(A) $H_2CO_3$ is a weak acid that dissociates as: $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$. Given concentration $C = 10^{-3} \ M$ and degree of diss

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(A) $H_2CO_3$ is a weak acid that dissociates as: $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$. Given concentration $C = 10^{-3} \ M$ and degree of dissociation $\alpha = 10\% = 0.1$. The concentration of $H^+$ ions is given by $[H^+] = C \times \alpha$. $[H^+] = 10^{-3} \times 0.1 = 10^{-4} \ M$. The $pH$ of the solution is calculated as: $pH = -\log[H^+]$. $pH = -\log(10^{-4}) = 4$.

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

Medium

Degree of dissociation is $10\%$ for $10^{-3} \ M$ solution of $H_2CO_3$. Then,the $pH$ of the solution is:

A
$4$
B
$2.7$
C
$3.7$
D
$3.3$

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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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Degree of dissociation is $10\%$ for $10^{-3} \ M$ solution of $H_2CO_3$. Then,the $pH$ of the solution is:

Similar Questions

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

$0.01 \, M$ acetic acid solution is $1 \%$ ionised,then $pH$ of this acetic acid solution is :

At $298 \ K$ temperature,calculate the $pH$ of a $0.25 \ M$ solution of $(CH_3)_2NH$ given that its $K_b$ is $5.4 \times 10^{-4}$.

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:

Acetic acid dissociates to $1.20 \%$ in its $0.01 \ M$ solution. What is the value of its dissociation constant?

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