Multiple Choice
What is the pH of a solution formed by mixing 140.0 mL of 0.25 M HF with 225.0 mL of 0.31 M NaF, given that the Ka of hydrofluoric acid (HF) is 6.8 x 10^-4?
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17. Acid and Base Equilibrium
pH of Weak Acids
Multiple Choice
Calculate the pH of a 0.03 M aqueous solution of a weak acid, HA, with a pKa of 3.39.
A
pH = 4.21
B
pH = 2.54
C
pH = 3.39
D
pH = 1.82
Verified step by step guidance1
Identify the given values: the concentration of the weak acid [HA] is 0.03 M, and the pKa of the acid is 3.39.
Use the relationship between pKa and Ka: \( \text{pKa} = -\log_{10}(K_a) \). Rearrange this to find \( K_a \) by calculating \( K_a = 10^{-\text{pKa}} \).
Set up the expression for the dissociation of the weak acid: \( HA \rightleftharpoons H^+ + A^- \). Write the expression for the acid dissociation constant: \( K_a = \frac{[H^+][A^-]}{[HA]} \).
Assume that the initial concentration of \( H^+ \) and \( A^- \) is negligible compared to \( [HA] \), and let \( x \) be the concentration of \( H^+ \) and \( A^- \) at equilibrium. Thus, \( [HA] = 0.03 - x \), \( [H^+] = x \), and \( [A^-] = x \).
Substitute these expressions into the \( K_a \) expression: \( K_a = \frac{x^2}{0.03 - x} \). Since \( x \) is small, approximate \( 0.03 - x \approx 0.03 \). Solve for \( x \) to find \( [H^+] \), and then calculate the pH using \( \text{pH} = -\log_{10}([H^+]) \).
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