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

Given a solution of 0.100 M propanoic acid (HC3H5O2) and its dissociation constant, Ka = 1.3 × 10^(-5). This is a weak acid, and we can set up an equilibrium expression to find its ionization degree: \[HC_3H_5O_2 \rightleftharpoons H^+ + C_3H_5O_2^-\] To find the ionization degree (x), we can write the expression for Ka: \[Ka = \frac{[H^+][C_3H_5O_2^-]}{[HC_3H_5O_2]}\] Since the initial concentration of propanoic acid is 0.100 M and assuming the ionization degree is small, we can write: \[Ka = \frac{x^2}{0.100 - x}\] Now, we'll solve for x, which is also the [H+] concentration: \[x^2 = Ka(0.100 - x) = 1.3 \times 10^{-5}(0.100)\] \[x = \sqrt{1.3 \times 10^{-6}}\] Since x is very small compared to 0.100, x ≈ 1.14 × 10^(-3) M Now we can calculate the pH: \[pH = -\log[H^+]\] \[pH = -\log(1.14 \times 10^{-3}) ≈ 2.94\]

Since sodium propanoate (NaC3H5O2) completely dissociates in water, the solution will be 0.100 M in C3H5O2^- ions. We can use the Kb value to determine the pH: \[Kb = \frac{Kw}{Ka}\] Where Kw is the ion product of water (1.0 × 10^(-14)). \[Kb = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-5}} = 7.69 \times 10^{-10}\] Now, we can write the equilibrium expression of sodium propanoate's conjugate base (C3H5O2^-): \[C_3H_5O_2^- + H_2O \rightleftharpoons HC_3H_5O_2 + OH^-\] Using Kb, we can write the ionization expression as: \[Kb = \frac{[HC_3H_5O_2][OH^-]}{[C_3H_5O_2^-]}\] We have the initial concentration of C3H5O2^- (0.100 M), and we'll assume that the hydroxide concentration is x: \[Kb = \frac{x^2}{0.100 - x} = 7.69 \times 10^{-10}\] \[x = \sqrt{7.69 \times 10^{-11}} ≈ 2.77 \times 10^{-6}\] Now, we can calculate the pOH: \[pOH = -\log[OH^-]\] \[pOH = -\log(2.77 \times 10^{-6}) ≈ 5.56\] Finally, we can find the pH: \[pH = 14 - pOH = 14 - 5.56 ≈ 8.44\]

For pure water, the concentration of H+ and OH- ions is equal and can be obtained using Kw: \[Kw = [H^+][OH^-]\] \[1.0 \times 10^{-14} = [H^+][H^+]\] \[H^+ = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\] The pH of pure water is: \[pH = -\log[H^+] = -\log(1.0 \times 10^{-7}) = 7\]

For this step, we can use the Henderson-Hasselbalch equation: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] \[pH = -\log(1.3 \times 10^{-5}) + \log \frac{0.100}{0.100}\] \[pH = 4.89 + \log(1) = 4.89\] In conclusion, the pH values for each solution are: a. Propanoic acid: pH = 2.94 b. Sodium propanoate: pH = 8.44 c. Pure water: pH = 7.00 d. Mixture of propanoic acid and sodium propanoate: pH = 4.89