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Chapter 22: Problem 53

Calculate \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of chloroacetic acid \(\left(K_{\mathrm{a}}=1.5 \times 10^{-3}\right)\)

Short Answer

Expert verified
Answer: The concentration of hydrogen ions (\(\mathrm{H}^+\)) is approximately \(0.012 \mathrm{M}\), and the pH of the 0.10 M chloroacetic acid solution is approximately 1.92.

Step by step solution

01

Write the dissociation reaction and equilibrium expression

Chloroacetic acid (HA) dissociates in water to form the hydrogen ion (H+) and the chloride ion (A-): $$HA \rightleftharpoons H^+ + A^-$$ The equilibrium expression for the dissociation reaction is given by: $$K_a = \frac{[H^+][A^-]}{[HA]}$$
02

Create an equilibrium table

Now, we will create an equilibrium table to find the concentrations of all species (HA, H+, and A-) present in the solution at equilibrium: | | Initial | Change | Equilibrium | |-------------|---------|--------|-------------| | HA | 0.10 M | -x | 0.10-x M | | H+ | 0 | +x | x M | | A- | 0 | +x | x M | Where x is the change in concentration of hydrogen ions at equilibrium.
03

Solve for the concentration of H+ ions using equilibrium expression

Substitute the concentrations of the species into the equilibrium expression and solve for x: $$K_a = \frac{[H^+][A^-]}{[HA]}$$ $$1.5 \times 10^{-3} = \frac{x \cdot x}{0.10-x}$$ This expression can be solved either by the quadratic equation or by making an assumption that the change in concentration of chloroacetic acid is negligible compared to its initial concentration. Here we will make that assumption because the given Ka value is relatively small: $$1.5 \times 10^{-3} \approx \frac{x^2}{0.10}$$ $$x = \sqrt{1.5 \times 10^{-3} \cdot 0.10}$$ $$x \approx 0.012$$ Thus, the equilibrium concentration of hydrogen ions [H+] is approximately 0.012 M.
04

Calculate the pH value

Now that we have the concentration of hydrogen ions, we can easily calculate the pH of the solution using the following formula: $$\mathrm{pH} = -\log_{10}([H^+])$$ Plug in the value of the H+ concentration, and we find: $$\mathrm{pH} = -\log_{10}(0.012)$$ $$\mathrm{pH} \approx 1.92$$ So, the hydrogen ion concentration (\(\mathrm{H}^+\)) is approximately \(0.012 \mathrm{M}\), and the pH of the 0.10 M chloroacetic acid solution is approximately 1.92.

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