Question

The $pH$ of a $0.1M$ solution of $N{H}_{4}OH$ (having $K_b=1.0\times {10}^{-5}$) is equal to

• A. 10
• B. 6
• C. 11
• D. 12

Answer 1

The correct option is C 11

$\left[O{H}^{-}\right]=\sqrt{K_b\times C}$
$=\sqrt{1\times {10}^{-5}\times {10}^{-1}}$
$=\sqrt{{10}^{-6}}={10}^{-3}$
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
${10}^{-14}=\left[H^{+}\right]\left[{10}^{-3}\right]$
$\left[H^{+}\right]={10}^{-11}$
Hence, $pH=-\log H^{+}$
$=-\log (1\times {10}^{-11})=11$