Question

What molar concentration of $N{H}_3$ provides a $\left[O{H}^{-}\right]$ of $1.5\times {10}^{-3}$ ?
$(K_b=1.8\times {10}^{-5})$

• A. $0.125$
• B. $(0.125+1.5\times {10}^{-3})$
• C. $(0.125-1.5\times {10}^{-3})$
• D. $(1.5\times {10}^{-3})$

Answer 1

The correct option is D $(1.5\times {10}^{-3})$

Given,

$\left[O{H}^{-}\right]=1.5\times {10}^{-1}$

$K_b=1.8\times {10}^{-5}$

The reaction involved is

$N{H}_3+H_{2}O⟶N{H}_4^{+}+O{H}^{-}$

$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}⟶\left(1\right)$
From above reaction it is clear that equilibrium concentration of $\left[O{H}^{-}\right]$ is same as equilibrium concentration of $\left[N{H}_4^{+}\right]$

$\Rightarrow$ From (1) $K_b=\frac{\left[O{H}^{-}\right]\left[O{H}^{-}\right]}{N{H}_3]}$

$\Rightarrow \left[N{H}_3\right]=\frac{[O{H}^{-}{]}^2}{K_b}$

$=\frac{(1.5\times {10}^{-1})}{1.8\times {10}^{-5}}$

$=\frac{2.25\times {10}^{-2}}{1.8\times {10}^{-5}}$

$=1.25\times {10}^3$

$=0.125\times {10}^4$