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Question

# A sample of AgCl was treated with 500 mL of 1.5M Na2CO3 solution to give Ag2CO3. The remaining solution contained 0.0026 g of Cl− per litre. The solubility product of AgCl is if Ksp(Ag2CO3)=8.2×10−12

A

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B

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C

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D

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Solution

## The correct option is B The double displacement reaction is: 2AgCl + Na2CO3→ Ag2CO3 + 2NaCl Ksp(Ag2CO3) = 8.2 × 10−12 (given) = [Ag+]2[CO2−3] Sodium Carbonate will completely dissociate into Na+ and CO2−3 ions. [CO2−3] = [Na2CO3] = 1.5M From this , we can arrive at [Ag+] = (Ksp(Ag2CO3)[CO2−3])0.5 = 2.34 × 10−6 M Amount of Cl− remaining = 00026 g = 0.002635.5 molL−1 = 7.33 × 10−5M We need to figure out Ksp (AgCl) = [Ag+][Cl−] = 2.34 × 10−6 M × 7.33 × 10−5M = 1.71 × 10−10

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