Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be \(1/30\) of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If \(\lambda_{\mathrm{Q}^{-}}^{0}\) happened to be equal with \(\lambda_{\mathrm{Z}^{-}}^{0}\), then the difference of the \(\mathrm{pK}_{\mathrm{a}}\) values of the two weak acids (\(\mathrm{pK}_{\mathrm{a}}(\mathrm{HQ}) - \mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})\)) is \_\_\_\_ (Nearest integer). [Given: degree of dissociation \((\alpha) \ll 1\) for both weak acids, \(\lambda^{\circ}\): limiting molar conductivity of ions]

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Redox Reactions and Electrochemistry

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be

1/30

of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If

λ_{Q^{-}}^{0}

happened to be equal with

λ_{Z^{-}}^{0}

, then the difference of the

pK_{a}

values of the two weak acids (

pK_{a} (HQ) - pK_{a} (HZ)

) is ____ (Nearest integer).
[Given: degree of dissociation

(α) ≪ 1

for both weak acids,

λ^{\circ}

: limiting molar conductivity of ions]

Answer: 2

For a weak acid with

α ≪ 1

K_{a} = C α^{2}

and

α = \frac{Λ _{m}}{Λ _{m}^{\circ}}

. Since

λ_{Q^{-}}^{\circ} = λ_{Z^{-}}^{\circ}

and both acids share

H^{+}

, their limiting molar conductivities are equal, so

\frac{α _{H Q}}{α _{H Z}} = \frac{Λ _{m} ( H Q )}{Λ _{m} ( H Z )} = \frac{1}{30}

. Then

\frac{K _{a} ( H Q )}{K _{a} ( H Z )} = \frac{C _{H Q}}{C _{H Z}} (\frac{α _{H Q}}{α _{H Z}})^{2} = \frac{0.18}{0.02} \times (\frac{1}{30})^{2} = 9 \times \frac{1}{900} = \frac{1}{100}

. Taking

- lo g

p K_{a} (H Q) - p K_{a} (H Z) = - lo g \frac{1}{100} = 2

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Integer answer