Multiple Choice
What is the pH of a solution that contains 0.200 M HF and 0.200 M HCN, given that the Ka of HF is 3.5 x 10^-4 and the Ka of HCN is 4.9 x 10^-10?
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.050 M solution of HNO2 (Ka = 4.6 × 10⁻⁴)?
A
3.45
B
2.34
C
3.01
D
2.68
Verified step by step guidance1
Start by writing the dissociation equation for HNO2: \( \text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^- \).
Use the expression for the acid dissociation constant \( K_a \): \( K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \).
Assume that the initial concentration of \( \text{HNO}_2 \) is 0.050 M and that \( x \) is the concentration of \( \text{H}^+ \) and \( \text{NO}_2^- \) at equilibrium. Therefore, \( [\text{HNO}_2] = 0.050 - x \).
Substitute the equilibrium concentrations into the \( K_a \) expression: \( 4.6 \times 10^{-4} = \frac{x^2}{0.050 - x} \).
Solve the equation for \( x \), which represents the \( [\text{H}^+] \) concentration, and then calculate the pH using the formula \( \text{pH} = -\log[\text{H}^+] \).
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