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Chapter 14: Q83 E (page 834)

Calculate the concentration of each species present in a \(0.010M\) solution of phthalic acid, \({C_6}{H_4}{\left( {C{O_2}H} \right)_2}\).

\(\begin{array}{*{20}{c}}{{C_6}{H_4}{{\left( {C{O_2}H} \right)}_2}(aq) + {H_2}O(l) \rightleftharpoons {H_3}{O^ + }(aq) + {C_6}{H_4}\left( {C{O_2}H} \right){{\left( {C{O_2}} \right)}^ - }(aq)}&{{K_a} = 1.1 \times 1{0^{ - 3}}} \\ {{C_6}{H_4}\left( {C{O_2}H} \right)\left( {C{O_2}} \right)(aq) + {H_2}O(l) \rightleftharpoons {H_3}{O^ + }(aq) + {C_6}{H_4}{{\left( {C{O_2}} \right)}_2}^{2 - }(aq)}&{{K_a} = 3.9 \times 1{0^{ - 6}}} \end{array}\)

Short Answer

Expert verified

\(\begin{align}{}\left( {{C_6}{H_4}\left( {C{O_2}H} \right)\left( {C{O_2}H} \right)} \right) &= 0.010 M.\\\left( {{C_6}{H_4}\left( {C{O_2}H} \right){{\left( {C{O_2}} \right)}^ - }} \right) &= 3.3 \times 1{0^{ - 3}} M.\\\;\;\;\;\;\;\;\;\;\;\left( {{C_6}{H_4}\left( {C{O_2}} \right)_2^{2 - }} \right) &= 1.1 \times 1{0^{ - 4}} M.\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{H_3}{O^ + }} \right) &= 3.4 \times 1{0^{ - 3}} M.\end{align}\)

Step by step solution

01

Calculation of the concentration

In the initial time, we can make two assumptions. The first is that the ions created have similar concentrations, and the second is that the concentration change of the acid molecule is minimal due to the low\({{\bf{K}}_{\bf{1}}}\)value.

\(\begin{align}{K_1} &= \frac{{{x^2}}}{{0.010}}\\1.1 \times 1{0^{ - 3}} &= \frac{{{x^2}}}{{0.010}}\\ x &= \left( {{H_3}{O^ + }} \right)\\ &= \left( {{C_6}{H_4}\left( {C{O_2}H} \right){{\left( {C{O_2}} \right)}^ - }} \right)\\ &= 3.3 \times 1{0^{ - 3}} M.\end{align}\)

We can compute the concentrations of the products of the second ionization reaction now that we know the concentration of the \({C_6}{H_4}\left( {C{O_2}H} \right){\left( {C{O_2}} \right)^ - }\) ion. We assume that the change in the concentration of the \({C_6}{H_4}\left( {C{O_2}H} \right){\left( {C{O_2}} \right)^ - }\) ion is small and that the concentrations of the ions generated are equal once again.

\(\begin{align}{K_2} &= \frac{{{x^2}}}{{3.3 \times 1{0^{ - 3}}}}\\3.9 \times 1{0^{ - 6}} &= \frac{{{x^2}}}{{3.3 \times 1{0^{ - 3}}}}\\ x &= \left( {{H_3}{O^ + }} \right)\\ &= \left( {{C_6}{H_4}\left( {C{O_2}} \right)_2^{2 - }} \right)\\ &= 1.1 \times 1{0^{ - 4}} M.\end{align}\)

To compute the overall concentration of the hydronium ions, add the ones created in the first ionization reaction to the ones formed in the second.

\(\begin{align}{\left( {{H_3}{O^ + }} \right)_{total }} &= \left( {3.3 \times 1{0^{ - 3}} + 1.1 \times 1{0^{ - 4}}} \right) M\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= 3.4 \times 1{0^{ - 3}} M.\end{align}\)

The phthalic acid molecule has a concentration of \(0.010M,\) the hydronium ions have a concentration of \(3.4 \times 1{0^{ - 3}}M\), the \({C_6}{H_4}\left( {C{O_2}H} \right){\left( {C{O_2}} \right)^ - }\) ion has a concentration of \(3.3 \times 1{0^{ - 3}}M\), and the \({C_6}{H_4}\left( {C{O_2}} \right)_2^{2 - }\) ion has a concentration of \(1.1 \times 1{0^{ - 4}}M\).

02

 Step 2: Results

\(\begin{align}{}\left( {{C_6}{H_4}\left( {C{O_2}H} \right)\left( {C{O_2}H} \right)} \right) &= 0.010 M.\\\left( {{C_6}{H_4}\left( {C{O_2}H} \right){{\left( {C{O_2}} \right)}^ - }} \right) &= 3.3 \times 1{0^{ - 3}} M.\\\;\;\;\;\;\;\;\;\;\;\left( {{C_6}{H_4}\left( {C{O_2}} \right)_2^{2 - }} \right) &= 1.1 \times 1{0^{ - 4}} M.\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{H_3}{O^ + }} \right) &= 3.4 \times 1{0^{ - 3}} M.\end{align}\)

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Most popular questions from this chapter

Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?

(a) Addition of \(NaOH\)

(b) Addition of \(HCl\)

(c) Addition of \(NaF\)

Identify and label the Br酶nsted-Lowry acid, its conjugate base, the Br酶nsted-Lowry base, and its conjugate acid in each of the following equations:

\({\rm{\;(a)\;NO}}_2^ - + {{\rm{H}}_2}{\rm{O}} \to {\rm{HN}}{{\rm{O}}_2} + {\rm{O}}{{\rm{H}}^ - }\).

\({\rm{\;(b)\;HBr}} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{B}}{{\rm{r}}^ - }\)

\({\rm{\;(c)\;H}}{{\rm{S}}^ - } + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_2}{\rm{S}} + {\rm{O}}{{\rm{H}}^ - }\)

\({\rm{\;(d)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{O}}{{\rm{H}}^ - } \to {\rm{HP}}{{\rm{O}}_4}^{2 - } + {{\rm{H}}_2}{\rm{O}}\)

\({\rm{\;(e)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{HCl}} \to {{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} + {\rm{C}}{{\rm{l}}^ - }\)

\({\rm{\;(f)\;}}{\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} \to {\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }}\)

\({\rm{\;(g)\;C}}{{\rm{H}}_3}{\rm{OH}} + {{\rm{H}}^ - } \to {\rm{C}}{{\rm{H}}_3}{{\rm{O}}^ - } + {{\rm{H}}_2}\)

Calculate the concentration of each species present in a \(0.050 - M\) solution of \({H_2}\;S. \)

Explain why the ionization constant, \({K_a}\), for\({H_2}S{O_4}\) is larger than the ionization constant for \({H_2}S{O_3}\).

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?
See all solutions

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