Equilibrium Question 280

Question: What is $ [{H^{+}}] $ of a solution that is $ 0.01,M $ in $ HCN $ and $ 0.02,M $ in $ NaCN $

$ (K_{a} $ for $ HCN=6.2\times {10^{-10}}) $ [MP PMT 2000]

Options:

A) $ 3.1\times 10^{10} $

B) $ 6.2\times 10^{5} $

C) $ 6.2\times {10^{-10}} $

D) $ 3.1\times {10^{-10}} $

Show Answer

Answer:

Correct Answer: D

Solution:

$ K_{a}=\frac{[{H^{+}}][C{N^{-}}]}{[HC{N^{-}}]} $

$ 6.2\times {10^{-10}}=\frac{[{H^{+}}][0.02]}{[0.01]} $

$ [{H^{+}}]=\frac{6.2\times {10^{-10}}\times 0.01}{0.02}=3.1\times {10^{-10}} $



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