Applications of Aqueous Equilibria

1
Applications of Aqueous Equilibria
2
Common Ions

• When we dissolve acetic acid in water, the
  following equilibrium is established
• CH3COOH ?? CH3COO- H
• If sodium acetate were dissolved in solution,
  which way would this equilibrium shift?

3
Common Ions

• When we dissolve acetic acid in water, the
  following equilibrium is established
• CH3COOH ?? CH3COO- H
• If sodium acetate were dissolved in solution,
  which way would this equilibrium shift?
• Adding acetate ions from a strong electrolyte
  would shift this equilibrium left (Le Chateliers
  principle).

4
Common Ion Effect

• Whenever a weak electrolyte (acetic acid) and a
  strong electrolyte (sodium acetate) share a
  common ion, the weak electrolyte ionizes less
  than it would if it were alone. (Le Chateliers)
• This is called the common-ion effect.

5
Steps for Common-Ion Problems

• 1. Consider which solutes are strong electrolyte
  and weak electrolytes.
• 2. Identify the important equilibrium (weak) that
  is the source of H and therefore determines pH.
• 3. Create an ICE chart using the equilibrium and
  strong electrolyte concentrations.
• 4. Use the equilibrium constant expression to
  calculuate H and pH.

6

• What is the concentration of silver and chromate
  in a solution with silver chromate in 0.1 M
  silver nitrate
• Ag2CrO4(s) ? 2Ag1 CrO4-2
• I 0.1 0
• C 2x x
• E 0.1 2x x
• Ksp 9.0 x 10-12 Ag12 CrO4-2
• 9.0 x 10-12 .1 2x2 x
• 9.0 x 10-12 0.12 x
• CrO4-2 x 9.0 x 10-10 M
• Ag1 .1 2x .1M
• Ag2CrO4 9.0 x 10-10

What are the sources of Ag1?
7
Sample Problem

• What is the pH of a solution made by adding 0.30
  mol of acetic acid and 0.30 mol of sodium acetate
  to enough water to make 1.0 L of solution?
• Ka for acetic acid 1.8 x 10-5
• pH 4.74

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Sample Problem

• Calculate the fluoride concentration and pH of a
  solution that is 0.20 M in HF and 0.10 M in HCl
• Ka for HF 6.8 x 10-4.
• F- 1.2x10-3M
• pH 1.00

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Buffers

• Solutions that resist pH change when small
  amounts of acid or base are added.
• Two types
• weak acid and its salt
• weak base and its salt
• HX(aq) H2O(l) H(aq) X-(aq)
• Add OH- Add H
• shift to right shift to
  left
• Based on the common ion effect.

10
Buffers
buffered unbuffered
pH
ml HCl added
11
Buffers and blood

• Control of blood pH
• Oxygen is transported primarily by hemoglobin in
  the red blood cells.
• CO2 is transported both in plasma and the red
  blood cells.

CO2 (aq) H2O
H2CO3 (aq)
H(aq) HCO3-(aq)
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Buffers

• Composition and Action of Buffered Solutions
• The Ka expression is
• A buffer resists a change in pH when a small
  amount of OH- or H is added.

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Buffered Solutions

• Addition of Strong Acids or Bases to Buffers
• With the concentrations of HX and X- (note the
  change in volume of solution) we can calculate
  the pH from the Henderson-Hasselbalch equation

Equation not necessary, since you know Ka
X- H HX
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Buffers

• Action of Buffered Solutions

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Buffers
Buffer Capacity and pH Buffer capacity is the
amount of acid or base neutralized by the buffer
before there is a significant change in pH. The
greater the amounts(molarity) of the conjugate
acid-base pair, the greater the buffer
capacity. The pH of the buffer depends on Ka
16
Buffered Solutions
Addition of Strong Acids or Bases to Buffers
17
Buffer
Addition of Strong Acids or Bases to
Buffers Break the calculation into two parts
stoichiometric and equilibrium. The amount of
strong acid or base added results in a
neutralization reaction X- H ? HX H2O HX
OH- ? X- H2O. By knowing how much H or OH-
was added (stoichiometry) we know how much HX or
X- is formed.
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Buffers

