Question

When $0.15mol$ of Ammonia​ is dissolved in water to make $1L$ of solution, the $[OH{]}^{-}$ of solution is $1.34\times {10}^{-3}M$. Calculate $K_b$​ for $N{H}_3$​.

• A. $11.97\times {10}^{-6}$
• B. $12.80\times {10}^{-4}$
• C. $12.80\times {10}^{-8}$
• D. $11.97\times {10}^{-4}$

Answer 1

The correct option is A $11.97\times {10}^{-6}$

Reaction involved-
$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

$C=\frac{0.15}{1}=0.15mol{L}^{-1}$
$\left[N{H}_4^{+}\right]=\left[O{H}^{-}\right]=1.34\times {10}^{-3}M$
$\left[O{H}^{-}\right]=1.34\times {10}^{-3}M$
we know that,
$\left[O{H}^{-}\right]=C\alpha$
$\left[O{H}^{-}\right]=C\times \sqrt{\frac{K_b}{C}}=\sqrt{K_b\times C}$
$\left[O{H}^{-}\right]=\sqrt{K_b\times C}$
squaring both side,
$K_b\times C=[O{H}^{-}{]}^2$
$K_b=\frac{[O{H}^{-}{]}^2}{C}$
$K_b=\frac{(1.34\times {10}^{-3}{)}^2}{0.15}$
$K_b=\frac{1.79\times {10}^{-6}}{0.15}$
$K_b=11.97\times {10}^{-6}$