Question

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

• A. 3
• B. 10.5
• C. 11
• D. 7.5

Answer 1

Answer: (C)

$\left[O{H}^{-}\right]=\sqrt{K_b.C}$
$=\sqrt{1.0\times 10^{-5}\times 0.1}=10^{-3}$
$pOH=-\log \left[O{H}^{-}\right]$
$=-\log 10^{-3}=3pH$
$=14-pOH$
$=14-3=11$