(C) Given: Concentration $C = 10^{-3} \ M$,degree of dissociation $\alpha = 10\% = 0.1 = 10^{-1}$.$(i)$ Calculation of $K_a$:$K_a = C\alpha^2 = (10^{-

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(C) Given: Concentration $C = 10^{-3} \ M$,degree of dissociation $\alpha = 10\% = 0.1 = 10^{-1}$.$(i)$ Calculation of $K_a$:$K_a = C\alpha^2 = (10^{-3}) \times (10^{-1})^2 = 10^{-3} \times 10^{-2} = 10^{-5}$.(ii) Calculation of $pH$:$[H^+] = C\alpha = 10^{-3} \times 10^{-1} = 10^{-4} \ M$.$pH = -\log[H^+] = -\log(10^{-4}) = 4$.Thus,$K_a = 10^{-5}$ and $pH = 4$.

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

Medium

For a $10^{-3} \ M \ HCN$ solution,if the degree of dissociation $\alpha = 10\%$,find the $K_a$ and $pH$ of the solution.

A
$10^{-3}, 9$
B
$10^{-4}, 6$
C
$10^{-5}, 4$
D
$10^{-6}, 7$

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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 $10^{-3} \ M \ HCN$ solution,if the degree of dissociation $\alpha = 10\%$,find the $K_a$ and $pH$ of the solution.

Similar Questions

If a $0.1 \ N$ solution of $CH_3COOH$ is $1.3\%$ ionized,what will be the $pH$ of the solution? (Given: $\log 1.3 = 0.11$)

$A$ monoprotic acid in $1.00 \ M$ solution is $0.01\%$ ionized. The dissociation constant of this acid is

An organic monobasic acid has a dissociation constant of $1.96 \times 10^{-8}$. What is its percentage dissociation in a $0.01 \text{ M}$ solution (in $\%$)?

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