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Ionic Equilibrum question

2003 · Q121
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Ionic Equilibrum question

2003 · Q121

NEETChemistryIonic EquilibrumMCQ+4 / −1
The solubility product of AgI at 25oC is 1.0 ×\times× 10−-−16 mol2 L−-−2. The solubility of AgI in 10−-−4 N solution of KI at 25oC is approximately (in mol L−-−1
  1. A
    1.0 ×\times× 10−-−16
  2. B
    1.0 ×\times× 10−-−12
  3. C
    1.0 ×\times× 10−-−10
  4. D
    1.0 ×\times× 10−-−8
View written solutionFree

Correct answer: B

AgI⇌Ag++I-
sss


Ksp = s2

$ \Rightarrow $ 1.0 × 10–16 = s2

$ \Rightarrow $ s = 1.0 × 10–8 mol L–1

$ \therefore $ [Ag+] = 1.0 × 10–8 mol L–1

Also, in 10–4 N KI solution,

[I–1] = (10–4 + 1.0 × 10–8) mol L–1

$ \Rightarrow $ [I–1] = (10–4) mol L–1

[As 1.0 × 10–8 mol L–1 << 1.0 × 10–4 mol L–1]

$ \therefore $ Ksp of AgI = [Ag+][I–]

= (1.0 × 10–8)(10–4)

= 1.0 × 10–12 mol L–1
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