• The final HX and X- after the neutralization
  reaction are used as the initial concentrations
  for the equilibrium reaction.
• HX H X-
• Initial conc., M
• Change, DM
• Eq. Conc., M
• Then the equilibrium constant expression is used
  to find H and pH - log H
• Ka H X-
• HX

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Buffer Example

• Determine the initial pH of a buffer solution
  that has 0.10 M benzoic acid and 0.20 M sodium
  benzoate at 25 oC. Ka 6.5 x 10-5
• HBz(aq) H2O(l) H(aq) Bz-(aq)

HBz H Bz- Initial conc., M 0.10
0.00 0.20 Change, DM -x x
x Eq. Conc., M 0.10 - x x 0.
20 x
20
Buffer Example

• Solve the equilibrium equation in terms of x
• Ka 6.5 x 10-5
• x (6.5 x 10-5 )(0.10) / (0. 20)
• (assuming xltlt0.10)
• 3.2 x 10-5 M H
• pH - log (3.2 x 10-5 M) 4.5
• initial pH

21
Buffer Example

• Determine the pH of a buffer solution that has
  0.10 M benzoic acid and 0.20 M sodium benzoate at
  25 oC after 0.05 moles of HCl is added.

First, find the concentrations of the HBz and Bz-
after HCl is added.
HBz H Bz- Initial conc., M 0.15
0.00 0.15 Change, DM -x x
x Eq. Conc., M 0.15 - x x 0. 15
x
The 0.05 mol HCl reacts completely with 0.05 mol
Bz-(aq) to form 0.05 mol HBz(aq) Then equilibrium
will be re-established based on the new initial
concentrations of 0.15 M HBz(aq) and 0.15 M
Bz-(aq).
22
Buffer Example

• Solve the equilibrium equation in terms of x
• Ka 6.5 x 10-5
• x (6.5 x 10-5 )(0.15) / (0. 15)
• (assuming xltlt0.15)
• 6.5 x 10-5 M H
• pH - log (6.5 x 10-5 M) 4.2
• after 0.05 mole HCl added

23
Buffer Example

• Determine the pH of a buffer solution that has
  0.10 M benzoic acid and 0.20 M sodium benzoate at
  25 oC after 0.05 moles of NaOH is added.

First, find the concentrations of HBz and Bz-
after NaOH is added.
HBz H Bz- Initial conc., M 0.05
0.00 0.25 Change, DM -x x
x Eq. Conc., M 0.05 - x x 0. 25
x
The 0.05 mol NaOH reacts completely with 0.05 mol
HBz(aq) to form 0.05 mol Bz-(aq) Then
equilibrium will be re-established based on the
new initial concentrations of 0.05 M HBz(aq) and
0.25 M Bz-(aq).
24
Buffer Example

• Solve the equilibrium equation in terms of x
• Ka 6.5 x 10-5
• x (6.5 x 10-5 )(0.05) / (0. 25)
• (assuming xltlt0.05)
• 1.3 x 10-5 M H
• pH - log (1.3 x 10-5 M) 4.9
• after 0.05 mole NaOH added

25
Acid-Base Titrations
Strong Acid-Base Titrations

• The plot of pH versus volume during a titration
  is a titration curve.

