Try It

1. Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
2. Explain why the ionization constant, K_a, for H₂SO₄ is larger than the ionization constant for H₂SO₃.
3. What is the ionization constant at 25 °C for the weak acid [latex]{\text{CH}}_{3}{\text{NH}}_{3}^{+}[/latex], the conjugate acid of the weak base CH₃NH₂, K_b = 4.4 × 10⁻⁴.

[reveal-answer q=”314309″]Show Selected Solutions[/reveal-answer]
[hidden-answer a=”314309″]

1. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH⁻, which causes the solution to be basic. An example is NaCN. The CN⁻ reacts with water as follows:
   1. [latex]{\text{CN}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]
2. The oxidation state of the sulfur in H₂SO₄ is greater than the oxidation state of the sulfur in H₂SO₃.
3. K_w = K_a × K_b; thus,
   1. [latex]\begin{array}{rcl}{K}_{\text{a}}&=&\frac{{K}_{\text{w}}}{{K}_{\text{b}}}\\{K}_{\text{a}}&=&\frac{1.0\times {10}^{-14}}{4.4\times {10}^{-4}}=2.3\times {10}^{-11}\end{array}[/latex]

[/hidden-answer]

Think about It

Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F⁻ or CN⁻, is the stronger base? See Table 15.3.3.

Try It

1. Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO₂H and 0.10 M in HClO.
2. Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO₂ and 0.120 M in HBrO.
3. Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH₃NH₂ and 0.10 M in C₅H₅N (K_b = 1.7 × 10⁻⁹).
4. Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH₃ and 0.100 M in C₆H₅NH₂.
5. Using the K_a values in Ionization Constants of Weak Acids, place [latex]\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}^{3+}[/latex] in the correct location in Figure 15.3.3.

[reveal-answer q=”588268″]Show Solutions[/reveal-answer]
[hidden-answer a=”588268″]
2. The reactions and equilibrium constants are:

[latex]\begin{array}{l}{\text{HNO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{NO}}_{2}^{-}\left(aq\right){K}_{\text{a}}=4.5\times {10}^{-4}\\ \text{HBrO}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{BrO}}^{-}\left(aq\right){K}_{\text{a}}=2\times {10}^{-9}\end{array}[/latex]

As K_a is much larger for HNO₂ than for HBrO, the first equilibrium will dominate. The equilibrium expression is [latex]{K}_{\text{a}}=\frac{\left[{\text{NO}}_{2}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{HNO}}_{2}\right]}=4.5\times {10}^{-5}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

	[HNO2]	[H3O+]	[NO2−]
Initial concentration (M)	0.134	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.134 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.134 − x) ≈ 0.134 gives:

[latex]\frac{\left[{\text{NO}}_{2}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{HNO}}_{2}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.134-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.134}=4.5\times {10}^{-4}[/latex]

Solving for x gives 7.77 × 10⁻³M. Because this value is 5.8% of 0.134, our assumption is incorrect. Therefore, we must use the quadratic formula. Using the data gives the following equation: x² + 4.5 × 10⁻⁷x − 6.03 × 10⁻⁵ = 0

Using the quadratic formula gives (a = 1, b = 4.5 × 10⁻⁴, and c = −6.03 × 10⁻⁵)

[latex]\begin{array}{rcl}{ }x&=&\frac{-b\pm \sqrt{{b}^{\text{2+}}-4ac}}{2a}=\frac{-\left(4.5\times {10}^{-4}\right)\pm \sqrt{{\left(4.5\times {10}^{-4}\right)}^{\text{2+}}-4\left(1\right)\left(-6.03\times {10}^{-5}\right)}}{2\left(1\right)}\\{}&=&\frac{-\left(4.5\times {10}^{-4}\right)\pm \left(1.55\times {10}^{-2}\right)}{2}=7.54\times {10}^{-3}M\left(\text{positive root}\right)\end{array}[/latex]

The equilibrium concentrations are therefore:

• [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\left[{\text{NO}}_{2}^{-}\right]=7.54\times {10}^{-3}=7.5\times {10}^{-3}M[/latex]
• [HNO₂] = 0.134 − 7.54 × 10⁻³ = 0.1264 = 0.126

[OH⁻] can be calculated using K_w:

[latex]\left[{\text{OH}}^{-}\right]=\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{7.54\times {10}^{-3}}=1.33\times {10}^{-12}=1.3\times {10}^{-12}M[/latex]

Finally, use the other equilibrium to find the other concentrations. Assume for [HBrO] that (0.120 − x) ≈ 0.124 M.

