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Ionic Equilibrium question

2019 · 12 Jan · Shift 2 · Q9
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Ionic Equilibrium question

2019 · 12 Jan · Shift 2 · Q9

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
If Ksp of Ag2CO3Ag_2CO_3Ag2​CO3​ is 8 ×\times× 10–12, the molar solubility of Ag2CO3Ag_2CO_3Ag2​CO3​ in 0.1 M AgNO3AgNO_3AgNO3​ is -
  1. A
    8 ×\times× 10–12 M
  2. B
    8 ×\times× 10–10 M
  3. C
    8 ×\times× 10–13 M
  4. D
    8 ×\times× 10–11 M
View written solutionFree

Correct answer: B

  1. Write the dissolution equilibrium

Ag2CO3(s)⇌2Ag++CO32−Ag_2CO_3(s) \rightleftharpoons 2Ag^+ + CO_3^{2-}Ag2​CO3​(s)⇌2Ag++CO32−​

If the molar solubility of Ag2CO3Ag_2CO_3Ag2​CO3​ in 0.1 M AgNO30.1\,M\,AgNO_30.1MAgNO3​ is sss, then:

  • [CO32−]=s[CO_3^{2-}] = s[CO32−​]=s
  • [Ag+]=0.1+2s[Ag^+] = 0.1 + 2s[Ag+]=0.1+2s

Since sss will be very small compared to 0.10.10.1, we take:

[Ag+]≈0.1[Ag^+] \approx 0.1[Ag+]≈0.1

  1. Use the solubility product expression

For Ag2CO3Ag_2CO_3Ag2​CO3​,

Ksp=[Ag+]2[CO32−]K_{sp} = [Ag^+]^2[CO_3^{2-}]Ksp​=[Ag+]2[CO32−​]

Given:

Ksp=8×10−12K_{sp} = 8 \times 10^{-12}Ksp​=8×10−12

Substitute:

8×10−12=(0.1)2(s)8 \times 10^{-12} = (0.1)^2(s)8×10−12=(0.1)2(s)

  1. Solve for sss

s=8×10−12(0.1)2s = \frac{8 \times 10^{-12}}{(0.1)^2}s=(0.1)28×10−12​

s=8×10−1210−2s = \frac{8 \times 10^{-12}}{10^{-2}}s=10−28×10−12​

s=8×10−10 Ms = 8 \times 10^{-10}\,Ms=8×10−10M

  1. Match with the options

Thus, the molar solubility is:

8×10−10 M\boxed{8 \times 10^{-10}\,M}8×10−10M​

This corresponds to Option B.

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