Solution Found!
Succinic acid 1H2C4H6O42, which we will denote H2Suc, is a
Chapter , Problem 16.108(choose chapter or problem)
Succinic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{4}\right)\), which we will denote \(\mathrm{H}_{2} \mathrm{Suc}\), is a biologically relevant diprotic acid with the structure shown below. It is closely related to tartaric acid and malic acid (Figure 16.1). At \(25^{\circ} \mathrm{C}\), the acid-dissociation constants for succinic acid are \(K_{a 1}=6.9 \times 10^{-5}\) and \(K_{a 2}=2.5 \times 10^{-6}\) .
(a) Determine the pH of a 0.32 M solution of \(\mathrm{H}_{2} \mathrm{Suc}\) at \(25^{\circ} \mathrm{C}\), assuming that only the first dissociation is relevant.
(b) Determine the molar concentration of \(\mathrm{Suc}^{2-}\) in the solution in part (a).
(c) Is the assumption you made in part (a) justified by the result from part (b)?
(d) Will a solution of the salt NaHSuc be acidic, neutral, or basic?
Questions & Answers
QUESTION:
Succinic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{4}\right)\), which we will denote \(\mathrm{H}_{2} \mathrm{Suc}\), is a biologically relevant diprotic acid with the structure shown below. It is closely related to tartaric acid and malic acid (Figure 16.1). At \(25^{\circ} \mathrm{C}\), the acid-dissociation constants for succinic acid are \(K_{a 1}=6.9 \times 10^{-5}\) and \(K_{a 2}=2.5 \times 10^{-6}\) .
(a) Determine the pH of a 0.32 M solution of \(\mathrm{H}_{2} \mathrm{Suc}\) at \(25^{\circ} \mathrm{C}\), assuming that only the first dissociation is relevant.
(b) Determine the molar concentration of \(\mathrm{Suc}^{2-}\) in the solution in part (a).
(c) Is the assumption you made in part (a) justified by the result from part (b)?
(d) Will a solution of the salt NaHSuc be acidic, neutral, or basic?
ANSWER:Step 1 of 4
(a)The diprotic succinic acid is dissociated into its respective ions represented through an equilibrium reaction as follows:
\({H_2}Suc\left( {aq} \right) \leftrightharpoons {H^ + } + HSu{c^ - }(aq)\;{K_{a1}} = 6.9 \times {10^{ - 5}}\)
At the equilibrium \(\left[ {{H^ + }} \right] = \left[ {HSu{c^ - }} \right] = x\;M\)
The value x is calculated as follows:
\({K_{a1}} = \frac{{\left[ {{H^ + }} \right]\left[ {Hsu{c^ - }} \right]}}{{\left[ {{H_2}Suc} \right]}}\)
\(6.9 \times {10^{ - 5}} = \frac{{{x^2}}}{{0.32\;M}}\)
\({x^2} = 0.32\;M \times 6.9 \times {10^{ - 5}}\)
\({x^2} = 2.208 \times {10^{ - 5}}M\)
\(x = 4.6 \times {10^{ - 2}}\;M\)
The pH of the solution is calculated below:
\(pH = - \log \left[ {{H^ + }} \right]\)
\(pH= - \log \left[ {0.00469} \right]\)
\(pH= 2.33\)
Thus, the pH of the solution is 2.33.