Selected Answers

1. [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] and [latex]\left[{\text{HCO}}_{3}{}^{\text{-}}\right][/latex] are equal in a 0.134-M solution of H₂CO₃. K_a of H₂CO₃ is significantly larger than K_a for [latex]{\text{HCO}}_{3}{}^{\text{-}}[/latex]. Therefore, very little of [latex]{\text{HCO}}_{3}{}^{\text{-}}[/latex] ionizes to give hydronium ions and [latex]{\text{CO}}_{3}{}^{2-}[/latex] ions, and the concentrations of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] and [latex]{\text{HCO}}_{3}{}^{\text{-}}[/latex] are practically equal in an aqueous solution of H₂CO₃.

3. [latex]\begin{array}{l}{\text{C}}_{6}{\text{H}}_{4}{\left({\text{CO}}_{2}\text{H}\right)}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{4}\left({\text{CO}}_{2}\text{H}\right){\left({\text{CO}}_{2}\right)}^{\text{-}}\left(aq\right){K}_{\text{a}}=1.1\times {10}^{-3}\\ {\text{C}}_{6}{\text{H}}_{4}\left({\text{CO}}_{2}\text{H}\right)\left({\text{CO}}_{2}\right)\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{4}{\left({\text{CO}}_{2}\right)}_{2}{}^{2-}\left(aq\right){K}_{\text{a}}=3.9\times {10}^{-6}\end{array}[/latex]

[C₆H₄(CO₂H)₂] 7.2 [latex]\times[/latex] 10⁻³M, [C₆H₄(CO₂H)(CO₂)⁻] = [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right][/latex] 2.8 [latex]\times[/latex] 10⁻³M, [latex]\left[{\text{C}}_{6}{\text{H}}_{4}{\left({\text{CO}}_{2}\right)}_{2}{}^{2-}\right][/latex] 3.9 [latex]\times[/latex] 10⁻⁶M, [OH⁻] 3.6 [latex]\times[/latex] 10⁻¹²M

5.  (a) as an acid,

[latex]\begin{array}{l}\\ {\text{HTe}}^{\text{-}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{Te}}^{2-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)\\ {K}_{\text{a2}}=\frac{\left[{\text{Te}}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{HTe}}^{\text{-}}\right]}=1\times {10}^{-5}\end{array}[/latex];

(b) as a base,

[latex]\begin{array}{l}\\ {\text{HTe}}^{\text{-}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}\text{Te}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)\\ \\ {K}_{\text{b}}=\frac{\left[{\text{H}}_{2}\text{Te}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{HTe}}^{\text{-}}\right]}=\frac{{K}_{\text{w}}}{{K}_{\text{a1}}}=\frac{1.0\times {10}^{-14}}{2.3\times {10}^{-3}}=4.3\times {10}^{-12}\end{array}[/latex];

(c) The reactions and ionization constants are:

[latex]\begin{array}{l}\\ {\text{H}}_{2}\text{Te}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{HTe}}^{\text{-}}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right){K}_{\text{a1}}=2.3\times {10}^{-3}\\ {\text{HTe}}^{\text{-}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{Te}}^{2-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right){K}_{\text{a2}}=1\times {10}^{-5}\end{array}[/latex]

As a general rule, if the first ionization constant is larger than the second by a factor of at least 20, then the second ionization can be neglected. Since [latex]{K}_{\text{a1}}[/latex] is 230-times larger than [latex]{K}_{\text{a2}}[/latex], the assumption should hold true for HTe⁻. To test the assumptions, find [HTe⁻] from the first reaction. The equilibrium expression for this reaction is [latex]{K}_{\text{a1}}=\frac{\left[{\text{HTe}}^{\text{-}}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{2}\text{Te}\right]}=2.3\text{\times }{10}^{-3}[/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 (0.10 − x) ≈ 0.10, gives [latex]\frac{\left[{\text{HTe}}^{\text{-}}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{2}\text{Te}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.10}=2.3\times {10}^{-3}[/latex].

Solving for x gives 0.0152 M. Because this value is 15% of 0.10 M, our assumption is incorrect. Therefore, use the quadratic formula. Using the data gives the quadratic equation:

x² + 2.3 [latex]\times[/latex] 10⁻³x − 2.3 [latex]\times[/latex] 10⁻⁴ = 0

Using the quadratic formula gives (a = 1, b = 2.3 [latex]\times[/latex] 10⁻³, and c = −2.3 [latex]\times[/latex] 10⁻⁴)

[latex]\begin{array}{ll}\hfill x& =\frac{-b\pm \sqrt{{b}^{\text{2+}}-4ac}}{2a}=\frac{-\left(2.3\times {10}^{-3}\right)\pm \sqrt{{\left(2.3\times {10}^{-3}\right)}^{\text{2+}}-4\left(1\right)\left(\text{-2.3}\times {10}^{-4}\right)}}{2\left(1\right)}\hfill \\ & =\frac{-\left(2.3\times {10}^{-3}\right)\pm \left(0.0304\right)}{2}=0.0141M\text{(positive root)}\hfill \end{array}[/latex]

Thus [HTe⁻] = 0.014 M. For the second ionization, [latex]{K}_{\text{a2}}=\frac{\left[{\text{Te}}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{HTe}}^{\text{-}}\right]}=1\text{\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 assumptions that (0.0140 − x) ≈ and (0.0141 + x) ≈ 0.0141, gives:

[latex]\frac{\left[{\text{Te}}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[{\text{HTe}}^{\text{-}}\right]}=\frac{\left(x\right)\left(0.0141+x\right)}{\left(0.0141-x\right)}\approx \frac{\left(x\right)\left(0.0141\right)}{0.0141}=1\text{\times }{10}^{-5}[/latex]

Solving for x gives 1 [latex]\times[/latex] 10⁻⁵M. Therefore, compared with 0.014 M, this value is negligible (0.071%).