Question

Calculate $\left[{\mathbf{OH}}^{-}\right]$ in a solution obtained by adding 0.0100 mole of solid NaOH to 1.00 L of $\mathbf{15}.\mathbf{0}\mathbf{M}{\mathbf{NH}}_3.$

Short answer

[OH^-] in the given solution${\text{[OH}}^{-}\text{]}=0.02215$

Step-by-step solution

Concentration of [OH-] from NaOH 

NaOH is a strong base, so it gets completely dissociated Thus,

[OH^-] from NaOH =0.010

NH3 Dissociation in Water

NH₃ reacts with water as:

$\begin{array}{l}{\text{NH}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O\hspace{1em}}\rightleftharpoons {{\text{\hspace{1em}NH}}_{\text{4}}}^{\text{+}}{\text{+OH}}^{\text{-}}\\ 1500.01\text{\hspace{0.17em}(from\hspace{0.17em}NaOH)}\\ 15-xx0.01+x\end{array}$

$\begin{array}{l}{\mathrm{K}}_{\mathrm{b}}=\frac{\mathrm{x}(0.01+\mathrm{x})}{15-\mathrm{x}}=1.8\times {10}^{-5}\\ 0.01\mathrm{x}+{\mathrm{x}}^2=2.7\times {10}^{-4}-1.8\times {10}^{-5}\mathrm{x}\\ {\mathrm{x}}^2+0.01\mathrm{x}-2.7\times {10}^{-4}=0\end{array}$

$\begin{array}{l}\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \mathrm{x}=\frac{-(0.01)\pm \sqrt{{(0.01)}^2-4\left(1\right)(-2.7\times {10}^{-4})}}{2\left(1\right)}\\ \mathrm{x}=0.01215\end{array}$

Total [OH-] Concentration Calculation

$\begin{array}{c}{\text{[OH}}^{-}\text{]}=0.01+x\\ =0.01+0.01215\\ =0.02215\end{array}$