Question

An aqueous solution contains $0.01M$ $RN{H}_2$ $(K_b=2\times {10}^{-6})$ and ${10}^{-4}MNaOH$.The concentration of $O{H}^{-}$ is nearly:

• A. $2.414\times {10}^{-4}$
• B. ${10}^{-4}M$
• C. $1.414\times {10}^{-4}$
• D. $2\times {10}^{-4}$

Answer 1

Answer: (C)

$1.414\times {10}^{-4}$

$\left[O{H}^{-}\right]⟹RN{H}_2$ is weak base, it reacts with $H_{2}O$

$p{K}_b=-lo{g}_{1}0(2\times {10}^{-6})=6-log2$

$pH=14-\frac{1}{2}(p{K}_b-logC)$

$=14-\frac{1}{2}(6-log2-log{10}^{-2})$

$=14-3+\frac{log2}{2}+\frac{1}{2}(-2)$

$=10+\frac{log2}{2}$

$=10.150515$

From, $NaOH⟹\left[O{H}^{-}\right]={10}^{-4}$

$pOH=14-pH=0.0001414⟹1.414\times {10}^{-4}$