[latex]{K}_{\text{a}}=\frac{\left[{\text{BrO}}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[\text{HBrO}\right]}=2\times {10}^{-9}\approx \frac{\left[{\text{BrO}}^{-}\right]\left(7.54\times {10}^{-3}\right)}{0.120}[/latex]

Solving for [BrO⁻] gives:

• [BrO⁻] = 3.2 × 10⁻⁸ = 3.2 × 10⁻⁸M
• [HBrO] = 0.120 − 3.2 × 10⁻⁸ = 0.120 M

4. The reactions and equilibrium constants are:

[latex]\begin{array}{l}{}{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{OH}}^{-}\left(aq\right)+{\text{NH}}_{4}^{+}\left(aq\right){K}_{\text{b}}=1.8\times {10}^{-5}\\ {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{OH}}^{-}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\left(aq\right){K}_{\text{b}}=4.6\times {10}^{-10}\end{array}[/latex]

As K_b is much larger for NH₃ than for C₆H₅NH_2, the first equilibrium will dominate. The equilibrium expression is [latex]{K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=1.8\times {10}^{-5}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

	[NH3]	[OH−]	[NH4+]
Initial concentration (M)	0.115	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.115 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.115 − x) ≈ 0.115 gives:

[latex]\frac{\left[{\text{NH}}_{4}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.115-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.115}=1.8\times {10}^{-5}[/latex]

Solving for x gives 1.44 × 10⁻³M. Because this value is less than 5% of 0.115 M, our assumption is correct. The equilibrium concentrations are therefore:

• [OH⁻] = [latex]\left[{\text{NO}}_{4}^{+}\right][/latex] = 1.44 × 10⁻³ = 0.0014 M
• [NH₃] = 0.115 − 0.00144 = 0.1136 = 0.144 M

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] can be calculated using K_w:

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{w}}}{\left[{\text{OH}}^{-}\right]}=\frac{1.0\times {10}^{-14}}{0.00144}=6.94\times {10}^{-12}=6.9\times {10}^{-12}M[/latex]

Finally, use the other equilibrium to find the other concentrations. Assume for [C₆H₅NH₂] that (0.100 − x) ≈ 0.100 M:

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{w}}}{\left[{\text{OH}}^{-}\right]}=\frac{1.0\times {10}^{-14}}{0.00144}=6.94\times {10}^{-12}=6.9\times {10}^{-12}M[/latex]

Solving for [latex]\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right][/latex] gives:

• [latex]\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right][/latex] = 3.9 × 10⁻⁸M
• [C₆H₅NH₂] = 0.100 − 3.19 × 10⁻⁸ = 0.100 M

[/hidden-answer]

Try It

Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Ionization Constants of Weak Acids and Ionization Constants of Weak Bases.

Solution 1

0.0092 M HClO, a weak acid

[reveal-answer q=”936215″]Show Solution[/reveal-answer]
[hidden-answer a=”936215″]

The reaction is: [latex]\text{HClO}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{ClO}}^{-}\left(aq\right)[/latex]

The equilibrium expression is [latex]{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{ClO}}^{-}\right]}{\left[\text{HClO}\right]}=3.5\times {10}^{-8}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

	[HClO]	[H3O+]	[ClO−]
Initial concentration (M)	0.0092	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.0092 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0092 − x) ≈ 0.0092 gives:

[latex]\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{ClO}}^{-}\right]}{\left[\text{HClO}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0092-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.0092}=3.5\times {10}^{-8}[/latex]

Solving for x gives 1.79 × 10⁻⁵M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

• [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = [ClO] = 1.8 × 10⁻⁵M
• [HClO] = 0.0092 − 1.79 × 10⁻⁵ = 0.00918 = 0.00092 M
• [latex]\left[{\text{OH}}^{-}\right]=\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{1.79\times {10}^{-5}}=5.6\times {10}^{-10}M[/latex]

[/hidden-answer]

Solution 2

0.0784 M C₆H₅NH₂, a weak base

[reveal-answer q=”996941″]Show Solution[/reveal-answer]
[hidden-answer a=”996941″]

The reaction is [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]

The equilibrium expression is [latex]{K}_{\text{a}}=\frac{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right]}=4.6\times {10}^{-10}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

	[C6H5NH2]	[C5H5NH3+]	[OH−]
Initial concentration (M)	0.0784	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.0784 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0784 − x) ≈ 0.0784 gives:

[latex]\frac{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0784-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.0784}=4.6\times {10}^{-10}[/latex]

Solving for x gives 6.01 × 10⁻⁶M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

• [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right][/latex] = [OH⁻] = 6.0 × 10⁻⁶M
• [C₆H₅NH₂] = 0.0784 − 6.01 × 10⁻⁶ = 0.007839 = 0.00784
• [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{w}}}{\left[{\text{OH}}^{-}\right]}=\frac{1.0\times {10}^{-14}}{6.01\times {10}^{-6}}=1.7\times {10}^{-9}M[/latex]

[/hidden-answer]

Solution 3

0.0810 M HCN, a weak acid

[reveal-answer q=”342499″]Show Solution[/reveal-answer]
[hidden-answer a=”342499″]

The reaction is [latex]\text{HCN}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)[/latex].

