Skip to content
dronacharya.ai
← JEE Advanced 2023 · Paper 1
ChemistrySingle correct+3 / -1Question 40 of 51

JEE Advanced 2023 · Paper 1, Chemistry question 6

On decreasing the pHp\mathrm{H} from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from 10−4 mol L−110^{-4}\,\mathrm{mol\,L^{-1}} to 10−3 mol L−110^{-3}\,\mathrm{mol\,L^{-1}}. The pKapK_\mathrm{a} of HX is

Choose one option
Answer and worked solution

Official answer: B

Solubility of MX in acid. The XX−\ce{X-} ions are partly protonated:
Ksp=[MX+][XX−],s=[XX−]+[HX]=[XX−](1+[HX+]Ka).K_{sp}=[\ce{M+}][\ce{X-}],\qquad s=[\ce{X-}]+[\ce{HX}]=[\ce{X-}]\left(1+\frac{[\ce{H+}]}{K_a}\right).
So s2=Ksp(1+[HX+]Ka)s^2=K_{sp}\left(1+\dfrac{[\ce{H+}]}{K_a}\right).

Ratio between pH 2 and pH 7:
(10−3)2(10−4)2=100=1+10−2/Ka1+10−7/Ka.\frac{(10^{-3})^2}{(10^{-4})^2}=100=\frac{1+10^{-2}/K_a}{1+10^{-7}/K_a}.

Solve. This gives 99Ka≈10−299K_a\approx10^{-2}, so Ka≈10−4K_a\approx10^{-4} and pKa=4pK_a=4.

Topics:Solubility productHydrolysis of salts
← Previous questionNext question →