26
Remember the Acid-Base Titration Curves?
Buffer Zone
HC2H3O2
HCl
27
Acetic Acid/Acetate Ion Buffer Lab
For this experiment, you will prepare a buffer
that contains acetic acid and its conjugate base,
the acetate ions. The equilibrium equation for
the reaction is shown below HC2H3O2(aq)
H2O(l) ltgt H (aq) C2H3O2- (aq) The equilibrium
expression for this reaction, Ka, has a value of
1.8 x 10-5 at 25ºC.
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Acetic Acid/Acetate Ion Buffer Lab

• The ratio between the molarity of the acetate
  ions to the molarity of the acetic acid in your
  buffer must equal the ratio between the Ka value
  and 10- assigned pH.
• This ratio should be reduced , so that either the
  HC2H3O2 or C2H3O2- has a concentration of
  0.10 M, and the concentration of the other
  component must fall within a range from 0.10 M to
  1.00 M.
• Complete the calculations only that are needed to
  prepare 100.0 mL of your assigned buffer solution
  that has these specific concentrations. Can you
  predict the final pH when a strong acid or base
  is added to the buffer solution?

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Acetic Acid/Acetate Ion Buffer Lab
30
Acetic Acid/Acetate Ion Buffer Lab
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Making a Buffer Calculations
You want to prepare 500.0 mL of a buffer with a
pH 10.00, with both the acid and conjugate base
having molarities between 0.10 M to 1.00 M. You
may choose from any of the acids listed below
You must select an acid with a Ka value close to
10- assigned pH. The only two options are
ammonium or the hydrogen carbonate ions.
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Making a Buffer Calculations
33
Making a Buffer Calculations
Place 250 mL of distilled water in a 500 mL
volumetric flask. Add 2.68 g NH4Cl and dissolve.
Then mix in 18.9 mL of 14.8 M NH3. Fill with
distilled water to the 500 mL mark on the flask.
If there is no concentrated NH3 available, the
NH3 can be produced by neutralizing additional
NH4Cl with 1.00 M NaOH.
Dissolve 19.06 g NH4Cl (2.68 g 16.38 g) in 280.
mL of 1.00 M NaOH. Then dilute with distilled
water and fill to the 500 mL mark on the flask.
34
Making a Buffer Calculations
Place 250 mL of distilled water in a 500 mL
volumetric flask. Add 2.68 g NH4Cl and dissolve.
Then mix in 18.9 mL of 14.8 M NH3. Fill with
distilled water to the 500 mL mark on the flask.
If there is no NH4Cl available, the NH4 can be
produced by neutralizing additional NH3 with 1.00
M HCl.
Place 250 mL distilled water in volumetric
flask. Add 50.0 mL of 1.00 M HCl and mix. Then
add 22.3 mL of concentrated NH3 (18.9 mL 3.4
mL). Mix and fill with distilled water to the
500 mL mark on the flask.
35
Making a Buffer Calculations
Prepare 500. mL of the buffer that has CO32-
0.100 M and HCO31- 0.179 M.
36
Making a Buffer Calculations
Prepare 500. mL of the buffer that has CO32-
0.100 M and HCO31- 0.179 M.
Place 250 mL distilled water in a 500 mL
volumetric flask. Add 5.30 g Na2CO3 and 7.52 g
NaHCO3 and dissolve. Fill with distilled water
to the 500 mL mark on the flask.
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Acid-Base Titrations
Strong Acid-Base Titrations

• The plot of pH versus volume during a titration
  is a titration curve.

38
Indicator examples

• Acid-base indicators are weak acids that undergo
  a color change at a known pH.

pH
phenolphthalein
39
Indicator examples
Select the indicator that undergoes a color
change closest to the pH at the equivalence
point, where all of the acid has been neutralized
by the base.
bromthymol blue
methyl red
40
Acid-Base Titrations

• Strong Acid-Base Titrations
• Consider adding a strong base (e.g. NaOH) to a
  solution of a strong acid (e.g. HCl).
• Before any base is added, the pH is given by the
  strong acid solution. Therefore, pH lt 7.
• When base is added, before the equivalence point,
  the pH is given by the amount of strong acid in
  excess. Therefore, pH lt 7.
• At equivalence point, the amount of base added is
  stoichiometrically equivalent to the amount of
  acid originally present. Therefore, the pH is
  determined by the salt solution. Therefore, pH
  7.