The equilibrium expression is [latex]{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}=4\times {10}^{-10}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

	[HClO]	[H3O+]	[CN−]
Initial concentration (M)	0.0810	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.0810 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0810 − x) ≈ 0.0810 gives:

[latex]\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0810-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.0810}=4\times {10}^{-10}[/latex]

Solving for x gives 5.69 × 10⁻⁶M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

• [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = [CN⁻] = 5.7 × 10⁻⁶M
• [HCN] = 0.0810 − 5.69 × 10⁻⁶ = 0.08099 = 0.0810 M
• [latex]\left[{\text{OH}}^{-}\right]=\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{5.69\times {10}^{-6}}=1.8\times {10}^{-9}M[/latex]

[/hidden-answer]

Solution 4

0.11 M (CH₃)₃N, a weak base

[reveal-answer q=”210337″]Show Solution[/reveal-answer]
[hidden-answer a=”210337″]

The reaction is [latex]{\left({\text{CH}}_{3}\right)}_{3}\text{N}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]

The equilibrium expression is [latex]{K}_{\text{b}}=\frac{\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{CH}}_{3}\right)}_{3}\text{N}\right]}=7.4\times {10}^{-5}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

	[(CH3)3N]	[(CH3)3NH+]	[OH−]
Initial concentration (M)	0.11	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.11 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.11 − x) ≈ 0.11 gives:

[latex]\frac{\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{CH}}_{3}\right)}_{3}\text{N}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.11-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.11}=7.4\times {10}^{-5}[/latex]

Solving for x gives 2.85 × 10⁻³M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

• [latex]\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\right][/latex] = [OH⁻] = 2.9 × 10⁻³M
• [(CH₃)₃N] = 0.11 − 2.85 × 10⁻³ = 0.107 = 0.11 M
• [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{w}}}{\left[{\text{OH}}^{-}\right]}=\frac{1.0\times {10}^{-14}}{2.85\times {10}^{-3}}=3.5\times {10}^{-12}M[/latex]

[/hidden-answer]

Solution 5

0.120 M [latex]\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}^{2+}[/latex] a weak acid, K_a = 1.6 × 10⁻⁷

[reveal-answer q=”874526″]Show Solution[/reveal-answer]
[hidden-answer a=”874526″]

The reaction is [latex]\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons \text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{+}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)[/latex]

The equilibrium expression is [latex]{K}_{\text{a}}=\frac{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{+}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}^{2+}\right]}=1.6\times {10}^{-7}[/latex]

The initial and equilibrium concentrations for this system can be written as follows:

	[Fe(H2O)62+]	[H3O+]	[Fe(H2O)5(OH)+]
Initial concentration (M)	0.120	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.120 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.120 − x) ≈ 0.120 gives:

[latex]\frac{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{+}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}^{2+}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.120-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.120}=1.6\times {10}^{-7}[/latex]

Solving for x gives 1.39 × 10⁻⁴M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

• [latex]\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}{\left(\text{OH}\right)}^{+}\right][/latex] = [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 1.4 × 10⁻⁴M
• [latex]\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}^{2+}\right][/latex] = 0.120 − 1.39 × 10⁻⁴ = 0.1199 = 0.120 M
• [latex]\left[{\text{OH}}^{-}\right]=\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{1.39\times {10}^{-4}}=7.2\times {10}^{-11}M[/latex]

[/hidden-answer]

Try It

1. Propionic acid, C₂H₅CO₂H (K_a = 1.34 × 10⁻⁵), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698-M solution of C₂H₅CO₂H?
2. White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm³, what is the pH?
3. The ionization constant of lactic acid, CH₃CH(OH)CO₂H, an acid found in the blood after strenuous exercise, is 1.36 × 10⁻⁴. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?
4. Nicotine, C₁₀H₁₄N₂, is a base that will accept two protons (K₁ = 7 × 10⁻⁷, K₂ = 1.4 × 10⁻¹¹). What is the concentration of each species present in a 0.050-M solution of nicotine?
5. The pH of a 0.20-M solution of HF is 1.92. Determine K_a for HF from these data.
6. The pH of a 0.15-M solution of [latex]{\text{HSO}}_{4}^{-}[/latex] is 1.43. Determine K_a for [latex]{\text{HSO}}_{4}^{-}[/latex] from these data.
7. The pH of a 0.10-M solution of caffeine is 11.16. Determine K_b for caffeine from these data:
8. The pH of a solution of household ammonia, a 0.950 M solution of NH_3, is 11.612. Determine K_b for NH₃ from these data.

