NEETChemistryIonic EquilibrumMCQ+4 / −1
The solubility product of AgI at 25oC is 1.0 1016 mol2 L2. The solubility of AgI in 104 N solution of KI at 25oC is approximately (in mol L1
- A1.0 1016
- B1.0 1012
- C1.0 1010
- D1.0 108
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Correct answer: B
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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