Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate Ka∘=1.0×10−5 will be

# Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate $\left({K}_{\mathrm{a}}^{\circ }=1.0×{10}^{-5}\right)$ will be

1. A

5.0

2. B

6.0

3. C

8.0

4. D

9.0

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### Solution:

The acetate ion undergoes hydrolysis as ${\mathrm{CH}}_{3}{\mathrm{COO}}^{-}+{\mathrm{H}}_{2}\mathrm{O}⇌{\mathrm{CH}}_{3}\mathrm{COOH}+{\mathrm{OH}}^{-}$

${K}_{\mathrm{h}}=\frac{\left[{\mathrm{CH}}_{3}\mathrm{COOH}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\right]}\equiv \frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]/\left[{\mathrm{CH}}_{3}\mathrm{COOH}\right]}=\frac{{K}_{\mathrm{w}}}{{K}_{\mathrm{a}}}=\frac{1.0×{10}^{-14}{\mathrm{M}}^{2}}{1.0×{10}^{-5}\mathrm{M}}=1.0×{10}^{-9}\mathrm{M}$also,$\left[{\mathrm{CH}}_{3}\mathrm{COOH}\right]=\left[{\mathrm{OH}}^{-}\right]$  Hence,   O$\begin{array}{r}\left[{\mathrm{OH}}^{-}\right]={\left[\left(1.0×{10}^{-9}\mathrm{M}\right)\left(0.1\mathrm{M}\right)\right]}^{1/2}=1.0×{10}^{-5}\mathrm{M}\\ \left[{\mathrm{H}}^{+}\right]={K}_{\mathrm{w}}/\left[{\mathrm{OH}}^{-}\right]=1.0×{10}^{-14}{\mathrm{M}}^{2}/\left(1.0×{10}^{-5}\mathrm{M}\right)=1.0×{10}^{-9}\mathrm{M}\\ \mathrm{pH}=-\mathrm{log}\left(\left[{\mathrm{H}}^{+}\right]/\mathrm{M}\right)=9\end{array}$