The ionization constant of $HF$, $HCOOH$ and $HCN$ at $298\, K$ are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
It is known that,
$K_{b}=\frac{K_{w}}{K_{a}}$
Given $K_{a}$ of $HF =6.8 \times 10^{-4}$
Hence, $K_{b}$ of its conjugate base $F^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{6.8 \times 10^{-4}}$
$=1.5 \times 10^{-11}$
Given,
$K_{a}$ of $HCOOH =1.8 \times 10^{-4}$
Hence, $K_{b}$ of its conjugate base $HCOO ^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{1.8 \times 10^{-4}}$
$=5.6 \times 10^{-11}$
Given,
$K_{a}$ of $HCN =4.8 \times 10^{-9}$
Hence, $K_{b}$ of its conjugate base $CN ^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{4.8 \times 10^{-9}}$
$=2.08 \times 10^{-6}$
If $pK_a =\, -\,log K_a=4$ for a weak acid $HX$ and $K_a= C\alpha ^2$ then Van't Haff factor when $C = 0.01\,M$ is
Accumulation of lactic acid $(HC_3H_5O_3),$ a monobasic acid in tissues leads to pain and a feeling of fatigue. In a $0.10\, M$ aqueous solution, lactic acid is $3.7\%$ dissociates. The value of dissociation constant, $K_a,$ for this acid will be
A $0.1\, M$ solution of $HF$ is $1\%$ ionized. What is the $K_a$
The first ionization constant of $H _{2} S$ is $9.1 \times 10^{-8}$. Calculate the concentration of $HS ^{-}$ ion in its $0.1 \,M$ solution. How will this concentration be affected if the solution is $0.1\, M$ in $HCl$ also? If the second dissociation constant of $H _{2} S$ is $1.2 \times 10^{-13}$, calculate the concentration of $S^{2-}$ under both conditions.
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