The expression to compute pH of NH4Cl solution is

# The expression to compute pH of solution is

1. A

$\mathrm{pH}=\frac{1}{2}\mathrm{p}{K}_{\mathrm{w}}^{\circ }-\frac{1}{2}\mathrm{p}{K}_{\mathrm{b}}^{\circ }\left({\mathrm{NH}}_{4}\mathrm{OH}\right)-\frac{1}{2}\mathrm{log}\left(\left[\mathrm{salt}\right]/\mathrm{M}\right)$

2. B

3. C

4. D

$\mathrm{pH}=\frac{1}{2}\mathrm{p}{K}_{\mathrm{b}}^{\circ }\left({\mathrm{NH}}_{4}\mathrm{OH}\right)-\frac{1}{2}\mathrm{p}{K}_{\mathrm{w}}^{\circ }+\frac{1}{2}\mathrm{log}\left(\left[\mathrm{salt}\right]/\mathrm{M}\right)$

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### Solution:

$\begin{array}{l}{\mathrm{NH}}_{4}^{+}+{\mathrm{H}}_{2}\mathrm{O}⇌{\mathrm{NH}}_{4}\mathrm{OH}+{\mathrm{H}}^{+}\\ {K}_{\mathrm{h}}=\frac{{K}_{\mathrm{w}}}{{K}_{\mathrm{b}}}=\frac{\left[{\mathrm{NH}}_{4}\mathrm{OH}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{NH}}_{4}^{+}\right]}\simeq \frac{{\left[{\mathrm{H}}^{+}\right]}^{2}}{{\left[{\mathrm{NH}}_{4}^{+}\right]}_{0}}\end{array}$

or ${\left[{\mathrm{H}}^{+}\right]}^{2}=\frac{{K}_{\mathrm{w}}}{{K}_{\mathrm{b}}}{\left[{\mathrm{NH}}_{4}\right]}_{0}$

$2\mathrm{log}\left(\left[{\mathrm{H}}^{+}\right]/\mathrm{M}\right)=\mathrm{log}{K}_{\mathrm{w}}^{\circ }-\mathrm{log}{K}_{\mathrm{b}}^{\circ }+\mathrm{log}\left(\left[{\mathrm{NH}}_{4}^{+}\right]/\mathrm{M}\right)$ or $-\mathrm{log}\left(\left[{\mathrm{H}}^{+}\right]/\mathrm{M}\right)=-\frac{1}{2}\mathrm{log}{K}_{\mathrm{W}}^{\circ }+\frac{1}{2}\mathrm{log}{K}_{\mathrm{b}}^{\circ }-\frac{1}{2}\mathrm{log}\left(\left[{\mathrm{NH}}_{4}^{+}\right]/\mathrm{M}\right)$

$\mathrm{pH}=\frac{1}{2}\mathrm{p}{K}_{\mathrm{w}}^{\circ }-\frac{1}{2}\mathrm{p}{K}_{\mathrm{b}}^{\circ }-\frac{1}{2}\mathrm{log}\left(\left[\mathrm{salt}\right]/\mathrm{M}\right)$

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