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QJEEAdv18Chedb8: JEE Advanced 2018 Chemistry - Class 11 JEEAdv18Chedb8 Solution: The solubility of a salt of weak acid at is The value of is…

The solubility of a salt of weak acid (AB)(AB) at pH pH\, 33 is Y×10−3Y \times {10^{ - 3}} mol L−1.mol\,{L^{ - 1}}. The value of YY is ________________.

(Given that the value of solubility product of ABAB (Ksp)=2×10−10\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}} and the value of ionization constant of HBHB (Ka)=1×10−8\left( {{K_a}} \right) = 1 \times {10^{ - 8}} )

Step-by-Step Verified Solution

Let solubility of AB in the buffer of pH 3 = x

JEE Advanced 2018 Paper 1 Offline Chemistry - Ionic Equilibrium Question 10 English Explanation

K3 = KspKa{{{K_{sp}}} \over {{K_a}}}

K3=[HB][A+][H+]=KspKa{K_3} = {{[HB][{A^ + }]} \over {[{H^ + }]}} = {{{K_{sp}}} \over {{K_a}}}

∴\therefore x2(10−3)=2×10−101×10−8{{{x^2}} \over {({{10}^{ - 3}})}} = {{2 \times {{10}^{ - 10}}} \over {1 \times {{10}^{ - 8}}}}

∴\therefore x = 4.47 ×\times 10 −- 3 M = y ×\times 10 −- 3 M

∴\therefore y = 4.47

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