19-1

AQUEOUS EQUILIBRIA II

• Common Ion Effect• Buffer Solutions• Net Ionic Equations• Chapter 16.1-16.9 (MF)• Chapter 4.2; 19.1-19.2 (Silberberg)

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Chapter 19

Ionic Equilibria in Aqueous Systems

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Ionic Equilibria in Aqueous Systems

19.1 Equilibria of Acid-Base Buffers

19.2 Acid-Base Titration Curves

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Goals & Objectives

• See the following Learning Objectives on page 870.

• Understand these Concepts:• 19.1-8

• Master these Skills:• 19.1-6

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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.

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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.

19-7

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-).

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Buffers and the Common-ion Effect

A buffer works through the common-ion effect.

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

acetic acid acetate ion

Acetic acid in water dissociates slightly to produce some 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.

19-9

Buffer Solutions

• Buffer solutions resist changes in pH.

• They are special cases of the common ion effect.

• The two most common types– weak acid + soluble salt of the weak acid

• HF + NaF

– weak base + soluble salt of the weak base• NH3 + NH4Cl

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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

0.10 0.050 0.036 3.6x10-5 4.44

0.10 0.10 0.018 1.8x10-5 4.74

0.10 0.15 0.012 1.2x1015 4.92

* % Dissociation = [CH3COOH]dissoc

[CH3COOH]init

x 100

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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 H3O+ reacts with CH3COO-, causing a shift to the left.

Added OH- reacts with CH3COOH, causing a shift to the right.

The shift in equilibrium position absorbs the change in [H3O+] or [OH-], and the pH changes only slightly.

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Figure 19.3 How a buffer works.

Buffer has equal concentrations of A- and HA.

H3O+

Buffer has more HA after addition of H3O+.

H2O + CH3COOH ← H3O+ + CH3COO-

OH-

Buffer has more A- after addition of OH-.

CH3COOH + OH- → CH3COO- + H2O

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Relative Concentrations of Buffer Components

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Ka =[CH3COO-][H3O+]

[CH3COOH]

If the ratio increases, [H3O+] decreases.[HA]

[A-]

Since Ka is constant, the [H3O+] of the solution depends on the ratio of buffer component concentrations.

[H3O+] = Ka x[CH3COOH]

[CH3COO-]

If the ratio decreases, [H3O+] increases.[HA]

[A-]

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Sample Problem 19.1 Calculating the Effect of Added H3O+ or OH- on Buffer pH

Calculate the pH:

PLAN: We can calculate [CH3COOH]init and [CH3COO-]init from the given information. From this we can find the starting pH. For (b) and (c) we assume that the added OH- or H3O+ reacts completely with the buffer components. We write a balanced equation in each case, set up a reaction table, and calculate the new [H3O+].

PROBLEM:

(a) Of a buffer solution consisting of 0.50 M CH3COOH and 0.50 M CH3COONa

(b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution

(c) After adding 0.020 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.)

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Sample Problem 19.1

SOLUTION: (a)

Initial 0.50 -

0.50 0

Change -−x +x +x

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M)

Equilibrium - 0.50 - x x0.50 + x

1.8x10-5 M

0.50 Mx 100 = 3.6x10-3% (< 5%; assumption is justified.)

Since Ka is small, x is small, so we assume

[CH3COOH] = 0.50 – x ≈ 0.50 M and [CH3COO-] = 0.50 + x ≈ 0.50 M

x = [H3O+] = Ka x[CH3COOH]

[CH3COO-]≈ 1.8x10-5 x 0.50

0.50= 1.8x10-5 M

Checking the assumption: pH = -log(1.8x10-5) = 4.74

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Sample Problem 19.1

Initial 0.50 0.020 0.50 -

Change -0.020-0.020 +0.020 -

CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Concentration (M)

(b) [OH−]added = 0.020 mol

1.0 L soln= 0.020 M OH−

Equilibrium 00.48 -0.52

Setting up a reaction table for the stoichiometry:

Initial 0.48 -

0.52 0

Change -−x +x +x

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M)

