For a weak acid $HA,$ Ostwald's dilution law is represented by the equation
${K_a} = \frac{{\alpha c}}{{1 - {\alpha ^2}}}$
${K_a} = \frac{{{\alpha ^2}c}}{{1 - \alpha }}$
$\alpha = \frac{{{K_a}c}}{{1 - c}}$
${K_a} = \frac{{{\alpha ^2}c}}{{1 - {\alpha ^2}}}$
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)
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 $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )
$2\, gm$ acetic acid and $3\, gm$ sodium acetate are present in $100\, ml$. aqueous solution then what will be the $pH$ of solution if ionisation constant of acetic acid is $1.8 \times 10^{-5}$
$50\ ml$ of $0.02\ M$ $NaHSO_4$ is mixed with $50$ $ml$ of $0.02\ M\ Na_2SO_4$. Calculate $pH$ of the resulting solution.$[pKa_2 (H_2SO_4) = 2]$
If the $pKa$ of lactic acid is $5$,then the $pH$ of $0.005$ $M$ calcium lactate solution at $25^{\circ}\,C$ is $........\times 10^{-1}$ (Nearest integer)