[reveal-answer q=”864039″]Show Solutions[/reveal-answer]
[hidden-answer a=”864039″]

2. First, find the mass of acetic acid. d = 1.007 g/cm³. Take 1.0 L of solution to have the quantities on a mole basis. Then, since 1000 cm³ = 1.0 L, 1000 cm³ × 1.007 g/cm³ = 1007 g in 1.0 L. Then, 5.00% of this is the mass of acetic acid:

[latex]\text{Mass (acetic acid)}=\text{1007 g}\times \frac{5.0%}{100%}=\text{50.35 g}[/latex]

Now calculate the number of moles of acetic acid present. The molar mass of acetic acid is 60.053 g/mol:

[latex]\text{mol acetic acid}=\frac{50.35\cancel{\text{g}}}{60.053\cancel{\text{g}}{\text{mol}}^{-1}}=\text{0.838 mol}[/latex]

From the moles of acetic acid and K_a, calculate [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]:[/latex]

[latex]{K}_{\text{a}}=1.8\times {10}^{-5}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}[/latex]

	[CH3CO2H]	[H3O+]	[CH3CO2−]
Initial concentration (M)	0.838	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.838 − x	x	x

Substitution gives: [latex]{K}_{\text{a}}=1.8\times {10}^{-5}=\frac{{x}^{\text{2}}}{0.838-x}[/latex]

Drop x because it is small in comparison with 0.838 M.

• x² = 0.838(1.8 × 10⁻⁵) = 1.508 × 10⁻⁵ = 3.88 × 10⁻³ = 2.41
• pH = −log(3.88 × 10⁻³) = 2.41

4.

[latex]\begin{array}{l}{ }{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\left({K}_{\text{b1}}=7\times {10}^{-7}\right)\\ {\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}^{2+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\left({K}_{\text{b2}}=1.4\times {10}^{-11}\right)\end{array}[/latex]

First set up a concentration table:

	[C10H14N2]	[C10H14N2H+]	[OH−]
Initial concentration (M)	0.050	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.050 − x	x	x

Substituting the equilibrium concentrations into the equilibrium equation and making the assumption that (0.050 − x) = 0.050, we get:

[latex]\begin{array}{rcl}{K}_{\text{b1}}&=&\frac{\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}\right]}=7\times {10}^{-7}\\{}&=&\frac{\left(x\right)\left(x\right)}{\left(0.050-x\right)}=\frac{{x}^{\text{2+}}}{0.050}=7\times {10}^{-7}\end{array}[/latex]

Solving for x gives 1.87 × 10⁻⁴ = 2 × 10⁻⁴M = [OH⁻]

Because x is less than 5% of 0.050 and [OH⁻] is greater than 4.5 × 10⁻⁷ M, our customary assumptions are justified. We can calculate

• [C₁₀H₁₄N₂] = 0.050 − x = 0.050 − 2 × 10⁻⁴ = 0.048 M
• [OH⁻] = [latex]\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}\right][/latex] = x = 2 × 10⁻⁴M

Now calculate the concentration of [latex]{\text{C}}_{10}{\text{H}}_{2}{\text{N}}_{2}{\text{H}}_{2}^{2+}[/latex] in a solution with [OH⁻] and [latex]\left[{\text{C}}_{10}{\text{H}}_{2}{\text{N}}_{2}{\text{H}}_{2}^{2+}\right][/latex] equal to 2 × 10⁻⁴M. The equilibrium between these species is

[latex]{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}^{\text{2+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]

We know [C₁₀H₁₄N₂H⁺] and [OH⁻], so we can calculate the concentration of [latex]{\text{C}}_{10}{\text{H}}_{2}{\text{N}}_{2}{\text{H}}_{2}^{2+}[/latex] from the equilibrium expression:

[latex]\begin{array}{rcl}{}{K}_{\text{b2}}&=&\frac{\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}^{2+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}\right]}=1.4\times {10}^{-11}\\{}&=&\frac{\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}^{2+}\right]\left[2\times {10}^{-4}\right]}{\left[2\times {10}^{-4}\right]}\\ \left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}^{2+}\right]&=&1.4\times {10}^{-11}M\end{array}[/latex]

The concentration of OH⁻ produced in this ionization is equal to the concentration of [latex]{\text{C}}_{10}{\text{H}}_{2}{\text{N}}_{2}{\text{H}}_{2}^{2+}[/latex], 1.4 × 10⁻¹¹M, which is much smaller than the 2 ×10⁻⁴M produced in the first ionization; therefore, we are justified in neglecting the OH⁻ formed from [latex]{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}[/latex].

We can now calculate the concentration of H₂O⁺ present from the ionization of water:

K_a = 1 × 10⁻¹⁴ = [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] [OH⁻]

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{1\times {10}^{-14}}{\left[{\text{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{1.9\times {10}^{-4}}=5.3\times {10}^{-11}M[/latex]

We can now summarize the concentrations of all species in solution as follows:

• [C₁₀H₁₄N₂] = 0.049 M
• [latex]\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}^{+}\right][/latex] = 1.9 × 10⁻⁴ M
• [latex]\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}^{2+}\right]=1.4\times {10}^{-11}M[/latex]
• [OH⁻] = 1.9 × 10⁻⁴M
• [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 5.3 × 10⁻¹¹M

6. The reaction is [latex]{\text{HSO}}_{4}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{SO}}_{2}^{2-}\left(aq\right)[/latex].