Equilibrium - 0.48 - x x0.52 + x

Setting up a reaction table for the acid dissociation, using new initial [ ]:

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Sample Problem 19.1

Since Ka is small, x is small, so we assume

[CH3COOH] = 0.48 – x ≈ 0.48 M and [CH3COO-] = 0.52 + x ≈ 0.52 M

x = [H3O+] = Ka x[CH3COOH]

[CH3COO-]

pH = -log(1.7x10-5) = 4.77

≈ 1.8x10-5 x 0.48

0.52= 1.7x10-5 M

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Sample Problem 19.1

Initial 0.50 0.020 0.50 -

Change -0.020-0.020 +0.020 -

CH3COO-(aq) + H3O+(aq) → CH3COOH(aq) + H2O(l) Concentration (M)

(c) [H3O+]added = 0.020 mol

1.0 L soln= 0.020 M H3O+

Equilibrium 00.48 -0.52

Setting up a reaction table for the stoichiometry:

Initial 0.52 -

0.48 0

Change -−x +x +x

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M)

Equilibrium - 0.52 - x x0.48 + x

Setting up a reaction table for the acid dissociation, using new initial [ ]:

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Sample Problem 19.1

Since Ka is small, x is small, so we assume

[CH3COOH] = 0.52 – x ≈ 0.52 M and [CH3COO-] = 0.48 + x ≈ 0.48 M

x = [H3O+] = Ka x[CH3COOH]

[CH3COO-]

pH = -log(2.0x10-5) = 4.70

≈ 1.8x10-5 x 0.52

0.48= 2.0x10-5 M

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The Common Ion Effect

• Determine the [H+] and the pH of a solution that is 0.15M in CH3COOH and 0.15M in NaCH3COO (a strong electrolyte). Ka = 1.8x10-5

• Partial list of soluble salts:– All K+, Na+ and NH4

+ salts

– All NO3- salts

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The Common Ion Effect

• Calculate the [OH-] and the pH of a solution that is 0.15M in NH3 and 0.15M in NH4Cl.

• Kb for NH3 = 1.8x10-5

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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]

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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.

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Figure 19.4 The relation between buffer capacity and pH change.

When strong base is added, the pH increases least for the most concentrated buffer.

This graph shows the final pH values for four different buffer solutions after the addition of strong base.

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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 one component is more than 10 times 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.

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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]

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Sample Problem 19.3 Preparing a Buffer

SOLUTION:

PROBLEM: An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on limsetone-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.

PLAN: The conjugate pair is HCO3- (acid) and CO3

2- (base), and we know both the buffer volume and the concentration of HCO3

-. We can calculate the ratio of components that gives a pH of 10.00, and hence the mass of Na2CO3 that must be added to make 1.5 L of solution.

[H3O+] = 10-pH = 10-10.00 = 1.0x10-10 M

HCO3-(aq) + H2O(l) H3O+(aq) + CO3

2-(aq) Ka =[CO3

2-][H3O+]

[HCO3-]

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Sample Problem 19.3 Preparing a Buffer

[CO32-] =

Ka[HCO3-]

[H3O+]=

(4.7x10-11)(0.20)

1.0x10-10= 0.094 M

Amount (mol) of CO32- needed = 1.5 L soln x 0.094 mol CO3

2-

1 L soln

= 0.14 mol CO32-

105.99 g Na2CO3

1 mol Na2CO3

0.14 mol Na2CO3 x = 15 g Na2CO3

The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 M NaHCO3

and add more 0.20 M NaHCO3 to make 1.5 L. Using a pH meter, she can then adjust the pH to 10.00 by dropwise addition of concentrated strong acid or base.

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• Determine the number of moles of NH4Cl that must be used to prepare 1.00L of a buffer solution that is 0.10M in NH3 and has a pH of 9.15. Kb for NH3 = 1.8x10-5

Buffer Solutions

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Acid-Base Indicators

An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In-).

The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

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.

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pH

Figure 19.5 Colors and approximate pH range of some common acid-base indicators.