41
Acid-Base Titrations

• Strong Acid-Base Titrations
• Consider adding a strong base (e.g. NaOH) to a
  solution of a strong acid (e.g. HCl).

42
Acid-Base Titrations

• Strong Acid-Base Titrations
• We know the pH at equivalent point is 7.00.
• To detect the equivalent point, we use an
  indicator that changes color somewhere near 7.00.
• Usually, we use phenolphthalein that changes
  color between pH 8.3 to 10.0.
• In acid, phenolphthalein is colorless.
• As NaOH is added, there is a slight pink color at
  the addition point.
• When the flask is swirled and the reagents mixed,
  the pink color disappears.
• At the end point, the solution is light pink.
• If more base is added, the solution turns darker
  pink.

43
Acid-Base Titrations

• Strong Acid-Base Titrations
• The equivalence point in a titration is the point
  at which the acid and base are present in
  stoichiometric quantities.
• The end point in a titration is the observed
  point.
• The difference between equivalence point and end
  point is called the titration error.
• The shape of a strong base-strong acid titration
  curve is very similar to a strong acid-strong
  base titration curve.

44
Acid-Base Titrations
Strong Acid-Base Titrations
45
Acid-Base Titrations

• Strong Acid-Base Titrations
• Initially, the strong base is in excess, so the
  pH gt 7.
• As acid is added, the pH decreases but is still
  greater than 7.
• At equivalence point, the pH is given by the salt
  solution (i.e. pH 7).
• After equivalence point, the pH is given by the
  strong acid in excess, so pH lt 7.

46
Acid-Base Titrations

• Weak Acid-Strong Base Titrations
• Consider the titration of acetic acid, HC2H3O2
  and NaOH.
• Before any base is added, the solution contains
  only weak acid. Therefore, pH is given by the
  equilibrium calculation.
• As strong base is added, the strong base consumes
  a stoichiometric quantity of weak acid
• HC2H3O2(aq) NaOH(aq) ? C2H3O2-(aq) H2O(l)

47
Acid-Base Titrations
Weak Acid-Strong Base Titrations
48
Acid-Base Titrations

• Weak Acid-Strong Base Titrations
• There is an excess of acetic acid before the
  equivalence point.
• Therefore, we have a mixture of weak acid and its
  conjugate base.
• The pH is given by the buffer calculation.
• First the amount of C2H3O2- generated is
  calculated, as well as the amount of HC2H3O2
  consumed. (Stoichiometry.)
• Then the pH is calculated using equilibrium
  conditions. (Henderson-Hasselbalch.)

49
Acid-Base Titrations

• Weak Acid-Strong Base Titrations
• At the equivalence point, all the acetic acid has
  been consumed and all the NaOH has been consumed.
  However, C2H3O2- has been generated.
• Therefore, the pH is given by the C2H3O2-
  solution.
• This means pH gt 7.
• More importantly, pH ? 7 for a weak acid-strong
  base titration.
• After the equivalence point, the pH is given by
  the strong base in excess.

50
Acid-Base Titrations

• Weak Acid-Strong Base Titrations
• For a strong acid-strong base titration, the pH
  begins at less than 7 and gradually increases as
  base is added.
• Near the equivalence point, the pH increases
  dramatically.
• For a weak acid-strong base titration, the
  initial pH rise is more steep than the strong
  acid-strong base case.
• However, then there is a leveling off due to
  buffer effects.

51
Acid-Base Titrations

• Weak Acid-Strong Base Titrations
• The inflection point is not as steep for a weak
  acid-strong base titration.
• The shape of the two curves after equivalence
  point is the same because pH is determined by the
  strong base in excess.
• Two features of titration curves are affected by
  the strength of the acid
• the amount of the initial rise in pH, and
• the length of the inflection point at equivalence.

52
Acid-Base Titrations
Weak Acid-Strong Base Titrations

• The weaker the acid, the smaller the equivalence
  point inflection.
• For very weak acids, it is impossible to detect
  the equivalence point.

