Calculate the 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 \(0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 M\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

To determine the pH for \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\), we can first write down the acid dissociation equation: \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} \leftrightharpoons \mathrm{H}^{+}+\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}\) Let the concentration of dissociated ions be \(x\). We can write the equilibrium concentrations in terms of the initial concentration: \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}: 0.100-x\) \(\mathrm{H}^{+}: x\) \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}: x\) Now, we can write the \(K_\mathrm{a}\) expression: \(K_\mathrm{a}=\dfrac{[\mathrm{H}^{+}][\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}]}{[\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}]}=\dfrac{x^2}{0.100-x}\) We can approximate \(0.100-x \approx 0.100\), because \(K_\mathrm{a}\) is very small (weak acid). Therefore, \(x^2 = K_\mathrm{a} \times 0.100\) \(x = \sqrt{1.3 \times 10^{-5} \times 0.100}=1.14×10^{-3}\) Finally, we calculate the pH: \(pH=-\log[\mathrm{H}^{+}]=- \log(1.14×10^{-3})=2.94\)

Sodium propanoate is a salt of a weak acid and a strong base, which will undergo hydrolysis and cause an increase in pH. Since the sodium ion is a spectator ion, we focus on the equilibrium of the propanoate ion, \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}\), in the water, which contributes \(\mathrm{OH}^-\) ions: \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-} + \mathrm{H}_{2} \mathrm{O} \leftrightharpoons \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} + \mathrm{OH}^{-}\) We can find the \(K_\mathrm{b}\) value for the propanoate ion by using the relationship between \(K_\mathrm{a}\), \(K_\mathrm{b}\) and \(K_\mathrm{w}\): \(K_\mathrm{b} = \dfrac{K_\mathrm{w}}{K_\mathrm{a}}\) Let \(x\) be the concentration of propanoate ions that underwent hydrolysis forming hydroxide ions. At equilibrium, the concentrations are: \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}: 0.100-x \) \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}: x\) \(\mathrm{OH}^{-}: x \) Now, we can write the \(K_\mathrm{b}\) expression: \(K_\mathrm{b}=\dfrac{[\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}]}\) Using the equilibrium concentrations and calculating \(K_\mathrm{b}\) from \(K_\mathrm{a}\) and \(K_\mathrm{w}\), we can solve for \(x\) to get the hydroxide ion concentration. Then, we can calculate the \(pOH\) and \(pH\) values accordingly.

Pure water acts as both an acid and a base and undergoes auto-ionization, resulting in very low concentrations of both hydroxide and hydronium ions. \(\mathrm{H}_{2} \mathrm{O} \leftrightharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}\) At 25 °C, the \(K_\mathrm{w}\) value is \(1.0 \times 10^{-14}\). Since in pure water the concentrations of hydronium and hydroxide ions are equal, we can write: \(K_\mathrm{w}=[\mathrm{H}^{+}]^2\) \(\mathrm{pH}=-\log\left( [\mathrm{H}^{+}] \right)\) We can substitute the \(K_\mathrm{w}\) value into the equations to find the pH of pure water, which should be close to 7.

We now need to analyze a mixture containing both the weak acid and its conjugate base, \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\). This is a buffer system. The buffer system's pH can be found using the Henderson-Hasselbalch equation: \(pH = pK_\mathrm{a} + \log \left(\dfrac{[\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}]}\right)\) We are given the initial concentrations of both species, so we can simply substitute them into the equation to compute the mixture's pH.