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Figure 19.6 The color change of the indicator bromthymol blue.

pH < 6.0 pH > 7.5pH = 6.0-7.5

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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.

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Figure 19.7 Curve for a strong acid–strong base titration.

The initial pH is low.

The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00.

The pH increases gradually when excess base has been added.

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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

19-40

pH at the equivalence point

pH = 7.00 for a strong acid-strong base titration.

Calculating the pH during 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 = 14.00 - pOH

mol OH-excess

Vacid + Vbase

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Example:

40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH.

The initial pH is simply the pH of the HCl solution:

[H3O+] = [HCl]init = 0.1000 M and pH = -log(0.1000) = 1.00

To calculate the pH after 20.00 mL of NaOH solution has been added:

OH- added = 0.02000 L NaOH x 0.1000 mol

1 L= 2.000x10-3 mol OH-

Initial mol of H3O+ = 0.04000 L HCl x 0.1000 mol

1 L= 4.000x10-3 mol H3O+

H3O+ remaining = 4.000x10-3 – 2.000x10-3 = 2.000x10-3 mol H3O+

The OH- ions react with an equal amount of H3O+ ions, so

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[H3O+] = 2.000x10-3 mol

0.04000 L + 0.02000 L= 0.03333 M

pH = -log(0.03333) = 1.48

To calculate the pH after 50.00 mL of NaOH solution has been added:

The equivalence point occurs when mol of OH- added = initial mol of HCl, so when 40.00 mL of NaOH has been added.

OH- added = 0.05000 L NaOH x 0.1000 mol

1 L= 5.000x10-3 mol OH-

OH- in excess = 5.000x10-3 – 4.000x10-3 = 1.000x10-3 mol OH-

[OH-] = 1.000x10-3 mol

0.04000 L + 0.05000 L= 0.01111 M

pOH = -log(0.01111) = 1.95 pH = 14.00 – 1.95 = 12.05

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Figure 19.8 Curve for a weak acid–strong base titration.

The initial pH is higher than for the strong acid solution.

The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution.

The pH increases slowly beyond the equivalence point.

The pH at the equivalent point is > 7.00 due to the reaction of the conjugate base with H2O.

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Calculating the pH during a weak acid–strong base titration

Initial pH

[H3O+] =

pH = -log[H3O+]

[H3O+][A-]

[HA]Ka = Ka x [HA]init)

pH before equivalence point

[H3O+] = Ka x or

pH = pKa + log[base]

[acid]

[HA]

[A-]

19-45

Calculating the pH during a weak acid–strong base titration

A-(aq) + H2O(l) HA(aq) + OH-(aq)

pH at the equivalence point

[OH- ] =

where [A-] = andKw

Ka

Kb =mol HAinit

Vacid + Vbase

Kb x [A-]

[H3O+] ≈ and pH = -log[H3O+]Kw

Kb x [A-])

pH beyond the equivalence point

[OH-] =

pH = -log[H3O+]

mol OH-excess

Vacid + Vbase

[H3O+] = Kw

[OH-]

19-46

Sample Problem 19.4 Finding the pH During a Weak Acid–Strong Base Titration

PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 M NaOH:

PLAN: The initial pH must be calculated using the Ka value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH- added. Once we know the volume of base required to reach the equivalence point we can calculate the pH based on the species present in solution.

(a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL

SOLUTION:

(a) [H3O+] = Ka x [HPr]init) = (1.3x10-5)(0.100) = 1.1x10-3 M

pH = -log(1.1x10-3) = 2.96

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Sample Problem 19.4

(b) 30.00 mL of 0.1000 M NaOH has been added.

