The percentage of ionization of 1 L of x M | vedclass

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Question

(B) For a weak acid,the degree of ionization $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$,where $C$ is the molar concentration.Given $K_a = 1.

Vedclass · Accepted Answer

$0.015$

Vedclass · Answer

(B) For a weak acid,the degree of ionization $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$,where $C$ is the molar concentration.Given $K_a = 1.8 	imes 10^{-5}$.For solution "$A$": $\alpha_A = \frac{4.242}{100} = 0.04242$. Thus,$0.04242 = \sqrt{\frac{1.8 	imes 10^{-5}}{x}}$. Squaring both sides: $0.0018 = \frac{1.8 	imes 10^{-5}}{x}$,which gives $x = \frac{1.8 	imes 10^{-5}}{1.8 	imes 10^{-3}} = 0.01 \ M$.For solution "$B$": $\alpha_B = \frac{3}{100} = 0.03$. Thus,$0.03 = \sqrt{\frac{1.8 	imes 10^{-5}}{y}}$. Squaring both sides: $0.0009 = \frac{1.8 	imes 10^{-5}}{y}$,which gives $y = \frac{1.8 	imes 10^{-5}}{9 	imes 10^{-4}} = 0.02 \ M$.When $1 \ L$ of solution "$A$" $(0.01 \ M)$ is mixed with $1 \ L$ of solution "$B$" $(0.02 \ M)$,the total volume becomes $2 \ L$.The total moles of acetic acid = $(1 \ L 	imes 0.01 \ M) + (1 \ L 	imes 0.02 \ M) = 0.03 \ 	ext{moles}$.The resultant concentration = $\frac{0.03 \ 	ext{moles}}{2 \ L} = 0.015 \ M$.

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