QuestionApril 30, 2025

Use the generic form for the acid: HAarrow H^++A^- Fill in the Blank 50 points Find the pH of a 0.0650 M solution of formic acid. The acid dissociation constant (Ka) for formic acid is 1.8times 10^-4

Use the generic form for the acid: HAarrow H^++A^- Fill in the Blank 50 points Find the pH of a 0.0650 M solution of formic acid. The acid dissociation constant (Ka) for formic acid is 1.8times 10^-4
Use the generic form for the acid: HAarrow H^++A^- Fill in the Blank 50 points Find the pH of a 0.0650 M solution of formic acid. The acid dissociation constant (Ka) for formic acid is 1.8times 10^-4

Solution
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Answer

pH = 2.26 Explanation 1. Write the equilibrium expression For HA \rightarrow H^+ + A^-, the equilibrium expression is K_a = \frac H^+][A^- HA . 2. Set up the ICE table Initial concentrations: [HA] = 0.0650, [H^+] = 0, [A^-] = 0. Change: [HA] = -x, [H^+] = +x, [A^-] = +x. Equilibrium: [HA] = 0.0650 - x, [H^+] = x, [A^-] = x. 3. Substitute into the equilibrium expression K_a = \frac{x^2}{0.0650 - x} = 1.8 \times 10^{-4}. 4. Simplify and solve for x Assume x is small compared to 0.0650, so 0.0650 - x \approx 0.0650. Then x^2 = 1.8 \times 10^{-4} \times 0.0650. Solve for x: x = \sqrt{1.8 \times 10^{-4} \times 0.0650}. 5. Calculate pH pH = -\log_{10}(x).

Explanation

1. Write the equilibrium expression<br /> For $HA \rightarrow H^+ + A^-$, the equilibrium expression is $K_a = \frac{[H^+][A^-]}{[HA]}$.<br />2. Set up the ICE table<br /> Initial concentrations: $[HA] = 0.0650$, $[H^+] = 0$, $[A^-] = 0$. Change: $[HA] = -x$, $[H^+] = +x$, $[A^-] = +x$. Equilibrium: $[HA] = 0.0650 - x$, $[H^+] = x$, $[A^-] = x$.<br />3. Substitute into the equilibrium expression<br /> $K_a = \frac{x^2}{0.0650 - x} = 1.8 \times 10^{-4}$.<br />4. Simplify and solve for x<br /> Assume $x$ is small compared to 0.0650, so $0.0650 - x \approx 0.0650$. Then $x^2 = 1.8 \times 10^{-4} \times 0.0650$. Solve for $x$: $x = \sqrt{1.8 \times 10^{-4} \times 0.0650}$.<br />5. Calculate pH<br /> $pH = -\log_{10}(x)$.
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