(C) For a weak monoprotic acid,the concentration of hydrogen ions $[H^{+}]$ is given by the product of the initial concentration $C$ and the degree of

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(C) For a weak monoprotic acid,the concentration of hydrogen ions $[H^{+}]$ is given by the product of the initial concentration $C$ and the degree of ionization $\alpha$.Given: $C = 0.1 \ M$ and $\alpha = 1\% = 0.01$.$[H^{+}] = C \times \alpha = 0.1 \times 0.01 = 0.001 \ M = 10^{-3} \ M$.The $pH$ is calculated as $pH = -\log[H^{+}]$.$pH = -\log(10^{-3}) = 3$.

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

Medium

The $pH$ of a $0.1 \ M$ solution of a weak monoprotic acid that is $1\%$ ionized is:

A
$1$
B
$2$
C
$3$
D
$4$

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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 the ionization of a weak acid $HA$ as $HA \rightleftharpoons H^{+} + A^{-}$,the $pH$ of a $1.0 \ M$ solution is $5$. The dissociation constant $K_a$ is:

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The $pH$ of a $0.1 \ M$ solution of a weak monoprotic acid that is $1\%$ ionized is:

Similar Questions

$A$ weak monobasic acid is $20 \%$ dissociated in its $0.01 \ M$ solution. What is the dissociation constant of the weak acid?

The $K_a$ of propanoic acid is $1.4 \times 10^{-5}$. Calculate the $pH$ of a $0.1 \ M$ solution.

The $pH$ of a $0.1 \ M$ aqueous solution of a very weak acid $(HA)$ is $3$. What is its degree of dissociation in percentage?

At what concentration of $0.2 \ M$ $CH_3COOH$ will its degree of dissociation become double (in $M$)? (Given: $K_a = 1.8 \times 10^{-5}$ for $CH_3COOH$)

For the ionization of a weak acid $HA$ as $HA \rightleftharpoons H^{+} + A^{-}$,the $pH$ of a $1.0 \ M$ solution is $5$. The dissociation constant $K_a$ is:

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