Equilibrium Question 503
Question: The dissociation constant of HCN is $ 5\times {10^{-10}} $ . The $ pH $ of the solution prepared by mixing 1.5 mole of HCN and 0.15 moles of KCN in water and making up the total volume to $ 0.5,dm^{3} $ is [JIPMER 2001]
Options:
A) 7.302
B) 9.302
C) 8.302
D) 10.302
Show Answer
Answer:
Correct Answer: C
Solution:
$ pH=-\log K_{a}+\log \frac{[KCN]}{[HCN]} $
$ pH=-\log ,[5\times {10^{-10}}]+\log ( \frac{0.15}{1.5} )=8.302 $