53
Acid-Base Titrations

• Weak Acid-Strong Base Titrations
• Titration of weak bases with strong acids have
  similar features to weak acid-strong base
  titrations.

54
Acid-Base Titrations

• Titrations of Polyprotic Acids
• In polyprotic acids, each ionizable proton
  dissociates in steps.
• Therefore, in a titration there are n equivalence
  points corresponding to each ionizable proton.
• In the titration of Na2CO3 with HCl there are two
  equivalence points
• one for the formation of HCO3-
• one for the formation of H2CO3.

55
Acid-Base Titrations
Titrations of Polyprotic Acids
56
Acid-Base Titrations
Titrations of Polyprotic Acids Ka H X-
HX At the equivalence point, H OH- At ½
the equivalence point, X- HX SOOOO. At ½
equivalence point, Ka H So pKa pH
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Solubility

• What happens when an ionic compound is dissolved
  in water?
• What forces does water have to overcome?
• Are all ionic compounds soluble?
• Remember the solubility rules?
• Its time to review them!

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Properties of aqueous solutions

• There are two general classes of solutes.
• Electrolytic
• ionic compounds in polar solvents
• dissociate in solution to make ions
• conduct electricity
• may be strong (100 dissociation) or weak (less
  than 10)
• Nonelectrolytic
• do not conduct electricity
• solute is dispersed but does not dissociate

61
Heterogeneous Equilibria

• A(s) H2O ? B(aq) C (aq)
• Ksp B C
• Why is this not divided by A?
• Why is H2O not included?
• Known as the Solubility Product Ksp

62
Solubility Products, KSP

• KSP expressions are used for ionic materials that
  are only slightly soluble in water.
• Their only means of dissolving is by
  dissociation.
• AgCl(s) Ag (aq) Cl- (aq)
• KSP

Ag Cl-
63
Solubility Products, KSP

• At equilibrium, the system is a saturated
  solution of silver and chloride ions.
• The only way to know that it is saturated it to
  observe some AgCl at the bottom of the solution.
• As such, AgCl is a constant and KSP
  expressions do not include the solid form in the
  equilibrium expression. The H2O for solvation
  process is also excluded from the KSP expression.

64
Solubility Products, KSP

• Determine the solubility of AgCl in water at 20
  oC in terms of grams / 100 mL
• KSP Ag Cl- 1.7 x 10-10
• At equilibrium, Ag Cl- so
• 1.7 x 10-10 x 2
• Ag 1.3 x 10-5 M
• g AgCl 1.3 x 10-5 mol/L 0.10 L 143.32
  g/mol
• Solubility 1.9 x 10-4 g / 100 mL

65
Calculating Ksp

• Calculate the Ksp for Bismuth Sulfide which has a
  solubility of 1.0 x 10-15 .
• Bi2S3 ? 2Bi3 3S-2
• Ksp
• For every Bi2S3 dissolved
• 2 Bi3 and 3S-2 are formed
• Ksp 2 x 1.0 x 10-15 2 3 x 1.0 x 10-15
  3
• 1.1 x 10-73

66
Calculating Solubility
What is the molarity of a saturated solution of
Cu(IO3)2?

• Calculate the concentration of ions
• for copper (II) iodate.
• ksp 1.4 x 10-7 at 25 C
• Cu(IO3)2(s) ? Cu2 2 IO3-1
• 1.4 x 10-7 x 2x2 4x3
• x 3.3 x 10-3 mol/L Cu2
• 2x 6.6 x 10-3 mol/L IO3-1

3.3 x 10-3 mol/L Cu(IO3)2
67
Solubility Products, KSP

• Another example
• Calculate the silver ion concentration when
  excess silver chromate is added to a 0.010 M
  sodium chromate solution.
• KSP Ag2CrO4 1.1 x 10-12
• Ag2CrO4 (s) 2Ag CrO42-

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Solubility Products, KSP

• KSP 1.1 x 10-12 Ag 2 CrO42-
• CrO42- CrO42-Ag2CrO4 CrO42-Na2CrO4
• With such a small value for KSP, we can assume
  that the CrO42- from Ag2CrO4 is negligible.
• If were wrong, our silver concentration will be
  significant (gt1 of the chromate concentration.)
• Then youd use the quadratic approach.

