1. Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change
Seventh Edition
Martin S. Silberberg and Patricia G. Amateis
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2. Chapter 19
Ionic Equilibria in Aqueous Systems

3. Ionic Equilibria in Aqueous Systems
19.1 Equilibria of Acid-Base Buffers
19.2 Acid-Base Titration Curves
19.3 Equilibria of Slightly Soluble Ionic Compounds
19.4 Equilibria Involving Complex Ions

4. Acid-Base Buffers
An acid-base buffer is a solution that lessens the impact of pH from the addition of acid or base.
An acid-base buffer usually consists of a conjugate acid-base pair where both species are present in appreciable quantities in solution.
An acid-base buffer is therefore a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid.

5. The effect of adding acid or base to an unbuffered solution.
Figure 19.1
The effect of adding acid or base to an unbuffered solution.
A 100-mL sample of dilute HCl is adjusted to pH 5.00.
The addition of 1 mL of strong acid (left) or strong base (right) changes the pH by several units.

6. The effect of adding acid or base to a buffered solution.
Figure 19.2
The effect of adding acid or base to a buffered solution.
A 100-mL sample of an acetate buffer is adjusted to pH 5.00.
The addition of 1 mL of strong acid (left) or strong base (right) changes the pH very little.
The acetate buffer is made by mixing 1 M CH3COOH ( a weak acid) with 1 M CH3COONa (which provides the conjugate base, CH3COO-).

7. Buffers and the Common-ion Effect
A buffer works through the common-ion effect.
Acetic acid in water dissociates slightly to produce some acetate ion:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
acetic acid
acetate ion
If NaCH3COO is added, it provides a source of CH3COO– ion, and the equilibrium shifts to the left. CH3COO-– is common to both solutions.
The addition of CH3COO– reduces the % dissociation of the acid.

8. Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
[CH3COOH]init
[CH3COO-]added
% Dissociation*
[H3O+] pH
0.10
0.00
1.3
1.3x10-3
2.89
0.050
0.036
3.6x10-5
4.44
0.018
1.8x10-5
4.74
0.15
0.012
1.2x10-5
4.92
* % Dissociation =
[CH3COOH]dissoc
[CH3COOH]init
x 100

9. How a Buffer Works
The buffer components (HA and A-) are able to consume small amounts of added OH- or H3O+ by a shift in equilibrium position.
CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O+(aq)
Added OH– reacts with CH3COOH, causing a shift to the right.
Added H3O+ reacts with CH3COO–, causing a shift to the left.
The shift in equilibrium position absorbs the change in [H3O+] or [OH–], and the pH changes only slightly.

10. Figure 19.3 How a buffer works. H3O+
Buffer has more HA after addition of H3O+.
H2O + CH3COOH ← H3O+ + CH3COO-
Buffer has equal concentrations of A– and HA.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
OH-
Buffer has more A- after addition of OH–.
CH3COOH + OH- → CH3COO- + H2O

11. Relative Concentrations of Buffer Components
CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O+(aq)
Ka =
[CH3COO–][H3O+]
[CH3COOH]
[H3O+] = Ka x
[CH3COOH]
[CH3COO–]
Since Ka is constant, the [H3O+] of the solution depends on the ratio of buffer component concentrations.
If the ratio increases, [H3O+] increases.
[HA]
[A–]
If the ratio decreases, [H3O+] decreases.
[HA]
[A–]

12. Sample Problem 19.1
Calculating the Effect of Added H3O+ or OH- on Buffer pH
PROBLEM:
(a) Of a buffer solution consisting of 0.50 M CH3COOH and 0.50 M CH3COONa
(b) After adding mol of solid NaOH to 1.0 L of the buffer solution in (a).
(c) After adding mol of HCl to 1.0 L of the buffer solution in (a).
Ka of CH3COOH = 1.8 x 10–5. (Assume the additions cause a negligible change in volume.)
Calculate the pH:

15. Figure 19.4
Effect of added acid or base on concentrations of buffer components

16. The Henderson-Hasselbalch Equation
HA(aq) + H2O(l) A-(aq) + H3O+(aq)
Ka =
[H3O+][A–]
[HA]
[H3O+] = Ka x
[HA]
[A–]
-log[H3O+] = -logKa – log
[HA]
[A–]
pH = pKa + log
[base]
[acid]

17. Buffer Capacity
The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base.
The greater the concentrations of the buffer components, the greater its capacity to resist pH changes.
The closer the component concentrations are to each other, the greater the buffer capacity.

