The conductivity of a weak acid HA of concentration 0.001 mol L - 1 is 2.0 \times 10 - 5 S cm - 1. If \Lambda _m^o (HA) = 190 S cm2 mol - 1, the ionization constant (Ka) of HA is equal to \underline\hspace2cm\times 10 - 6. (Round off to the Nearest Integer)

Electrochemistry PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q15 with Solution

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Correct answer: 12

Concentration: $c = 0.001\ \text{mol L}^{-1}$
Conductivity: $\kappa = 2.0 \times 10^{-5}\ \text{S cm}^{-1}$
Limiting molar conductivity: $\Lambda_m^\circ(\text{HA}) = 190\ \text{S cm}^2\text{ mol}^{-1}$

We need to find $K_a$ .

For a solution of concentration $c$ in mol L $^{-1}$ ,

\Lambda_m = \frac{\kappa \times 1000}{c}

Substitute the values:

\Lambda_m = \frac{2.0 \times 10^{-5} \times 1000}{0.001}

\Lambda_m = \frac{2.0 \times 10^{-2}}{10^{-3}} = 20\ \text{S cm}^2\text{ mol}^{-1}

For a weak electrolyte,

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

So,

\alpha = \frac{20}{190} = \frac{2}{19} \approx 0.1053

For weak acid $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ ,

K_a = \frac{c\alpha^2}{1-\alpha}

Substitute $c = 0.001$ and $\alpha = \frac{2}{19}$ :

K_a = \frac{0.001\left(\frac{2}{19}\right)^2}{1-\frac{2}{19}}

K_a = \frac{0.001\cdot \frac{4}{361}}{\frac{17}{19}}

K_a = 0.001 \cdot \frac{4}{361} \cdot \frac{19}{17}

K_a = 0.001 \cdot \frac{76}{6137}

K_a \approx 0.001 \cdot 0.012384

K_a \approx 1.2384 \times 10^{-5}

Write in the form $x \times 10^{-6}$ :

K_a = 12.384 \times 10^{-6}

Rounded to nearest integer:

x = 12

The required integer is:

\boxed{12}