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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2018 · 8 April · Chemistry · Ionic Equilibrium

An aqueous solution contains an unknown concentration of Ba ^ 2 + ....

An aqueous solution contains an unknown concentration of Ba2+{Ba}^{2 +}. When 50mLofa1M50mLofa1M solution of Na2SO4{Na}_{2}{SO}_{4} is added, BaSO4{BaSO}_{4} just begins to precipitate. The final volume is 500mL500mL. The solubility product of BaSO4{BaSO}_{4} is 1×10−101 \times {10}^{- 10}. What is the original concentration of Ba2+{Ba}^{2 +}?

Options

  1. A1.0×10−10M1 .0 \times {10}^{- 10}M
  2. B5×10−9M5 \times {10}^{- 9}M
  3. C2×10−9M2 \times {10}^{- 9}M
  4. D1.1×10−9M1 .1 \times {10}^{- 9}M

Correct answer

D. 1.1×10−9M1 .1 \times {10}^{- 9}M

Step-by-step solution

Ba+2+SO42−⇌BaSO4(s)ppt{Ba}^{+ 2} + {SO}_{4}^{2 -} ⇌ \mathop{{BaSO}_{4( s )}}_{ppt}

Final conc. of [SO42−]=MV1V1+V2=1×50500=0.1M[ {SO}_{4}^{2 -} ] = \dfrac{{MV}_{1}}{{V}_{1} + {V}_{2}} = \dfrac{1 \times 50}{500} = 0 .1 M

Final conc.of [Ba+2][ {Ba}^{+ 2} ] When BaSO4{BaSO}_{4} start precipitating

KSP=QSP=[Ba+2][SO42−]{K}_{SP} = {Q}_{SP} = [ {Ba}^{+ 2} ][ {SO}_{4}^{2 -} ]

10−10=[Ba+2](0.1M){10}^{- 10} = [ {Ba}^{+ 2} ]( 0 .1 M )

Initial conc. [Ba+2][ {Ba}^{+ 2} ]; initial volume was =500−50=450ml= 500 - 50 = 450ml
M1V1=M2V2{M}_{1}{V}_{1} = {M}_{2}{V}_{2}

M1=M2V2V1=10−9×500450{M}_{1} = \dfrac{{M}_{2}{V}_{2}}{{V}_{1}} = \dfrac{{10}^{- 9} \times 500}{450}

M1=1.1×10−9M{M}_{1} = 1 .1 \times {10}^{- 9}M
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