JEE Advanced 2022 Paper 2 · Chemistry · Electrochemistry · medium

A weak monobasic acid HA is studied at 298 K. The limiting molar ionic conductivities are λ^°(H^+)=350 and λ^°(A^-)=40 S cm^2 mol^-1. A 0.0100 M…

Numerical answer

A weak monobasic acid

\mathrm{HA}

is studied at

298\ \mathrm{K}

. The limiting molar ionic conductivities are

\lambda^\circ(\mathrm{H^+})=350

and

\lambda^\circ(\mathrm{A^-})=40\ \mathrm{S\,cm^2\,mol^{-1}}

. A

0.0100\ \mathrm{M}

solution of

\mathrm{HA}

has conductivity

\kappa=7.80\times10^{-4}\ \mathrm{S\,cm^{-1}}

. Ignore ions from water. If the dissociation constant is

K_a=x\times10^{-4}

, the value of

x

is _____.

Show answer and solution

Answer: 5

Kohlrausch's law of independent migration gives the limiting molar conductivity of the acid:

\Lambda_m^\circ=\lambda^\circ(\mathrm{H^+})+\lambda^\circ(\mathrm{A^-})=390\ \mathrm{S\,cm^2\,mol^{-1}}

.

The measured molar conductivity follows from

\Lambda_m=\dfrac{1000\,\kappa}{c}

with

c

in mol per litre:

\Lambda_m=\dfrac{1000\times7.80\times10^{-4}}{0.0100}=78\ \mathrm{S\,cm^2\,mol^{-1}}

.

For a weak electrolyte,

\alpha=\Lambda_m/\Lambda_m^\circ=78/390=0.20

. Because

\alpha

is not negligible, use the full Ostwald expression and not the

c\alpha^2

approximation:

K_a=\dfrac{c\alpha^2}{1-\alpha}=\dfrac{0.0100\times0.04}{0.80}=5.0\times10^{-4}

. Using

c\alpha^2

alone would give

4\times10^{-4}

, the usual wrong answer. Hence

x=5

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