Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

The ionization reaction of propanoic acid can be represented as: \[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2 (\mathrm{aq}) \rightleftharpoons \mathrm{H^+} (\mathrm{aq}) + \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- (\mathrm{aq})\]

Using the equilibrium constant, \(K_\mathrm{a}\), we can set up the expression as: \[K_\mathrm{a} = \frac{[\mathrm{H^+}] [\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]}{[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2]}\]

Let \(x\) be the concentration of \(\mathrm{H^+}\) ions in the solution. At equilibrium, the concentrations will be: \[ [\mathrm{H^+}] = x \\ {[\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]} = x \\ [\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2] = 0.100 - x \] Now, substitute these concentrations into the expression for \(K_\mathrm{a}\): \[1.3 \times 10^{-5} = \frac{x^2}{0.100 - x}\] Since \(K_\mathrm{a}\) is very small (\(1.3 \times 10^{-5}\)), the amount of ionization is very small, so \(x\) is much smaller than \(0.100\), and we can approximate \(0.100 - x \approx 0.100\). Now, we can solve for x: \(x^2 = 1.3 \times 10^{-5} \times 0.100 \\ x = \sqrt{1.3 \times 10^{-6}} \\ x \approx 1.14 \times 10^{-3}\) Therefore, the concentration of \(\mathrm{H^+}\) ions in the solution is approximately \(1.14 \times 10^{-3} M\).

Now that we have the concentration of \(\mathrm{H^+}\) ions, we can calculate the pH using the formula: \[pH = -\log [\mathrm{H^+}]\] Substitute the value of \([\mathrm{H^+}]\) into the formula: \(pH = -\log(1.14 \times 10^{-3}) \\ pH \approx 2.94\) The pH of the \(0.100 M\) propanoic acid solution is approximately 2.94. #b. pH of \(0.100 M\) sodium propanoate.#

The reaction of sodium propanoate in water can be represented as: \[\mathrm{NaC}_3\mathrm{H}_5\mathrm{O}_2 (\mathrm{aq}) \rightleftharpoons \mathrm{OH^-} (\mathrm{aq}) + \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- (\mathrm{aq})\]

Since sodium propanoate is a strong electrolyte, it will completely dissociate in water. Therefore, the concentration of \(\mathrm{OH^-}\) ions in the solution will be the same as the concentration of sodium propanoate, which is \(0.100 M\).

Now that we have the concentration of \(\mathrm{OH^-}\) ions in the solution, we can calculate the pOH of the solution using the formula: \[pOH = -\log [\mathrm{OH^-}]\] \[pOH = -\log (0.100) \\ pOH = 1\] The pOH of the \(0.100 M\) sodium propanoate solution is 1.

Now, we can determine the pH by using the relation between pH and pOH: \[pH = 14 - pOH\] \[pH = 14 - 1 \\ pH = 13\] The pH of the \(0.100 M\) sodium propanoate solution is 13. #c. pH of pure water. Since pure water is neutral, it has equal concentrations of \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) ions, both at a concentration of \(10^{-7} M\). The pH of a neutral aqueous solution like pure water is 7. #d. pH of a mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\).

The ionization reaction for the mixture can be represented as: \[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2 (\mathrm{aq}) + \mathrm{OH^-} (\mathrm{aq}) \rightleftharpoons \mathrm{H_2O} (\mathrm{l}) + \mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^- (\mathrm{aq})\]

To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation: \[pH = pK_\mathrm{a} + \log \frac{[\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]}{[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2]}\] \(pK_\mathrm{a} = -\log K_\mathrm{a} \\ pK_\mathrm{a} = -\log (1.3 \times 10^{-5}) \\ pK_\mathrm{a} \approx 4.89\) Since the concentrations of both the acid and the conjugate base are equal (\(0.100 M\)), the ratio in the equation becomes: \(\frac{[\mathrm{C}_3\mathrm{H}_5\mathrm{O}_2^-]}{[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_2]} = \frac{0.100}{0.100} = 1\) Now, we can use the values in the Henderson-Hasselbalch equation: \[pH = 4.89 + \log 1 \\ pH = 4.89\] The pH of the mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5}\mathrm{O}_{2}\) and \(0.100 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5}\mathrm{O}_{2}\) is 4.89.