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ChemistryNumerical+3Question 31 of 54

JEE Advanced 2018 · Paper 1, Chemistry question 13

The solubility of a salt of weak acid (AB) at pH 3 is Y×10−3\mathbf Y\times10^{-3} mol L⁻¹. The value of Y is ______.

(Given that the value of solubility product of AB (KspK_{sp}) = 2×10−102\times10^{-10} and the value of ionization constant of HB (KaK_a) = 1×10−81\times10^{-8})

Answer and worked solution

Official answer: 4.3

Equilibria. AB⇌AX++BX−\ce{AB <=> A+ + B-} and BX−+HX+⇌HB\ce{B- + H+ <=> HB}.

Mass balance. With solubility ss: [AX+]=s[\ce{A+}]=s and [BX−]+[HB]=s[\ce{B-}]+[\ce{HB}]=s.

Protonation. At pH 3,
[HB]=[BX−][HX+]Ka=10−310−8[BX−]=105[BX−]  ⟹  [BX−]≈s105.[\ce{HB}]=\frac{[\ce{B-}][\ce{H+}]}{K_a}=\frac{10^{-3}}{10^{-8}}[\ce{B-}]=10^5[\ce{B-}]\implies[\ce{B-}]\approx\frac s{10^5}.

Solubility.
Ksp=s⋅s105  ⟹  s2=2×10−10×105=2×10−5  ⟹  s=4.47×10−3 mol L−1.K_{sp}=s\cdot\frac s{10^5}\implies s^2=2\times10^{-10}\times10^5=2\times10^{-5}\implies s=4.47\times10^{-3}\ \mathrm{mol\,L^{-1}}.

Answer: 4.47 (key: 4.30 to 4.60)

Topics:Solubility productHydrolysis of saltsCommon ion effect
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