Chapter 17 Additional Aspects of Aqueous Equilibria

1
Chapter 17Additional Aspects of Aqueous
Equilibria
CHEMISTRY The Central Science 9th Edition
2
The Common Ion Effect

• The solubility of a weak electrolyte is decreased
  by the addition of a strong electrolyte that has
  a ion in common with the weak electrolyte.
• Consider the equilibrium established when acetic
  acid, HC2H3O2, is added to water.
• At equilibrium H and C2H3O2- are constantly
  moving into and out of solution, but the
  concentrations of ions is constant and equal.
• Consider the addition of C2H3O2-, which is a
  common ion. (The source of acetate could be a
  strong electrolyte such as NaC2H3O2.)
• Therefore, C2H3O2- increases and the system is
  no longer at equilibrium.
• So, H must decrease.

3
Class Practice Problem

• What is the pH of a solution made by adding 0.30
  mol of acetic acid (HC2H3O2) and 0.30 mol of
  sodium acetate (NaC2H3O2) to enough water to make
  1.0 L of solution?
• Calculate the pH of a solution containing 0.085 M
  nitrous acid, (HNO2 Ka 4.5 x 10-4), and 0.10 M
  potassium nitrite (KNO2)
• Calculate the fluoride ion concentration and pH
  of a solution that is 0.20 M in HF and 0.10 M in
  HCl.

4
Equilibria of Acid-Base Buffer Systems

• Why do some lakes become acidic when showered
  with acid rain, while others show no change in
  their pH?
• How does blood maintain a certain pH while in
  constant contact with countless cellular
  acid-base reactions that occur in the body?
• And, how can chemists maintain a certain pH level
  (i.e., constant H concentration) in a reaction
  that produces or consumes H or OH-?

5
How Due Buffer Solutions Work

• Buffer work thorough the phenomenon known as the
  Common-ion Effect.
• A buffer must contain an acidic component to
  react with the OH- ion and a basic component to
  react with the H ion.
• When OH- is added to the buffer, the OH- reacts
  with HA to produce A- and water. But, the
  HA/A- ratio remains more or less constant, so
  the pH is not significantly changed.
• When H is added to the buffer, A- is consumed to
  produce HA. Once again, the HA/A- ratio is
  more or less constant, so the pH does not change
  significantly.

6
Buffer Diagram, Figure 17.2
7
Buffered Solutions

• Buffers are solutions that resist changes in
  their pH caused by external force.
• A buffer consists of a mixture of a weak acid
  (HA) and its conjugate base (A-)
• The Ka expression is

HA (aq)
A- (aq)
8
Class Practice Problem

• Calculate the pH of a buffer solution (a)
  consisting of 0.50 M acetic acid (HC2H3O2) and
  0.50 mol of sodium acetate (NaC2H3O2) to enough
  water to make 1.0 L of solution.
• (b) After adding 0.020 M solid NaOH to the buffer
  solution in (a).
• (c) After adding 0.020 M HCl to the buffer
  solution in (a).

9
Buffered Capacity and pH

• Buffer capacity is the amount of acid or base
  neutralized by the buffer before there is a
  significant change in pH.
• Buffer capacity depends on the composition of the
  buffer.
• The greater the amounts of conjugate acid-base
  pair, the greater the buffer capacity.
• The pH of the buffer depends on Ka.

10
Adding Strong Acids or Bases to Buffers

• The amount of strong acid or base added results
  in a neutralization reaction
• A- H3O ? HA H2O
• HA OH- ? A- H2O.
• By knowing the amount of H3O or OH- added,
  (stoichiometry) we know how much HA or A- is
  formed.
• With the concentrations of HA and A- (note the
  change in volume of solution) we can calculate
  the pH from the Henderson-Hasselbalch equation.

11
Class Practice Problem

• Calculate the pH of a buffer solution (a)
  consisting of 0.50 M acetic acid (HC2H3O2) and
  0.50 mol of sodium acetate (NaC2H3O2) to enough
  water to make 1.0 L of solution.
• (b) After adding 0.020 M solid NaOH to the buffer
  solution in (a).
• (c) After adding 0.020 M HCl to the buffer
  solution in (a).

