Equilibrium Question 344
What is the $ pH $ of $ Ba{{(OH)}_2} $ if normality is 0.1 [CPMT 1996]
Options:
4
10
7
9
Show Answer
Answer:
Correct Answer: B
Solution:
$ Ba{{(OH)}_2} $ ⇌ $ Ba^{2+} + 2OH^{-} $
One molecule on dissociation furnishes $ 2OH^{-} $ ions. So, $ [OH^{-}]=2\times {10^{-4}}M $
$ N=M\times 2 $ ; $ M=\frac{N}{2}=\frac{2\times {10^{-4}}}{2}={10^{-4}} $ pOH $ =-\log [O{H^{-}}]=-\log (1\times {10^{-4}})=-4 $
$ pH+pOH=14 $ ; $ pH=14-4=10 $ .
BETA
