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}} $
