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JEE Main 201912 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2019 · 12 January · Evening Shift · Chemistry · Ionic Equilibrium

If K _ sp of Ag _ 2 CO _ 3 is 8 × 10 ^ - 12 , the molar solubility...

If Ksp{K}_{sp} of Ag2CO3{Ag}_{2}{CO}_{3} is 8×10−128 \times {10}^{- 12} , the molar solubility of Ag2CO3{Ag}_{2}{CO}_{3} in 0.1MAgNO30 .1 M{AgNO}_{3} is:If K _ sp of Ag _ 2 CO _ 3 is 8 × 10 ^ - 12 , the molar solubility of Ag _ 2 CO _ 3 in 0 .1 M AgNO _ 3 is:If K _ sp of Ag _ 2 CO _ 3 is 8 × 10 ^ - 12 , the molar solubility of Ag _ 2 CO _ 3 in 0 .1 M AgNO _ 3 is:

Options

  1. A8x10−13M8 x {10}^{- 13}M
  2. B8x10−11M8 x {10}^{- 11}M
  3. C8x10−12M8 x {10}^{- 12}M
  4. D8x10−10M8 x {10}^{- 10}M

Correct answer

D. 8x10−10M8 x {10}^{- 10}M

Step-by-step solution

(i) If K _ sp of Ag _ 2 CO _ 3 is 8 × 10 ^ - 12 , the molar solubility of Ag _ 2 CO _ 3 in 0 .1 M AgNO _ 3 is:

(ii) If K _ sp of Ag _ 2 CO _ 3 is 8 × 10 ^ - 12 , the molar solubility of Ag _ 2 CO _ 3 in 0 .1 M AgNO _ 3 is:

Ksp=8×10−12=(2S′+0.1)2S′{K}_{sp} = 8 \times {10}^{- 12} = {( 2{S}^{′} + 0.1 )}^{2} {S}^{′}

or S′=8×10−10M{S}^{′} = 8 \times {10}^{- 10} M

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