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JEE Advanced 2021 Paper 2 · Chemistry · Ionic Equilibrium · medium

JEE Advanced 2021 Paper 2 Chemistry: Paragraph 1

Calcium fluoride is a sparingly soluble salt with Ksp=3.2×10−11K_{sp}=3.2\times10^{-11} at 298 K298\ \mathrm{K}. Fluoride is the conjugate base of the weak acid HF\mathrm{HF} with Ka=6.6×10−4K_a=6.6\times10^{-4}. Assume ideal behaviour (activity coefficients equal to 1), that Ca2+\mathrm{Ca^{2+}} does not hydrolyse or form complexes, and that all dissolved fluorine exists as either F−\mathrm{F^-} or HF\mathrm{HF}. The solubility ss of CaF2\mathrm{CaF_2} (in mol L−1\mathrm{mol\,L^{-1}}) is to be found in two situations: in pure water, where the hydrolysis of F−\mathrm{F^-} is neglected, and in a solution buffered to a constant pH=3.00\mathrm{pH}=3.00.

Question 1

Numerical answer

The solubility of CaF2\mathrm{CaF_2} in pure water, in units of 10−4 mol L−110^{-4}\ \mathrm{mol\,L^{-1}}, is ______. Give the answer correct to two decimal places.
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Answer: 2.00

In pure water, with hydrolysis of fluoride neglected, CaF2(s)⇌Ca2++2F−\mathrm{CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-} gives [Ca2+]=s[\mathrm{Ca^{2+}}]=s and [F−]=2s[\mathrm{F^-}]=2s. Then Ksp=s(2s)2=4s3K_{sp}=s(2s)^2=4s^3, so s=(Ksp/4)1/3=(3.2×10−11/4)1/3=(8.0×10−12)1/3=2.00×10−4 mol L−1s=(K_{sp}/4)^{1/3}=(3.2\times10^{-11}/4)^{1/3}=(8.0\times10^{-12})^{1/3}=2.00\times10^{-4}\ \mathrm{mol\,L^{-1}}. A common slip is to use Ksp=s⋅2sK_{sp}=s\cdot 2s or s=Ksps=\sqrt{K_{sp}}, giving a wrong power; another is to forget the factor 44 and compute (Ksp)1/3(K_{sp})^{1/3} directly. The cube-root form follows from the 1:21:2 stoichiometry, and the value, expressed in units of 10−4 mol L−110^{-4}\ \mathrm{mol\,L^{-1}}, is 2.002.00. This value serves as the reference for judging how much the pH-dependent protonation of fluoride enhances solubility in the next part.

Question 2

Numerical answer

The solubility of CaF2\mathrm{CaF_2} in the solution buffered at pH=3.00\mathrm{pH}=3.00, in units of 10−4 mol L−110^{-4}\ \mathrm{mol\,L^{-1}}, is ______. Give the answer correct to two decimal places.
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Answer: 3.70

At constant pH, dissolution is accompanied by protonation of fluoride. Mass balance on fluorine gives 2s=[F−]+[HF]=[F−](1+[H+]/Ka)2s=[\mathrm{F^-}]+[\mathrm{HF}]=[\mathrm{F^-}](1+[\mathrm{H^+}]/K_a), so [F−]=2s/(1+[H+]/Ka)[\mathrm{F^-}]=2s/(1+[\mathrm{H^+}]/K_a). With [H+]=1.0×10−3[\mathrm{H^+}]=1.0\times10^{-3} and Ka=6.6×10−4K_a=6.6\times10^{-4}, the factor is 1+1.515=2.5151+1.515=2.515. Substitute into Ksp=[Ca2+][F−]2=s 4s2(2.515)2K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2=s\,\dfrac{4s^2}{(2.515)^2}, giving s3=Ksp(2.515)2/4=3.2×10−11×6.326/4=5.06×10−11s^3=K_{sp}(2.515)^2/4=3.2\times10^{-11}\times6.326/4=5.06\times10^{-11}. Hence s=3.70×10−4 mol L−1s=3.70\times10^{-4}\ \mathrm{mol\,L^{-1}} (checked by numerical root-finding), about 1.851.85 times the solubility in pure water. The enhancement arises because protonation of F−\mathrm{F^-} lowers the free fluoride concentration and shifts the dissolution equilibrium forward.
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