KVPY 2019 SB SX Exam Previous Papers

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Question : 104 of 120

Marks: +1, -0

Solubility products of CuI and

Ag_2CrO_4

have almost the same value (~4 × 10

^{–12}

). The ratio of solubilities of the two salts (CuI:Ag

_2

CrO

_4

) is closest to

0.01
0.02
0.03
0.10

Solution:

For

CuI⇒ CuI\ –––––▸ {Cu^{+}}↙{S_1 }+{I^{-}}↙{S_1 }

K_ {sp}(CuI) =S_1^2 = 4 × 10^{–12} ⇒ S_1 = 2 × 10^{–6}

For

Ag_2CrO_4 ⇒Ag_2CrO_4 \ –––––▸ 2Ag^{+}

+CrO_4^{2-}

K_{sp}(Ag_2CrO_4) = 4S_2^3= 4 × 10^{–12}

⇒ S_2 = 10^{ –4}

So,

S_1/S_2={2×10^{-6}}/10^{-4}=0.02

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