The solubility of solid silver chromate Ag2CrO4 , is determined in three solvents substance Ksp of Ag2CrO4=9×10−12 .I) Pure water II) 0.1M AgNO3 III) 0.1M Na2CrO4Predict the relative solubility of Ag2CrO4 in I , II and III.

Solubility in water is more than remaining two because, as common ion effect increases solubility decreasesAg2CrO4⇌ 2Ag+2s+CrO4−2s (i) 0.1M Ag+ ions is common ion 9×10−12=(0.1)2×s ⇒s=9×10−10M(ii) 0.1M CrO4−2 ions is common ion9×10−12=(2s)2(0.1) 9×10−12=4s2(0.1) s=94×10−11≃4.5×10−6M

The solubility of solid silver chromate Ag2CrO4 , is determined in three solvents substance Ksp of Ag2CrO4=9×10−12 .I) Pure water II) 0.1M AgNO3 III) 0.1M Na2CrO4Predict the relative solubility of Ag2CrO4 in I , II and III.

Solubility in water is more than remaining two because, as common ion effect increases solubility decreasesAg2CrO4⇌ 2Ag+2s+CrO4−2s (i) 0.1M Ag+ ions is common ion 9×10−12=(0.1)2×s ⇒s=9×10−10M(ii) 0.1M CrO4−2 ions is common ion9×10−12=(2s)2(0.1) 9×10−12=4s2(0.1) s=94×10−11≃4.5×10−6M

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The solubility of solid silver chromate $A g_{2} C r O_{4}$ , is determined in three solvents substance $K_{s p}$ of $A g_{2} C r O_{4} = 9 \times 10^{- 12}$ .
I) Pure water II) 0.1M $A g N O_{3}$ III) 0.1M $N a_{2} C r O_{4}$
Predict the relative solubility of $A g_{2} C r O_{4}$ in I , II and III.

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II < III < I

I = II = III

II = III < I

I < II < III

answer is D.

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Detailed Solution

Solubility in water is more than remaining two because, as common ion effect increases solubility decreases
$A g_{2} C r O_{4} \overset{}{⇌} \underset{2 s}{2 A g^{+}} + \underset{s}{C r O_{4}^{- 2}}$
(i) 0.1M $A g^{+}$ ions is common ion
$9 \times 10^{- 12} = {(0.1)}^{2} \times s \Rightarrow s = 9 \times 10^{- 10} M$
(ii) 0.1M $C r O_{4}^{- 2}$ ions is common ion