QuestionJune 15, 2025

16.56 If solution of HF (K_(a)=6.8times 10^-4) has a pH of 3.65, cal- culate the concentration of hydrofluoric acid.

16.56 If solution of HF (K_(a)=6.8times 10^-4) has a pH of 3.65, cal- culate the concentration of hydrofluoric acid.
16.56 If solution of HF (K_(a)=6.8times 10^-4) has a pH of 3.65, cal- culate the concentration of hydrofluoric acid.

Solution
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Answer

0.074 M Explanation 1. Calculate [H⁺] from pH Use the formula **[H^+] = 10^{-\text{pH}}**. For pH = 3.65, [H^+] = 10^{-3.65} \approx 2.24 \times 10^{-4} M. 2. Set up the equilibrium expression HF dissociates as HF \rightleftharpoons H^+ + F^-. The equilibrium expression is **K_a = \frac H^+][F^- HF **. 3. Substitute known values into the equilibrium expression Assume initial concentration of HF is C and at equilibrium, [H^+] = [F^-] = 2.24 \times 10^{-4} M. Then, K_a = \frac{(2.24 \times 10^{-4})^2}{C - 2.24 \times 10^{-4}}. 4. Solve for C Rearrange to find C: 6.8 \times 10^{-4} = \frac{(2.24 \times 10^{-4})^2}{C - 2.24 \times 10^{-4}}. Solving gives C \approx 0.074 M.

Explanation

1. Calculate [H⁺] from pH<br /> Use the formula **$[H^+] = 10^{-\text{pH}}$**. For pH = 3.65, $[H^+] = 10^{-3.65} \approx 2.24 \times 10^{-4}$ M.<br /><br />2. Set up the equilibrium expression<br /> HF dissociates as $HF \rightleftharpoons H^+ + F^-$. The equilibrium expression is **$K_a = \frac{[H^+][F^-]}{[HF]}$**.<br /><br />3. Substitute known values into the equilibrium expression<br /> Assume initial concentration of HF is $C$ and at equilibrium, $[H^+] = [F^-] = 2.24 \times 10^{-4}$ M. Then, $K_a = \frac{(2.24 \times 10^{-4})^2}{C - 2.24 \times 10^{-4}}$.<br /><br />4. Solve for C<br /> Rearrange to find $C$: $6.8 \times 10^{-4} = \frac{(2.24 \times 10^{-4})^2}{C - 2.24 \times 10^{-4}}$. Solving gives $C \approx 0.074$ M.
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