The $pH$ of $0.1$ $M$ solution of cyanic acid $(HCNO)$ is $2.34$. Calculate the ionization constant of the acid and its degree of ionization in the solution.
$c=0.1 \,M$
$pH =2.34$
$-\log \left[ H ^{+}\right]= pH$
$-\log \left[ H ^{+}\right]=2.34$
$\left[ H ^{+}\right]=4.5 \times 10^{-3}$
Also.
$\left[ H ^{+}\right]=c \alpha$
$4.5 \times 10^{-3}=0.1 \times \alpha$
$\frac{4.5 \times 10^{-3}}{0.1}=\alpha$
$\alpha=45 \times 10^{-3}=.045$
Then
$K_{a}=c \alpha^{2}$
$=0.1 \times\left(45 \times 10^{-3}\right)^{2}$
$=202.5 \times 10^{-6}$
$=2.02 \times 10^{-4}$
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
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.
Write characteristic and uses of weak base equilibrium constant ${K_b}$.
Sulphurous acid $\left( H _{2} SO _{3}\right)$ has $Ka _{1}=1.7 \times 10^{-2}$ and $Ka _{2}=6.4 \times 10^{-8} .$ The $pH$ of $0.588 \,M\, H _{2} SO _{3}$ is ..... . (Round off to the Nearest Integer)
For a concentrated solution of a weak electrolyte ( $K _{ eq }=$ equilibrium constant) $A _2 B _3$ of concentration ' $c$ ', the degree of dissociation " $\alpha$ ' is