Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be of the molar conductivity of another weak acid HZ with concentration 0.02 M. If happened to be equal to , then the difference of the values of the two weak acids is _______ (

Question

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be $\frac{1}{30}$ of the molar conductivity of another weak acid HZ with concentration 0.02 M. If $λ_{Q^{-}}^{\circ}$ happened to be equal to $λ_{Z^{-}}^{\circ}$ , then the difference of the $p K_{a}$ values of the two weak acids $(p K_{a} (HQ) - p K_{a} (HZ))$ is _______ (Nearest integer).

[Given: degree of dissociation $(α) ≪ 1$ for both weak acids, $λ °$ : limiting molar conductivity of ions] [2026]

Smart Achievers · Accepted Answer

(2)

$K_{a} (HQ) = C_{1} α_{1}^{2}$ $α_{1} = \frac{λ_{m} (HQ)}{λ_{m}^{\infty} (HQ)}$

$K_{a} (HZ) = C_{2} α_{2}^{2}$ $α_{2} = \frac{λ_{m} (HZ)}{λ_{m}^{\infty} (HZ)}$

$\frac{K_{a} (HQ)}{K_{a} (HZ)} = \frac{C_{1}}{C_{2}} . {(\frac{α_{1}}{α_{2}})}^{2} = \frac{0.18}{0.02} {[\frac{λ_{m} (HQ)}{λ_{m}^{\infty} (HZ)}]}^{2}$

$\frac{K_{a} (HQ)}{K_{a} (HZ)} = 9 \times {(\frac{1}{30})}^{2} = \frac{1}{100}$

$p K_{a} (HQ) - p K_{a} (HZ) = 2$

Solution

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Solution

Ans. (2) Ka(HQ)=C1α12 α1=λm(HQ)λm∞(HQ) Ka(HZ)=C2α22 α2=λm(HZ)λm∞(HZ) Ka(HQ)Ka(HZ)=C1C2.(α1α2)2=0.180.02[λm(HQ)λm∞(HZ)]2 Ka(HQ)Ka(HZ)=9×(130)2=1100 pKa(HQ)-pKa(HZ)=2

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