The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Asked by Pragya Singh | 11 months ago |  116

##### Solution :-

The Ka for chloroacetic acid (ClCH2COOH) is $$1.35\times 10^{-3}$$.

$$\Rightarrow K_{a} = c\alpha ^{2}$$

$$\alpha = \sqrt{\dfrac{K_{a}}{c}}$$

$$= \sqrt{\dfrac{1.35\times 10^{-3}}{0.1}}$$​​; (given concentration = 0.1M)

$$\alpha = \sqrt{1.35\times 10^{-2}}$$

= 0.116

$$[H^{+}] = c\alpha = 0.1×0.116 = 0.0116$$

$$pH = -\log [H^{+}] = 1.94$$

ClCH2COONa is a salt of strong base i.e. NaOH, and weak acid i.e. ClCH2COOH

$$ClCH_{2}COO^{-} + H_{2}O$$

$$\leftrightarrow ClCH_{2}COOH + OH^{-}$$

$$K_{h} = \dfrac{[ClCH_{2}COO][OH^{-}]}{[ClCH_{2}COO^{-}]}$$

Now, $$K_{h} = \dfrac{K_{w}}{K_{a}}$$

$$K_{h} = \dfrac{10^{-14}}{1.35\times 10^{-3}}$$

$$= 0.740\times 10^{-11}$$

Also, $$K_{h} = \dfrac{y^2}{0.1}\Rightarrow \; 0.740\times 10^{-11}$$

$$= \dfrac{y^2}{0.1}\Rightarrow \; 0.0740\times 10^{-11} = y^2$$

y =$$0.86\times 10^{-6}$$

$$[OH^{-}] = 0.86\times 10^{-6}$$

$$[H^{+}] =\dfrac{K_{w}}{0.86\times 10^{-6}}$$

$$Kw​​ =\dfrac{10^{-14}}{0.86\times 10^{-6}}$$

$$[H^{+}] = 1.162\times 10^{-3}$$

$$pH = - \log [H^{+}]= 7.94$$​​​​​

Answered by Abhisek | 11 months ago

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