Question

Gallic acid, \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{5},\) is an ingredient in some ointments used to treat psoriasis. Its \(K_{\mathrm{a}}\) is \(3.9 \times 10^{-5} .\) For a \(0.168 \mathrm{M}\) solution of gallic acid, calculate (a) \(\left[\mathrm{H}^{+}\right]\) (b) \(\left[\mathrm{OH}^{-}\right]\) (c) \(\mathrm{pH}\) (d) \% ionization

Short answer

Answer: The concentration of H+ ions is approximately \(2.03 \times 10^{-3} \mathrm{M}\), the concentration of OH- ions is approximately \(4.93 \times 10^{-12} \mathrm{M}\), the pH is approximately 2.69, and the percentage ionization is approximately 1.21 %.

Step-by-step solution

Write the ionization equation and Ka expression for Gallic acid

Gallic acid ionizes in water as follows: \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{5}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{5}^{-}(aq)\) The Ka expression for this ionization is: \(K_{a}=\frac{[\mathrm{H}^{+}][\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{5}^{-}]}{[\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{5}]}\)

Write the expressions for H+ concentration and percentage ionization

Let's assume x mol of HC7H5O5 ionizes. So, the final concentrations are: [\(\mathrm{H}^{+}\)] = x [\(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{5}^{-}\)] = x [\(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{5}\)] = 0.168 - x According to the problem, Ka = \(3.9 \times 10^{-5}\) Percentage ionization = \(\frac{x}{0.168}\) × 100

Solve for H+ concentration

We can now substitute the expressions in the Ka equation: \(3.9 \times 10^{-5} = \frac{x^{2}}{0.168-x}\) Since Ka is small, we can assume x << 0.168. So, we have: \(3.9 \times 10^{-5} = \frac{x^{2}}{0.168}\) Solve for x: \(x = \sqrt{(3.9 \times 10^{-5}) × 0.168} \approx 2.03 \times 10^{-3}\) Therefore, the concentration of H+ ions is approximately \(2.03 \times 10^{-3} \mathrm{M}\).

Calculate OH- concentration

Using the relationship between H+ and OH- concentrations: \(K_{w} = [\mathrm{H}^{+}] [\mathrm{OH}^{-}]\) where \(K_w = 1 \times 10^{-14}\), the ion product of water Solving for OH- concentration, we have: \([\mathrm{OH}^{-}] = \frac{1 \times 10^{-14}}{2.03 \times 10^{-3}} \approx 4.93 \times 10^{-12} \mathrm{M}\)

Calculate pH

pH is defined as: \(pH = -\log ([\mathrm{H}^{+}])\) \(pH = -\log (2.03 \times 10^{-3}) \approx 2.69\)

Calculate the percentage ionization

We can now find the percentage ionization using the formula defined in Step 2: Percentage ionization = \(\frac{2.03 \times 10^{-3}}{0.168}\) × 100 ≈ 1.21 % The answers to the exercise are as follows: (a) [\(\mathrm{H}^{+}\)] ≈ \(2.03 \times 10^{-3} \mathrm{M}\) (b) [\(\mathrm{OH}^{-}\)] ≈ \(4.93 \times 10^{-12} \mathrm{M}\) (c) pH ≈ 2.69 (d) \% ionization ≈ 1.21 %