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 $CN ^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{4.8 \times 10^{-9}}$
$=2.08 \times 10^{-6}$
A $0.1\,N $ solution of an acid at room temperature has a degree of ionisation $ 0.1$ . The concentration of $O{H^ - }$ would be
In aqueous solution the ionization constants for carbonic acid are
$K_1 = 4.2 \times 10^{-7}$ and $K_2 = 4.8 \times 10^{-11}$
Select the correct statement for a saturated $0.034\, M$ solution of the carbonic acid.
A weak acid $HA$ has a $K_a$ of $1.00 \times 10^{-5} $. If $0.100\,mol$ of this acid is dissolved in one litre of water the percentage of acid dissociated at equilibrium is closest to.....$\%$
It has been found that the $pH$ of a $0.01$ $M$ solution of an organic acid is $4.15 .$ Calculate the concentration of the anion, the ionization constant of the acid and its $p{K_a}$
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