JEE Advanced 2015 · Paper 2 · Chemistry Q26

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ChemistryInteger+4Question 26 of 60

The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If $\lambda^0_{\ce{X-}}\approx\lambda^0_{\ce{Y-}}$ , the difference in their $\mathrm pK_a$ values, $\mathrm pK_a(\ce{HX})-\mathrm pK_a(\ce{HY})$ , is (consider degree of ionization of both acids to be $\ll1$ )

Your answer

Answer and worked solution

Official answer: 3

Degree of ionisation. $\alpha=\Lambda_m/\Lambda_m^0$ . With equal $\Lambda^0$ , $\alpha_X=\alpha_Y/10$ .

Ka for each acid ( $\alpha\ll1$ ): $K_a=C\alpha^2$ .
$\frac{K_a(HX)}{K_a(HY)}=\frac{0.01\,\alpha_X^2}{0.10\,\alpha_Y^2}=\frac1{10}\cdot\frac1{100}=10^{-3}.$

Difference. $\mathrm pK_a(HX)-\mathrm pK_a(HY)=3$ .

Answer: 3

Topics:Conductance and Kohlrausch's law pH and ionisation

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JEE Advanced 2015 · Paper 2, Chemistry question 26