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  4. Calculate pH of 0.002 N NH 4 OH having 2 % dissociation

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Q. Calculate pH of $0.002\, N\, NH _{4} OH$ having $2 \%$ dissociation

Equilibrium

Solution:

$NH _{4} OH$ is a weak base and partially dissociated

image

$\therefore \left[O H^{-}\right] =C \alpha=2 \times 10^{-3} \times \frac{2}{100}=4 \times 10^{-5} M$

$ pOH =-\log \left[ OH ^{-}\right]$

$=-\log 4 \times 10^{-5}=4.4 $

$ pH =14-4.4=9.6$