How much water must be added to 300 mL of 0. | vedclass

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(B) For a weak acid,the degree of ionisation $(\alpha)$ is given by $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{K_a 	imes V}$,where $V$ is the dilution (v

Vedclass · Accepted Answer

$900$

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(B) For a weak acid,the degree of ionisation $(\alpha)$ is given by $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{K_a 	imes V}$,where $V$ is the dilution (volume per mole of solute).Since $K_a$ is constant,$\alpha \propto \sqrt{V}$.Let the initial volume be $V_1 = 300 \ mL$ and the final volume be $V_2$.Given that the degree of ionisation doubles,$\alpha_2 = 2\alpha_1$.Therefore,$\frac{\alpha_1}{\alpha_2} = \sqrt{\frac{V_1}{V_2}} \Rightarrow \frac{1}{2} = \sqrt{\frac{300}{V_2}}$.Squaring both sides,$\frac{1}{4} = \frac{300}{V_2}$,which gives $V_2 = 1200 \ mL$.The volume of water to be added is $V_2 - V_1 = 1200 \ mL - 300 \ mL = 900 \ mL$.

How much water must be added to $300 \ mL$ of $0.2 \ M$ solution of $CH_3COOH$ $(K_a = 1.8 \times 10^{-5})$ for the degree of ionisation $(\alpha)$ of the acid to double (in $mL$)?

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