A monoprotic acid in a $0.1\,\,M$ solution ionizes to $0.001\%$. Its ionisation constant is
A monoprotic acid in a $0.1\,\,M$ solution ionizes to $0.001\%$. Its ionisation constant is
- A
$1.0 \times {10^{ - 3}}$
- B
$1.0 \times {10^{ - 6}}$
- C
$1.0 \times {10^{ - 8}}$
- D
$1.0 \times {10^{ - 11}}$
Similar Questions
${K_b}$ of $N{H_4}OH = 1.8 \times {10^{ - 5}}$ calculate $pH$ of $0.15$ $mol$ $N{H_4}OH$ and $0.25$ $mol$ $N{H_4}OH$ containing solution.
${K_b}$ of $N{H_4}OH = 1.8 \times {10^{ - 5}}$ calculate $pH$ of $0.15$ $mol$ $N{H_4}OH$ and $0.25$ $mol$ $N{H_4}OH$ containing solution.
The solubility of a salt of weak acid $( A B )$ at $pH 3$ is $Y \times 10^{-3} mol L ^{-1}$. The value of $Y$ is
. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )
The solubility of a salt of weak acid $( A B )$ at $pH 3$ is $Y \times 10^{-3} mol L ^{-1}$. The value of $Y$ is
. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )
- [IIT 2018]
Values of dissociation constant, $K_a$ are given as follows
Acid
$K_a$
$HCN$
$6.2\times 10^{-10}$
$HF$
$7.2\times 10^{-4}$
$HNO_2$
$4.0\times 10^{-4}$
Correct order of increasing base strength of the base $CN^-,F^-$ and $NO_2^-$ will be
Values of dissociation constant, $K_a$ are given as follows
| Acid | $K_a$ |
| $HCN$ | $6.2\times 10^{-10}$ |
| $HF$ | $7.2\times 10^{-4}$ |
| $HNO_2$ | $4.0\times 10^{-4}$ |
Correct order of increasing base strength of the base $CN^-,F^-$ and $NO_2^-$ will be
- [JEE MAIN 2013]
The solution of $N{a_2}C{O_3}$ has $pH$
The solution of $N{a_2}C{O_3}$ has $pH$
The $pH $ of a $0.01\,M$ solution of acetic acid having degree of dissociation $12.5\%$ is
The $pH $ of a $0.01\,M$ solution of acetic acid having degree of dissociation $12.5\%$ is