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Q)

# The $pH$ of 0.1 M solution of $NH_4OH$ (having dissociation constant $K_b = 1.0 \times 10^{-5})$ is equal to

(a) 10

(b) 6

(c) 11

(d) 12

$[OH^-] = \sqrt{K_b \times (NH_4OH)} = (10^{-5} \times 10^{-1})^{\frac{1}{2}} = 1 \times 10^{-3}$
$[H^+] = \frac{10^{-14}}{10^{-3}} = 10^{-11}$
Hence, $pH = - log 10^{-11} = 11.0$