2. 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:

[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]

4. The oxidation state of the sulfur in H₂SO₄ is greater than the oxidation state of the sulfur in H₂SO₃.

6. K_w = K_a × K_b; thus,

[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]

Question 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:

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

Question 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:

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

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:

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]

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:

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]

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:

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]

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:

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]

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:

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]

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]

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

[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:

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

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]

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]

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]

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]

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

1. 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.
2. Assume we can neglect the contribution of water to the equilibrium concentration of [latex]{\text{H}}_{3}{\text{O}}^{+}[/latex].

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:

[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]

[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]

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].

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:

[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]

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].