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
$\left(\frac{ K _{ eq }}{108 c ^4}\right)^{\frac{1}{5}}$
$\left(\frac{ K _{ eq }}{6 c ^5}\right)^{\frac{1}{5}}$
$\left(\frac{K_{e q}}{5 c^4}\right)^{\frac{1}{5}}$
$\left(\frac{ K _{ eq }}{25 c ^2}\right)^{\frac{1}{5}}$
Given the two concentration of $HCN (K_a = 10^{-9})$ are $0.1\,M$ and $0.001\,M$ respectively. What will be the ratio of degree of dissociation ?
The concentration of $[{H^ + }]$ and concentration of $[O{H^ - }]$ of a $ 0.1$ aqueous solution of $2\%$ ionised weak acid is [Ionic product of water $ = 1 \times {10^{ - 14}}]$
When $100 \ mL$ of $1.0 \ M \ HCl$ was mixed with $100 \ mL$ of $1.0 \ M \ NaOH$ in an insulated beaker at constant pressure, a temperature increase of $5.7^{\circ} C$ was measured for the beaker and its contents (Expt. $1$). Because the enthalpy of neutralization of a strong acid with a strong base is a constant $\left(-57.0 \ kJ \ mol ^{-1}\right)$, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. $2$), $100 \ mL$ of $2.0 \ M$ acetic acid $\left(K_a=2.0 \times 10^{-5}\right)$ was mixed with $100 \ mL$ of $1.0 M \ NaOH$ (under identical conditions to Expt. $1$) where a temperature rise of $5.6^{\circ} C$ was measured.
(Consider heat capacity of all solutions as $4.2 J g ^{-1} K ^{-1}$ and density of all solutions as $1.0 \ g mL ^{-1}$ )
$1.$ Enthalpy of dissociation (in $kJ mol ^{-1}$ ) of acetic acid obtained from the Expt. $2$ is
$(A)$ $1.0$ $(B)$ $10.0$ $(C)$ $24.5$ $(D)$ $51.4$
$2.$ The $pH$ of the solution after Expt. $2$ is
$(A)$ $2.8$ $(B)$ $4.7$ $(C)$ $5.0$ $(D)$ $7.0$
Give the answer question $1$ and $2.$
Determine the degree of ionization and $pH$ of a $0.05 \,M$ of ammonia solution. The ionization constant of ammonia can be taken from Table $7.7 .$ Also, calculate the ionization constant of the conjugate acid of ammonia.
$pH$ of an aqueous solution $H_2CO_3$ is $3.3$. If ${K_{{a_1}}} = {10^{ - 3}}$and ${K_{{a_2}}} = {10^{ - 13}}$ then $[HCO_3^-]$ is