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)
$2$
$1$
$3$
$4$
$5\%$ ionization is occur in $0.01$ $M$ $C{H_3}COOH$ solution. Calculate its dissociation constant.
At $298$ $K$ temperature, the ${K_b}$ of ${\left( {C{H_3}} \right)_2}NH$ is $5.4 \times {10^{ - 4}}$ $0.25$ $M$ solution.
$pH$ of $0.1\,\, M$ $N{H_3}$ aqueous solution is $({K_b} = 1.8 \times {10^{ - 5}})$
Ionisation constant of $CH_3COOH$ is $1.7 \times 10^{-5}$ and concentration of $H^+$ ions is $3.4 \times 10^{-4}$. Then find out initial concentration of $CH_3COOH$ Molecules
What is the percent ionization $(\alpha)$ of a $0.01\, M\, HA$ solution ? .......$\%$ $(K_a = 10^{-6})$