The concentrations at equilibrium are [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 10^−pH = 10^−1.43 = 0.0372 M

• [HF] = 0.15 − 0.0372 M = 0.113 M
• [latex]{K}_{\text{a}}=\frac{\left[{\text{SO}}_{4}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{HSO}}_{4}^{-}\right]}=\frac{\left(0.0372\right)\left(0.0372\right)}{\left(0.113\right)}=1.2\times {10}^{-2}[/latex]

8. The reaction is [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_{4}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex].

The pOH can be determined from pOH = 14.000 − pH = 14.000 − 11.612 = 2.388. Therefore, the concentrations at equilibrium are [latex]\left[{\text{NH}}_{4}^{+}\right][/latex] = [OH⁻] = 10^−pOH = 10^−2.388 = 0.004093 M

• [NH₃] = 0.950 − 0.004093 = 0.9459 M
• [latex]{K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(0.004093\right)\left(0.004093\right)}{\left(0.9459\right)}=1.77\times {10}^{-5}[/latex]

[/hidden-answer]

Try It

1. The odor of vinegar is due to the presence of acetic acid, CH₃CO₂H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.
2. Household ammonia is a solution of the weak base NH₃ in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.
3. Which base, CH₃NH₂ or (CH₃)₂NH, is the strongest base? Which conjugate acid, [latex]{\left({\text{CH}}_{3}\right)}_{2}{\text{NH}}_{2}^{+}[/latex] or (CH₃)₂NH, is the strongest acid?
4. Which is the stronger acid, [latex]{\text{NH}}_{4}^{+}[/latex] or HBrO?
5. Which is the stronger base, (CH₃)₃N or [latex]{\text{H}}_{2}{\text{BO}}_{3}^{-}?[/latex]
6. Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
   1. H₂O or HF
   2. B(OH)₃ or Al(OH)₃
   3. [latex]{\text{HSO}}_{3}^{-}[/latex] or [latex]{\text{HSO}}_{4}^{-}[/latex]
   4. NH₃ or H₂S
   5. H₂O or H₂Te
7. Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
   1. [latex]{\text{HSO}}_{4}^{-}[/latex] or [latex]{\text{HSeO}}_{4}^{-}[/latex]
   2. NH₃ or H₂O
   3. PH₃ or HI
   4. NH₃ or PH₃
   5. H₂S or HBr
8. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
   1. acidity: HCl, HBr, HI
   2. basicity: H₂O, OH⁻, H⁻, Cl⁻
   3. basicity: Mg(OH)₂, Si(OH)₄, ClO₃(OH) (Hint: Formula could also be written as HClO₄).
   4. acidity: HF, H₂O, NH₃, CH₄
9. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
   1. acidity: NaHSO₃, NaHSeO₃, NaHSO₄
   2. basicity: [latex]{\text{BrO}}_{2}^{-}[/latex], [latex]{\text{ClO}}_{2}^{-}[/latex], [latex]{\text{IO}}_{2}^{-}[/latex]
   3. acidity: HOCl, HOBr, HOI
   4. acidity: HOCl, HOClO, HOClO₂, HOClO₃
   5. basicity: [latex]{\text{NH}}_{2}^{-}[/latex], HS⁻, HTe⁻, [latex]{\text{PH}}_{2}^{-}[/latex]
   6. basicity: BrO⁻, [latex]{\text{BrO}}_{2}^{-}[/latex], [latex]{\text{BrO}}_{3}^{-}[/latex], [latex]{\text{BrO}}_{4}^{-}[/latex]
10. The active ingredient formed by aspirin in the body is salicylic acid, C₆H₄OH(CO₂H). The carboxyl group (−CO₂H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C₆H₄OH(CO₂H).

[reveal-answer q=”929665″]Show Solution[/reveal-answer]
[hidden-answer a=”929665″]

