M
!  Calculate the molar solubility of calcium fluoride in
0.010 M sodium fluoride.
x = 3.5 x 10-7 M
Example 15.12
Factors Affecting Solubility - pH
Mg(OH)2(s)
⇌
Mg2+ (aq)
+
2OH–(aq)
Ksp = [Mg2+][OH–]2 = 1.2 x 10–11
x = 1.4 x 10–4 M
15.2 Lewis Acids and Bases
A Lewis base is a substance that can donate a pair of electrons.
A Lewis acid is a substance that can accept a pair of electrons.
The bond to form a coordinate covalent bond.
At equilibrium:
[OH–] = 2(1.4 x 10–4 M ) = 2.8 x 10–4 M
pH = 14.00 – 3.55 = 10.45
In a solution with a pH of less than 10.45, the solubility of
Mg(OH)2 increases. Why?
Acid added will react with OH- and remove it from solution.
This will shift the equilibrium to the right and increase
solubility.
Example 15.10
Complex Ions
Complex ion formation:
A complex ion is an ion containing a central metal cation bonded
to one or more molecules or ions.
AgCl (s) + 2 NH3 (aq) ! Ag(NH3)2+ (aq) + Cl- (aq)
Ammonia is called a ligand in this compound. Ligands are small
neutral molecules or ions that often act as Lewis bases.
Adding acid to the product will break the coordinate covalent bond
between silver and ammonia to reprecipitate AgCl.
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Formation Constant (Kf) values
Table 15.2: Common Complex Ions and Their Formation
Constants
Substance
[Cd(CN)4]2Ag(NH3)2+
[AlF6]3-
Kf at 25oC
1.3x107
1.7x107
7.0x1019
The larger the formation constant, the more stable the complex ion
is. More values can be found in Appendix K.
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