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 $



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