69
Solubility Products, KSP

• KSP 1.1 x 10-12 Ag 2 CrO42-
• CrO42- 0.010 M
• Ag ( KSP / CrO42- ) 1/2
• (1.1 x 10-12 / 0.010 M ) 1/2
• 1.1 x 10 -5 M
• Ag ltlt CrO42-
• so our assumption was valid.

70
Relative Solubility

• Be careful when comparing solubility products
  (Ksp)
• Compare
• AgI Ksp 1.5 x 10-16
• CuI Ksp 5.0 x 10-12
• Which is more soluble?
• Copper (I) iodide is more soluble. Each compound
  produces the same number ions

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Relative Solubility

• CuS Ksp 8.5 x 10-45
• Ag2S Ksp 1.6 x 10-49
• Bi2S3 Ksp 1.1 x 10-73
• Which is more soluble?
• Each has a different number of ions,
• so calculate the solubility of each.
• Estimate (guess)
• three groups and calculate the solubility

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Factors influencing solubility

• Common ion and salt effects.
• As with other equilibria weve discussed, adding
  a common ion will result in a shift of a
  solubility equilibrium.
• AgCl (s) Ag (aq) Cl- (aq)
• KSP Ag Cl-
• Adding either Ag or Cl- to our equilibrium
  system will result in driving it to the left.

73
Factors That Affect Solubility
Common-Ion Effect
74
Factors influencing solubility

• Complex ion formation.
• The solubility of slightly soluble salts can be
  increased by complex ion formation.
• Example. Addition of excess Cl-
• AgCl(s) Ag(aq) Cl-(aq)
• 2Cl-(aq)
• AgCl2-(aq)

A large excess of chloride results in the
formation of the complex. More AgCl will
dissolve as a result.
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Factors That Affect Solubility
Formation of Complex Ions
76
Factors That Affect Solubility

• Formation of Complex Ions
• Consider the addition of ammonia to AgCl (white
  precipitate)

• The overall reaction is

• Effectively, the Ag(aq) has been removed from
  solution.
• By Le Châteliers principle, the forward reaction
  (the dissolving of AgCl) is favored.
• Knet KspKf (1.8 x 10-10)(1.7 x 107) 3.1 x
  10-3 at 25ºC

77
Factors That Affect Solubility

• Amphoterism
• Amphoteric oxides will dissolve in either a
  strong acid or a strong base.
• Examples hydroxides and oxides of Al3, Cr3,
  Zn2, and Sn2.
• The hydroxides generally form complex ions with
  four hydroxide ligands attached to the metal
• Hydrated metal ions act as weak acids. Thus, the
  amphoterism is interrupted

78
Factors That Affect Solubility

• Amphoterism
• Hydrated metal ions act as weak acids. Thus, the
  amphoterism is interrupted

79
Factors influencing solubility

• Hydrolysis.
• If the anion of a weak acid, or cation of a weak
  base is part of a KSP, solubilities are greater
  than expected.
• AgCN(s) Ag(aq) CN-(aq)
• H2O(l)
• HCN(aq) OH-(aq)

This competing equilibrium causes the CN- to be
lower than expected. More AgCN will dissolve as
a result.
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Factors That Affect Solubility

• Solubility and pH
• Again we apply Le Châteliers principle
• If the F- is removed, then the equilibrium shifts
  towards the decrease and CaF2 dissolves.
• F- can be removed by adding a strong acid
• As pH decreases, H increases and solubility
  increases.
• The effect of pH on solubility is dramatic.

