(B) For a weak acid,the degree of ionization $(\alpha)$ is given by the formula: $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 10^{-6}$ and $C = 0.01 \

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(B) For a weak acid,the degree of ionization $(\alpha)$ is given by the formula: $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 10^{-6}$ and $C = 0.01 \ M = 10^{-2} \ M$.Substituting the values: $\alpha = \sqrt{\frac{10^{-6}}{10^{-2}}} = \sqrt{10^{-4}} = 10^{-2}$.To find the percent ionization: $\text{Percent Ionization} = \alpha \times 100 = 10^{-2} \times 100 = 1\%$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of weak Acids and weak Bases

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What is the percent ionization $(\alpha)$ of a $0.01 \ M \ HA$ solution? $......\%$ $(K_a = 10^{-6})$

A
$0.995$
B
$1$
C
$10.5$
D
None

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of weak Acids and weak Bases

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For a $0.1 \ M$ solution of a weak acid $HQ$,the $pH$ is $3$. The ionization constant $K_a$ of the acid is:

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What is the percent ionization $(\alpha)$ of a $0.01 \ M \ HA$ solution? $......\%$ $(K_a = 10^{-6})$

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The $pH$ of a $0.1 \ M$ solution of the acid $HA$ is $3$. The value of the ionization constant,$K_a$ of this acid is

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