Question

Rank the following 0.10 M solutions in order of increasing pH.

$\begin{array}{l}\mathbf{a}.{\mathbf{NH}}_3\\ \mathbf{b}.\mathbf{KOH}\\ \mathbf{c}.{\mathbf{HC}}_2{\mathbf{H}}_3{\mathbf{O}}_2\\ \mathbf{d}.\mathbf{KCl}\\ \mathbf{e}.\mathbf{HCl}\end{array}$

Short answer

The increasing order of pH: $\mathrm{HCl}<{\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2<\mathrm{KCl}<{\mathrm{NH}}_3<\mathrm{KOH}$

Step-by-step solution

pH Calculation of 0.1 M NH3 Solution 

For weak bases:

${\text{[OH}}^{\text{-}}\text{]=}\sqrt{{\mathrm{K}}_{\mathrm{b}}\times {\left[\mathrm{BOH}\right]}_0}$

$\begin{array}{c}{\text{[OH}}^{\text{-}}\text{]=}\sqrt{1.8\times {10}^{-5}\times 0.1}\\ =1.34\times {10}^{-3}\end{array}$

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[\mathrm{O}{\text{H}}^{-}\right]}\\ =\frac{1\times {10}^{-14}}{1.34\times {10}^{-3}}\\ =7.46\times {10}^{-12}\end{array}$

$\begin{array}{c}{\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]}\\ \text{=-log(}7.46\times {10}^{-12}\text{)}\\ \text{=11.13}\end{array}$

pH calculation of 0.1 M KOH solution

For strong bases like KOH:

${\text{[OH}}^{\text{-}}\text{]=}{\left[\mathrm{BOH}\right]}_0$

${\text{[OH}}^{\text{-}}\text{]=}0.1$

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[\mathrm{O}{\text{H}}^{-}\right]}\\ =\frac{1\times {10}^{-14}}{{10}^{-1}}\\ =1\times {10}^{-13}\end{array}$

$\begin{array}{c}{\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]}\\ \text{=-log(}1\times {10}^{-13}\text{)}\\ \text{=13}\end{array}$

pH calculation of 0.1 M HC2H3O2 solution

For weak acids like acetic acid:

${\text{[H}}^{+}\text{]=}\sqrt{{\mathrm{K}}_{\mathrm{a}}\times {\left[\mathrm{HA}\right]}_0}$

$\begin{array}{c}{\text{[H}}^{+}\text{]=}\sqrt{1.8\times {10}^{-5}\times 0.1}\\ =1.34\times {10}^{-3}\end{array}$

$\begin{array}{c}{\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]}\\ \text{=-log(}1.34\times {10}^{-3}\text{)}\\ \text{=2.9}\end{array}$

pH calculation of 0.1 M KCl solution

As KCl is a salt of strong acid and base, the pH of the solution will be 7.

pH calculation of 0.1 M HCl Solution 

For strong acids like HCl:

${\text{[H}}^{+}\text{]=}{\left[\mathrm{HA}\right]}_0$

${\text{[H}}^{+}\text{]=}0.\text{1\hspace{0.17em}M}$

$\begin{array}{c}{\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]}\\ \text{=-log(}0.1\text{)}\\ \text{=1}\end{array}$

Thus, the increasing order of pH: $\mathrm{HCl}<{\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2<\mathrm{KCl}<{\mathrm{NH}}_3<\mathrm{KOH}$