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Factors That Affect Solubility

• Solubility and pH

82
Solubility Equilibria
Solubility-Product Constant, Ksp Consider for
which Ksp is the solubility product. (BaSO4
is ignored because it is a pure solid so its
concentration is constant.)
83
Solubility Equilibria

• Solubility-Product Constant, Ksp
• In general the solubility product is the molar
  concentration of ions raised to their
  stoichiometric powers.
• Solubility is the amount (grams) of substance
  that dissolves to form a saturated solution.
• Molar solubility is the number of moles of solute
  dissolving to form a liter of saturated solution.

84
Solubility Equilibria

• Solubility and Ksp
• To convert solubility to Ksp
• solubility needs to be converted into molar
  solubility (via molar mass)
• molar solubility is converted into the molar
  concentration of ions at equilibrium (equilibrium
  calculation),
• Ksp is the product of equilibrium concentration
  of ions.

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Solubility Equilibria
Solubility and Ksp
86
Precipitation Reactions

• We will use Q and Ksp
• If Q is bigger than Ksp, what will happen?
• Smaller
• The Q in a solubilty problem is called the
• Ion Product

87
A solution prepared by adding 750 mL of 4.00 x
10-3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M
KIO3. Will Ce(IO3)3 precipitate? (Ksp 1.9 x
10-10)

• Step one Calculate the concentration of Ce3 and
  IO3-1 before a reaction occurs
• Ce30 (750.0 mL)(4.00x10-3 M) 2.86 x
  10-3 M (750. 300.mL)
• IO3-10 (300.0 mL)(2.00x10-3 M) 5.71 x
  10-3 M (750. 300.mL)
• Q (2.86 x 10-3 M) x (5.71 x 10-3 M)3 5.32
  x 10-10
• Q gt Ksp therefore the precipitation reaction
  will occur

88
Precipitation and Separation of Ions

• At any instant in time, Q Ba2SO42-.
• If Q lt Ksp, precipitation occurs until Q Ksp.
• If Q Ksp, equilibrium exists.
• If Q gt Ksp, solid dissolves until Q Ksp.
• Based on solubilities, ions can be selectively
  removed from solutions.
• Consider a mixture of Zn2(aq) and Cu2(aq). CuS
  (Ksp 6?10-37) is less soluble than
  ZnS(Ksp2?10-25), CuS will be removed from
  solution before ZnS.

89
Precipitation and Separation of Ions

• As H2S is added to the green solution, black CuS
  forms in a colorless solution of Zn2(aq).
• When more H2S is added, a second precipitate of
  white ZnS forms.

• Selective Precipitation of Ions
• Ions can be separated from each other based on
  their salt solubilities.
• Example if HCl is added to a solution containing
  Ag and Cu2, the silver precipitates(Ksp for
  AgCl is 1.8?10-10) while the Cu2 remains in
  solution, since CuCl2.
• Removal of one metal ion from a solution is
  called selective precipitation.

90
Qualitative Analysis for Metallic Elements

• Qualitative analysis is designed to detect the
  presence of metal ions.
• Quantitative analysis is designed to determine
  how much metal ion is present.

91
Qualitative Analysis for Metallic Elements

• We can separate a complicated mixture of ions
  into five groups
• Add 6 M HCl to precipitate insoluble chlorides
  (AgCl, Hg2Cl2, and PbCl2).
• To the remaining mix of cations, add H2S in 0.2 M
  HCl to remove acid insoluble sulfides (e.g. CuS,
  Bi2S3, CdS, PbS, HgS, etc.).
• To the remaining mix, add (NH4)2S at pH 8 to
  remove base insoluble sulfides and hydroxides
  (e.g. Al(OH)3, Fe(OH)3, ZnS, NiS, CoS, etc.).
• To the remaining mixture add (NH4)2HPO4 to remove
  insoluble phosphates (Ba3(PO4)2, Ca3(PO4)2,
  MgNH4PO4).
• The final mixture contains alkali metal ions and
  NH4.