1. [H₂O] > [CH₃CO₂H] > [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] ≈ [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right][/latex] > [OH⁻]
3. The strongest base or strongest acid is the one with the larger K_b or K_a, respectively. In these two examples, they are (CH₃)₂NH and [latex]{\text{CH}}_{3}{\text{NH}}_{3}^{+}[/latex].
5. Look up Ionization Constants of Weak Bases the value of K_b for (CH₃)₃N and the value of K_a for H₃BO₃. From the latter, calculate the value of K_b for [latex]{\text{H}}_{2}{\text{BO}}_{3}^{-}[/latex]. Then compare values:
   • K_b(CH₃)₃N = 7.4 × 10⁻⁵
   • [latex]{K}_{\text{a}}\left({\text{H}}_{3}{\text{BO}}_{3}\right)=5.8\times {10}^{-10}=\frac{{K}_{\text{w}}}{{K}_{\text{b}}}[/latex], [latex]{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{5.8\times {10}^{-10}}=1.7\times {10}^{-5}[/latex]
6. A comparison shows that the larger K_b is that of triethylamine.
7. The more acidic compounds are as follows:
   1. [latex]{\text{HSO}}_{4}^{-}[/latex]; higher electronegativity of the central ion.
   2. H₂O; NH₃ is a base and water is neutral, or decide on the basis of K_a values.
   3. HI; PH₃ is weaker than HCl; HCl is weaker than HI. Thus, PH₃ is weaker than HI.
   4. PH₃; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group.
   5. HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.
9. The correct ordered lists are as follows:
   1. NaHSeO₃ < NaHSO₃ < NaHSO₄; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner.
   2. [latex]{\text{ClO}}_{2}^{-}<{\text{BrO}}_{2}^{-}<{\text{IO}}_{2}^{-}[/latex]; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.
   3. HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.
   4. HOCl < HOClO < HOClO₂ < HOClO₃; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases).
   5. [latex]{\text{HTe}}^{-}<{\text{HS}}^{-}<<{\text{PH}}_{2}^{-}<{\text{NH}}_{2}^{-}[/latex]; [latex]{\text{PH}}_{2}^{-}[/latex] and [latex]{\text{NH}}_{2}^{-}[/latex] are anions of weak bases, so they act as strong bases toward H⁺. [latex]{\text{HTe}}^{-}[/latex] and HS⁻ are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion.
   6. [latex]{\text{BrO}}_{4}^{-}<{\text{BrO}}_{3}^{-}<{\text{BrO}}_{2}^{-}<{\text{BrO}}^{-}[/latex]; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.
10.  [latex]\left[\text{H}_{2}\text{O}\right]\gt\left[\text{C}_{6}\text{H}_{4}\text{OH}\left(\text{CO}_{2}\text{H}\right)\right]\gt\left[\text{H}^{+}\right]\text{O}\gt\left[\text{C}_{6}\text{H}_{4}\text{OH}\left(\text{CO}_{2}\right)^{-}\right]\gt\left[\text{C}_{6}\text{H}_{4}\text{O}\left(\text{CO}_{2}\text{H}\right)^{-}\right]\gt\left[\text{OH}^{-}\right][/latex]

[/hidden-answer]

Try It

1. Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO₃)₂, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO₃ with CuO.
2. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)₂ in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs
3. What do we represent when we write: [latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}^{-}\left(aq\right)?[/latex]
4. Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?
5. Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.
6. What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?
7. What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base?

[reveal-answer q=”405128″]Show Selected Solutions[/reveal-answer]
[hidden-answer a=”405128″]

2. [latex]\begin{array}{cccccc}\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)+& \text{HCl}\left(aq\right)& \longrightarrow & {\text{Mg}}^{2+}\left(aq\right)+& 2{\text{Cl}}^{-}\left(aq\right)+& {\text{2H}}_{2}\text{O}\left(l\right)\\ \text{BB}& \text{BA}& & \text{CB}& \text{CA}& \end{array}[/latex]
4. Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.
6. The two assumptions are
   • Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration.
   • Assume we can neglect the contribution of water to the equilibrium concentration of [latex]{\text{H}}_{3}{\text{O}}^{+}[/latex].

[/hidden-answer]

Try It

1. Which of the following will increase the percent of NH₃ that is converted to the ammonium ion in water (Hint: Use LeChâtelier’s principle.)?
   1. addition of NaOH
   2. addition of HCl
   3. addition of NH₄Cl
2. Which of the following will increase the percent of HF that is converted to the fluoride ion in water?
   1. addition of NaOH
   2. addition of HCl
   3. addition of NaF
3. What is the effect on the concentrations of [latex]{\text{NO}}_{2}^{-}[/latex], HNO₂, and OH⁻ when the following are added to a solution of KNO₂ in water:
   1. HCl
   2. HNO₂
   3. NaOH
   4. NaCl
   5. KNO

   The equation for the equilibrium is: [latex]{\text{NO}}_{2}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{HNO}}_{2}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]

4. What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?
   1. HCl
   2. KF
   3. NaCl
   4. KOH
   5. HF

   The equation for the equilibrium is: [latex]\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{F}}^{-}\left(aq\right)[/latex]

5. Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?
6. From the equilibrium concentrations given, calculate K_a for each of the weak acids and K_b for each of the weak bases.
   1. CH₃CO₂H: [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 1.34 × 10⁻³M; [latex]\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right][/latex] = 1.34 × 10⁻³M; [CH₃CO₂H] = 9.866 × 10⁻²M
   2. ClO⁻: [OH⁻] = 4.0 × 10⁻⁴M; [HClO] = 2.38 × 10⁻⁵M; [ClO⁻] = 0.273 M
   3. HCO₂H: [HCO₂H] = 0.524 M; [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 9.8 × 10⁻³M; [latex]\left[{\text{HCO}}_{2}^{-}\right][/latex] = 9.8 × 10⁻³M
   4. [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}:[/latex] [latex]\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right][/latex] = 0.233 M; [C₆H₅NH₂] = 2.3 × 10⁻³M; [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 2.3 × 10⁻³M
7. From the equilibrium concentrations given, calculate K_a for each of the weak acids and K_b for each of the weak bases.
   1. NH₃: [OH⁻] = 3.1 × 10⁻³M; [latex]\left[{\text{NH}}_{4}^{+}\right][/latex] = 3.1 × 10⁻³M; [NH₃] = 0.533 M
   2. HNO₂: [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 0.011 M; [latex]\left[{\text{NO}}_{2}^{-}\right][/latex] = 0.0438 M; [HNO₂] = 1.07 M
   3. (CH₃)₃N: [(CH₃)₃N] = 0.25 M; [latex]\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\right][/latex] = 4.3 × 10⁻³M; [OH⁻] = 4.3 × 10⁻³M
   4. [latex]{\text{NH}}_{4}^{+}:[/latex] [latex]\left[{\text{NH}}_{4}^{+}\right][/latex] = 0.100 M; [NH₃] = 7.5 × 10⁻⁶M; [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = 7.5 × 10⁻⁶M
8. Determine K_b for the nitrite ion, [latex]{\text{NO}}_{2}^{-}[/latex]. In a 0.10-M solution this base is 0.0015% ionized.
9. Determine K_a for hydrogen sulfate ion, [latex]{\text{HSO}}_{4}^{-}[/latex]. In a 0.10-M solution the acid is 29% ionized.
10. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
    1. F⁻
    2. [latex]{\text{NH}}_{4}^{+}[/latex]
    3. [latex]{\text{AsO}}_{4}^{3-}[/latex]
    4. [latex]{\left({\text{CH}}_{3}\right)}_{2}{\text{NH}}_{2}^{+}[/latex]
    5. [latex]{\text{NO}}_{2}^{-}[/latex]
    6. [latex]{\text{HC}}_{2}{\text{O}}_{4}^{-}[/latex] (as a base)
11. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
    1. HTe⁻ (as a base)
    2. [latex]{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}[/latex]
    3. [latex]{\text{HAsO}}_{4}^{3-}[/latex] (as a base)
    4. [latex]{\text{HO}}_{2}^{-}[/latex] (as a base)
    5. [latex]{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}[/latex]
    6. [latex]{\text{HSO}}_{3}^{-}[/latex] (as a base)
12. For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?
    1. 3 × 10⁻⁸M HNO₃
    2. 0.10 g HCl in 1.0 L of solution
    3. 0.00080 g NaOH in 0.50 L of solution
    4. 1 × 10⁻⁷M Ca(OH)₂
    5. 0.0245 M KNO₃
13. Even though both NH₃ and C₆H₅NH₂ are weak bases, NH₃ is a much stronger acid than C₆H₅NH₂. Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH₃ and 0.10 M in C₆H₅NH₂?
    1. [latex]\left[{\text{OH}}^{-}\right]=\left[{\text{NH}}_{4}^{+}\right][/latex]
    2. [latex]\left[{\text{NH}}_{4}^{+}\right]=\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right][/latex]
    3. [latex]\left[{\text{OH}}^{-}\right]=\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right][/latex]
    4. [NH₃] = [C₆H₅NH₂]
    5. both a and b are correct

[reveal-answer q=”789235″]Show Selected Solutions[/reveal-answer]
[hidden-answer a=”789235″]