Initial amount of HPr = 0.04000 L x 0.1000 M = 4.000x10-3 mol HPr

Amount of NaOH added = 0.03000 L x 0.1000 M = 3.000x10-3 mol OH-

Each mol of OH- reacts to form 1 mol of Pr-, so

HPr(aq) + OH-(aq) → Pr-(aq) + H2O(l)Concentration (M)

Initial 0.004000 0.003000

0 -

Change −0.003000 -0.003000

+0.003000

-

Equilibrium 0.001000 0 -0.003000

[H3O+] = Ka x [HPr]

[Pr-]

pH = -log(4.3x10-6) = 5.37

0.001000

0.003000= (1.3x10-5) x = 4.3x10-6 M

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Sample Problem 19.4

(c) 40.00 mL of 0.1000 M NaOH has been added.

This is the equivalence point because mol of OH- added = 0.004000 = mol of HAinit. All the OH- added has reacted with HA to form 0.004000 mol of Pr-.

0.004000 mol

0.04000 L + 0.04000 L[Pr-] = = 0.05000 M

Pr- is a weak base, so we calculate Kb =Kw

Ka

=1.0x10-14

1.3x10-5= 7.7x10-10

[H3O+] ≈Kw

Kb x [A-])=

(7.7x10-10)(0.05000)1.0x10-14

= 1.6x10-9 M

pH = -log(1.6x10-9) = 8.80

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Sample Problem 19.4

(d) 50.00 mL of 0.1000 M NaOH has been added.

Amount of OH- added = 0.05000 L x 0.1000 M = 0.005000 mol

Excess OH- = OH-added – HAinit = 0.005000 – 0.004000 = 0.001000 mol

[OH-] = mol OH-

excess

total volume=

0.001000 mol

0.09000 L= 0.01111 M

[H3O+] = Kw

[OH-]=

1x10-14

0.01111= 9.0x10-13 M

pH = -log(9.0x10-13) = 12.05

19-50

Figure 19.9 Curve for a weak base–strong acid titration.

The pH decreases gradually in the buffer region. The weak base and its conjugate acid are both present in solution.

The pH at the equivalence point is < 7.00 due to the reaction of the conjugate acid with H2O.

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Figure 19.10 Curve for the titration of a weak polyprotic acid.

Titration of 40.00 mL of 0.1000 M H2SO3 with 0.1000 M NaOH

pKa2 = 7.19

pKa1 = 1.85

19-52

Net Ionic Equations

• Include in the equation only those species that actually change concentration.

• Molecular reaction:– NaOH + HCl = H2O + NaCl

• Net Ionic Equation:– H+ + OH- = H2O

19-53

19-54

Net Ionic Equations

• Rules for writing:– A. List predominant species

• All soluble salts, strong acids and strong bases are written as their component ions. All others are written as the molecule.

– B. Combine ions of opposite charge and look for one or more of the following:

• 1. formation of a weak acid

• 2. formation of a weak base

• 3. formation of water

19-55

Net Ionic Equations

• Rules (continued):– C. If no reaction in B. above, then look at

each of the following in the order given as a source of secondary species:

• 1. weak acid

• 2. weak base

• 3. water, if no other possibility

19-56

Net Ionic Equations

• Write net ionic equations for each of the following for 0.10M solutions being mixed. Write the form of the equilibrium constant using Ka, Kb and Kw

– 1. NaOH + HCl

– 2. KF + HNO3

– 3. NH4NO3 + Ca(OH)2

– 4. HF + NaOH

– 5. H2S + KOH

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19-58

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19-60

Hydrolysis Equilibria

• Refers to the reaction of a substance with water or its ions.

• Write net ionic equations for each of the following:– 1. NaF + H2O

– 2. NH4Cl + H2O

– 3. NaCl + H2O

– 4. AlCl3 + H2O

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19-62

19-63

Hydrolysis Equilibria

Determine the [H+]. pH and percent hydrolysis for a 0.10M solution of NH4Br. Kb for NH3 is 1.8x 10-5

Calculate the pH of a 0.10M solution of NaCN. Ka for HCN is 4.0x10-10

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Buffering Action

If 0.20 mole of HCl is added to 1.00L of a solution that is 0.100M in NH3 and 0.200M in NH4Cl, determine the final pH of the solution. Assume no volume change from the addition of HCl(g). Kb for NH3 is 1.8x10-5

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