Question

The molar conductivity of $0.012{\text{mol dm}}^{-3}$ aqueous solution of chloroacetic acid is $100{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$. The ion conductance of chloroacetate and $H^{+}$ ions are $50{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$ and $350{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$, respectively.

• A. Degree of dissociation of chloroacetic acid is 0.5
• B. $p{K}_a$ of chloroacetic acid is 3
• C. $H^{+}$ ion concentration in the solution is $3\times {10}^{-3}$
• D. $\left(A\right)$ and $\left(B\right)$

Answer 1

Answer: (B), (C)

The correct options are
B $p{K}_a$ of chloroacetic acid is 3
C $H^{+}$ ion concentration in the solution is $3\times {10}^{-3}$

$\text{Molar conductivity}=100{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$
${\lambda }_{\left(ClC{H}_{2}CO{O}^{-}\right)}$ = $50{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$
${\lambda }_{\left(H^{+}\right)}$= $350{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$
$C=0.012{\text{mol dm}}^{-3}$
For the dissociation of $ClC{H}_{2}COOH$
${\lambda }_{\left(ClC{H}_{2}COOH\right)}^{\infty }={\lambda }_{\left(ClC{H}_{2}CO{O}^{-}\right)}^{\infty }+{\lambda }_{\left(H^{+}\right)}^{\infty }=400{\text{cm}}^2{\text{mol}}^{-1}$
$\alpha =\frac{{\lambda }^o}{{\lambda }^{\infty }}$
$\alpha =\frac{100}{400}=0.25$
Now ,
$\begin{array}{cccccc}ClC{H}_{2}COOH& \rightleftharpoons & C{H}_{2}ClCO{O}^{-}& +& H^{+}& \\ C& & -& & -& \text{(Initially)}\\ C(1-\alpha )& & C\alpha & & C\alpha & \text{(At Equilibrium)}\end{array}$
$K_a=\frac{(C\alpha {)}^2}{C(1-\alpha )}$
$\alpha <<1$
$K_a=\frac{0.012(0.25{)}^2}{(1-0.25)}={10}^{-3}$
$p{K}_a=-\log {10}^{-3}=3$
$[H{]}^{+}=C\alpha$
$[H{]}^{+}=0.012\times 0.25=3\times {10}^{-3}{\text{mol dm}}^{-3}$