1. The equilibrium is [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_{4}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]
   1. The addition of NaOH adds OH⁻ to the system and, according to Le Châtelier’s principle, the equilibrium will shift to the left. Thus, the percent of converted NH₃ will decrease.
   2. The addition of HCl will add [latex]{\text{H}}_{3}{\text{O}}^{+}[/latex] ions, which will then react with the OH⁻ ions. Thus, the equilibrium will shift to the right, and the percent will increase.
   3. The addition of NH₄Cl adds [latex]{\text{NH}}_{4}^{+}[/latex] ions, shifting the equilibrium to the left. Thus, the percent will decrease.
3. The effects are as follows:
   1. Adding HCl will add [latex]{\text{H}}_{3}{\text{O}}^{+}[/latex] ions, which will then react with the OH⁻ ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO₂, and decreasing the concentration of [latex]{\text{NO}}_{2}^{-}[/latex] ions.
   2. Adding HNO₂ increases the concentration of HNO₂ and shifts the equilibrium to the left, increasing the concentration of [latex]{\text{NO}}_{2}^{-}[/latex] ions and decreasing the concentration of OH⁻ ions.
   3. Adding NaOH adds OH⁻ ions, which shifts the equilibrium to the left, increasing the concentration of [latex]{\text{NO}}_{2}^{-}[/latex] ions and decreasing the concentrations of HNO₂.
   4. Adding NaCl has no effect on the concentrations of the ions.
   5. Adding KNO₂ adds [latex]{\text{NO}}_{2}^{-}[/latex] ions and shifts the equilibrium to the right, increasing the HNO₂ and OH⁻ ion concentrations.
5. The equations of the occurring chemical processes are:
6. [latex]\text{HCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)[/latex]
7. [latex]{\text{CH}}_{3}\text{COOH}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CH}}_{3}{\text{COO}}^{-}\left(aq\right)[/latex]
8. This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO₂H exists primarily as HCO₂H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO₂H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] produced by the stronger acid.
7. K_a and K_b are as follows:
   1. The reaction is [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{OH}}^{-}\left(aq\right)+{\text{NH}}_{4}^{+}\left(aq\right)[/latex]
      [latex]{K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(3.1\times {10}^{-3}\right)\left(3.1\times {10}^{-3}\right)}{\left(0.533\right)}=1.8\times {10}^{-5}[/latex]
   2. The reaction is [latex]{\text{HNO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{NO}}_{2}^{-}\left(aq\right)[/latex]
      [latex]{K}_{\text{a}}=\frac{\left[{\text{NO}}_{2}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{HNO}}_{2}\right]}=\frac{\left(0.0438\right)\left(0.011\right)}{\left(1.07\right)}=4.5\times {10}^{-4}[/latex]
   3. The reaction is [latex]{\left({\text{CH}}_{3}\right)}_{3}\text{N}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{OH}}^{-}\left(aq\right)+{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\left(aq\right)[/latex]
      [latex]{K}_{\text{b}}=\frac{\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{CH}}_{3}\right)}_{3}\text{N}\right]}=\frac{\left(4.3\times {10}^{-3}\right)\left(4.3\times {10}^{-3}\right)}{\left(0.25\right)}=7.4\times {10}^{-5}[/latex]
   4. The reaction is [latex]{\text{NH}}_{4}^{+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{NH}}_{3}\left(aq\right)[/latex]
      [latex]{K}_{\text{a}}=\frac{\left[{\text{NH}}_{3}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{NH}}_{4}^{+}\right]}=\frac{\left(7.5\times {10}^{-6}\right)\left(7.5\times {10}^{-6}\right)}{\left(0.100\right)}=5.6\times {10}^{-10}[/latex]
9. The reaction is [latex]{\text{HSO}}_{4}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{SO}}_{4}^{2-}\left(aq\right)[/latex].
10. The concentrations at equilibrium are:
    • [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] = [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = (0.29)(0.10 M) = 0.029 M
    • [latex]\left[{\text{HSO}}_{4}^{-}\right][/latex] = 0.10 M − 0.029 M = 0.071 M
    • [latex]{K}_{\text{a}}=\frac{\left[{\text{SO}}_{4}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{HSO}}_{4}^{-}\right]}=\frac{\left(0.029\right)\left(0.029\right)}{\left(0.071\right)}=1.2\times {10}^{-2}[/latex]
11. The ionization constants are as follows:
    1. [latex]{K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{\left(1.00\times {10}^{-14}\right)}{2.3\times {10}^{-3}}=4.3\times {10}^{-12}[/latex]
    2. [latex]{K}_{\text{a}}=\frac{{K}_{\text{w}}}{{K}_{\text{b}}}=\frac{\left(1.00\times {10}^{-14}\right)}{7.4\times {10}^{-5}}=1.4\times {10}^{-10}[/latex]
    3. [latex]{K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{\left(1.00\times {10}^{-14}\right)}{1\times {10}^{-7}}=1\times {10}^{-7}[/latex]
    4. [latex]{K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{\left(1.00\times {10}^{-14}\right)}{2.4\times {10}^{-12}}=4.2\times {10}^{-3}[/latex]
    5. [latex]{K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{\left(1.00\times {10}^{-14}\right)}{2.4\times {10}^{-12}}=4.2\times {10}^{-3}[/latex]
    6. [latex]{K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{\left(1.00\times {10}^{-14}\right)}{1.2\times {10}^{-2}}=8.3\times {10}^{-13}[/latex]
13. The reactions are:
14. [latex]\begin{array}{l}{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_{4}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\end{array}[/latex]
15. Because NH₃ is much stronger than C₆H₅NH₂, it dissociates more. As the initial concentrations of both bases are the same, at equilibrium, [NH₃] < [C₆H₅NH₂], [latex]\left[{\text{NH}}_{4}^{+}\right][/latex] > [latex]\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right][/latex], and [OH⁻] > [latex]\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right][/latex]. Therefore, (a) is the correct statement. The contribution to the total [OH⁻] at equilibrium from C₆H₅NH₂ is negligible compared to HN₃. Therefore [latex]\left[{\text{OH}}^{-}\right]=\left[{\text{NH}}_{4}^{+}\right][/latex].

[/hidden-answer]