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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2023 · 8 April · Evening Shift · Chemistry · Ionic Equilibrium

The solubility product of BaSO _ 4 is 1 × 10 ^ - 10 at 298K . The...

The solubility product of BaSO4{BaSO}_{4} is 1×10−101 \times {10}^{- 10} at 298K298K. The solubility of BaSO4{BaSO}_{4} in 0.1MK2SO4(aq)0 . 1M{K}_{2}{SO}_{4} ( aq ) solution is ------×10−9gL−1\times {10}^{- 9}g{L}^{- 1}(nearest integer).
Given: Molar mass of BaSO4{BaSO}_{4} is 233gmol−1233g{mol}^{- 1}

Correct answer

0

Step-by-step solution

For the insoluble salt barium sulphate, sulphate ion added is the common ion. Due to the common ion effect the solubility of barium sulphate decreases. Let us assume solubility of barium sulphate is SS.

BaSO4(s)⇌Ba2+S(aq)+SO42−S+0.1(aq){BaSO}_{4}\left(s\right) ⇌ \mathop{{Ba}^{2 +}}_{S}\left(aq\right) + \mathop{{SO}_{4}^{2 -}}_{S + 0 . 1}\left(aq\right)
K2SO40.1−0.1⟶K+0.1(aq)+SO42−0.1(aq)\mathop{{K}_{2}{SO}_{4}}_{0 . 1 - 0 . 1} ⟶ \mathop{{K}^{+}}_{0 . 1}\left(aq\right) + \mathop{{SO}_{4}^{2 -}}_{0 . 1}\left(aq\right)
KSP=[Ba2+][SO42−]{K}_{SP} = \left[{Ba}^{2 +}\right]\left[{SO}_{4}^{2 -}\right]
1×10−10=S(S+0.1)1 \times {10}^{- 10} = S ( S + 0 . 1 )
≃0.1S≃ 0 . 1S
S=10−9M=233×10−9gL−1S = {10}^{- 9}M = 233 \times {10}^{- 9}{gL}^{- 1}

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