Multiple Choice
What is the pH of a mixture of 200 mL of a 0.39 M solution of the weak acid hydronium ion (H₃O⁺) and 250 mL of a 0.62 M solution of hypoiodous acid (HIO, Kₐ = 2.3 × 10⁻¹¹)?
17. Acid and Base Equilibrium
pH of Weak Acids
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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?
A
pH = 5.20
B
pH = 4.75
C
pH = 3.46
D
pH = 2.08
Verified step by step guidance1
Identify the acids in the solution: HF and HCN. Both are weak acids, and their dissociation in water is characterized by their acid dissociation constants (Ka).
Compare the Ka values of HF and HCN. HF has a Ka of 3.5 x 10^-4, and HCN has a Ka of 4.9 x 10^-10. Since HF has a larger Ka, it is the stronger acid and will contribute more to the hydrogen ion concentration in the solution.
Assume that the contribution of HCN to the hydrogen ion concentration is negligible due to its much smaller Ka value compared to HF. Therefore, focus on the dissociation of HF to determine the pH.
Write the equilibrium expression for the dissociation of HF: HF ⇌ H⁺ + F⁻. The expression for the acid dissociation constant is Ka = [H⁺][F⁻]/[HF].
Set up an ICE (Initial, Change, Equilibrium) table to calculate the concentration of H⁺ at equilibrium. Use the initial concentration of HF (0.200 M) and the expression for Ka to solve for [H⁺]. Once [H⁺] is found, calculate the pH using the formula pH = -log[H⁺].
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