12
An Alternative Approach to Calculating pH
13
Strong Acid-Base Titrations

• A plot of pH versus volume of acid (or base)
  added is called a titration curve.
• Consider adding a strong base (NaOH) to a
  solution of a strong acid (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.
• To detect the equivalent point, we use an
  indicator that changes color somewhere near 7.00.

14
Strong Acid-Base Titration Curve
15
Strong Acid-Base Titrations Cont.

• 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.
• 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).

16
Class Practice Problem

• Calculate the pH when the following quantities of
  0.100 M NaOH solution have been added to 50.0 mL
  of 0.100 M HCL solution (a) 49.0 mL (b) 51.0
  mL
• Calculate the pH of the solution formed when 45.0
  mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M
  (weak acid) HC2H3O2. (Ka 1.8 x 10-5)

17
Weak Acid-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)

18
Before the Equivalence Point

• Weak Acid-Strong Base Titrations
• There is an excess of 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.)

19
Equivalence Point

• 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.
• The equivalence point is determined by Ka of the
  acid.

20
Acid Strength influence on Titration Curves
21
Strong-Weak Acid-Base Titration Comparison

• 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.

22
(No Transcript)
23
Strong-Weak Acid-Base Titration Comparison Cont.

• 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.

24
Class Practice Problem

• Calculate the pH when the following quantities of
  0.100 M NaOH solution have been added to 50.0 mL
  of 0.100 M HCL solution (a) 49.0 mL (b) 51.0
  mL
• Calculate the pH of the solution formed when 45.0
  mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M
  (weak acid) HC2H3O2. (Ka 1.8 x 10-5)

25
Polyprotic Acid-Base Titrations

• 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 H3PO3 with NaOH.
• The first proton dissociates to form H2PO3-.
• Then the second proton dissociates to form HPO32-.

26
(No Transcript)
27
Solubility Equilibria

• The Solubility-Product Constant, Ksp
• Consider the saturated solution
• for which
• Ksp is the solubility product. (BaSO4 is ignored
  because it is a pure solid so its concentration
  is constant.)
• The Ksp value for many ionic solids are tabulated
  in Appendix D.

28
Class practice problem

• Write the expression for the solubility-product
  constant for Ca3(PO4)2.
• Solid silver chloride is added to pure water at
  25 oC. Some of the solid remains undissolved at
  the bottom of the flask. The mixture is stirred
  for several days to ensure that equilibrium is
  achieved between the undissolved AgCl(s) and the
  solution. Analysis of the equilibrated solution
  shows that its silver-ion concentration is 1.34 x
  105 M. What is the Ksp for AgCl?
• The Ksp for CaF2 is 3.9 x 10-11 at 25 oC. What
  is the solubility of CaF2 in water in grams per
  liter?

29
Difference between Solubility and Solubility
Product, 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.
• The Ksp is the equilibrium constant for the
  equilibrium between and ionic solid and its
  saturated solution.

30
Factors that Affect Solubility Common-Ion Effects

• Solubility is decreased when a common ion is
  added.
• This is an application of Le Châteliers
  principle
• as F- (from NaF, say) is added, the equilibrium
  shifts away from the increase.
• Therefore, CaF2(s) is formed and precipitation
  occurs.
• As NaF is added to the system, the solubility of
  CaF2 decreases.

31
Factors that Affect Solubility pH Effects

• 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.

32
Factors that Affect Solubility Complex Ion Effects

• A Consider the formation of Ag(NH3)2
• The Ag(NH3)2 is called a complex ion.
• NH3 (the attached Lewis base) is called a ligand.
• The equilibrium constant for the reaction is
  called the formation constant, Kf

33
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.

34
Complex Ions Table 17.1
35
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

36
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.

37
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.
• Removal of one metal ion from a solution is
  called selective precipitation.

38

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

39
End of Chapter 17Additional Aspects of Aqueous
Equilibria