A solution of weak acid HA containing 0.01 m | vedclass

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Question

(A) Given,concentration $C = 0.01 \ mol/L = 10^{-2} \ mol/L$ and $pH = 4$. The concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-4} \ mol/L

Vedclass · Accepted Answer

$1\%, \ 10^{-6}$

Vedclass · Answer

(A) Given,concentration $C = 0.01 \ mol/L = 10^{-2} \ mol/L$ and $pH = 4$. The concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-4} \ mol/L$. The degree of ionisation $\alpha$ is given by $\alpha = \frac{[H^{+}]}{C} = \frac{10^{-4}}{10^{-2}} = 10^{-2}$. Percentage degree of ionisation = $\alpha 	imes 100 = 10^{-2} 	imes 100 = 1\%$. The ionisation constant $K_a$ is given by $K_a = C \alpha^2 = 10^{-2} 	imes (10^{-2})^2 = 10^{-2} 	imes 10^{-4} = 10^{-6}$.

$A$ solution of weak acid $HA$ containing $0.01 \ mol/L$ of acid has $pH = 4$. The percentage degree of ionisation of the acid and the ionisation constant of acid are respectively.

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