Equilibrium Question 739
Question: Calculate pH of 0.002 N $ NH_4 $ OH having 2% dissociation.
Options:
A) 7.6
B) 8.6
C) 9.6
D) 10.6
Show Answer
Answer:
Correct Answer: C
Solution:
$ NH_4OH $ is a weak base and partially dissociated
$ NH_4OH\rightarrow NH_4^{+}+O{H^{-}} $
Concentration before dissociation $ 1 $
$ 0 $
$ 0 $
Concentration after dissociation $ 1-\alpha $
$ \alpha $
$ \alpha $
$ \therefore [O{H^{-}}]=C\alpha =2\times {10^{-3}}\times \frac{2}{100}=4\times {10^{-5}}M $
$ pOH=-\log [O{H^{-}}] $
$ =-\log 4\times {10^{-5}}=4.4 $
$ pH=14-4.4=9.6 $