18. The relation between buffer capacity and pH change.
Figure 19.5
The relation between buffer capacity and pH change.
When strong base is added, the pH increases least for the most concentrated buffer.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
This graph shows the final pH values for four different buffer solutions after the addition of strong base.

19. Buffer Range
The buffer range is the pH range over which the buffer is effective.
Buffer range is related to the ratio of buffer component concentrations.
[HA]
[A–]
The closer is to 1, the more effective the buffer.
If the concentration of one component is more than 10 times the concentration of the other, buffering action is poor. Since log10 = 1, buffers have a usable range within ± 1 pH unit of the pKa of the acid component.

20. Sample Problem 19.2
Using Molecular Scenes to Examine Buffers
PROBLEM:
The molecular scenes below represent samples of four HA/A- buffers. (HA is blue and green, A- is green, and other ions and water are not shown.)
(a) Which buffer has the highest pH?
(b) Which buffer has the greatest capacity?
(c) Should we add a small amount of concentrated strong acid or strong base to convert sample 1 to sample 2 (assuming no volume changes)?

21. Preparing a Buffer Choose the conjugate acid-base pair.
The pKa of the weak acid component should be close to the desired pH.
Calculate the ratio of buffer component concentrations.
Determine the buffer concentration, and calculate the required volume of stock solutions and/or masses of components.
Mix the solution and correct the pH.
pH = pKa + log
[base]
[acid]

22. Sample Problem 19.3
Preparing a Buffer
PROBLEM:
An environmental chemist needs a carbonate buffer of pH to study the effects of acid rain on limestone-rich soils. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to make the buffer? Ka of HCO3– is 4.7x10–11.

23. Acid-Base Indicators
An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In–).
The ratio [HIn]/[In–] is governed by the [H3O+] of the solution. Indicators can therefore be used to monitor the pH change during an acid-base reaction.
The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

24. Colors and approximate pH range of some common acid-base indicators.
Figure 19.6
Colors and approximate pH range of some common acid-base indicators. pH
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

25. The color change of the indicator bromthymol blue.
Figure 19.7
The color change of the indicator bromthymol blue.
pH < 6.0
pH =
pH > 7.5

26. Acid-Base Titrations
In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration.
The equivalence point of the reaction occurs when the number of moles of OH– added equals the number of moles of H3O+ originally present, or vice versa.
The end point occurs when the indicator changes color.
- The indicator should be selected so that its color change occurs at a pH close to that of the equivalence point.

27. Curve for a strong acid–strong base titration.
Figure 19.8
Curve for a strong acid–strong base titration.
The pH increases gradually when excess base has been added.
The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00.
The initial pH is low.

28. Calculating the pH during a strong acid–strong base titration
Initial pH
[H3O+] = [HA]init
pH = –log[H3O+]
pH before equivalence point
initial mol H3O+ = Vacid x Macid
mol OH– added = Vbase x Mbase
mol H3O+remaining = (mol H3O+init) – (mol OH–added)
[H3O+] = pH = –log[H3O+]
mol H3O+remaining
Vacid + Vbase

29. Calculating the pH during a strong acid–strong base titration
pH at the equivalence point
pH = 7.00 for a strong acid-strong base titration.
pH beyond the equivalence point
initial mol H3O+ = Vacid x Macid
mol OH– added = Vbase x Mbase
mol OH–excess = (mol OH–added) – (mol H3O+init)
[OH–] =
pOH = –log[OH–] and pH = pOH
mol OH–excess
Vacid + Vbase

30. Example:
40.00 mL of M HCl is titrated with M NaOH.
The initial pH is simply the pH of the HCl solution:
[H3O+] = [HCl]init = M and pH = –log(0.1000) = 1.00
To calculate the pH after mL of NaOH solution has been added:
Initial mol of H3O+ = L HCl x
mol
1 L
= 4.000x10–3 mol H3O+
OH– added = L NaOH x
mol
1 L
= 2.000x10–3 mol OH–
The OH– ions react with an equal amount of H3O+ ions, so
H3O+ remaining = 4.000x10–3 – 2.000x10–3 = 2.000x10–3 mol H3O+