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NEET CHEMISTRYEquilibriumMedium

Question

In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+\text{Ag}^+ and Pb2+\text{Pb}^{2+} at a concentration of 0.10 M0.10 \text{ M}. Aqueous HCl\text{HCl} is added to this solution until the Cl−\text{Cl}^- concentration is 0.10 M0.10 \text{ M}. What will the concentration of Ag+\text{Ag}^+ and Pb2+\text{Pb}^{2+} at equilibrium? (KspK_{sp} for AgCl=1.8×10−10\text{AgCl} = 1.8 \times 10^{-10}, KspK_{sp} for PbCl2=1.7×10−5\text{PbCl}_2 = 1.7 \times 10^{-5})

A

[Ag+]=1.8×10−11 M;[Pb2+]=1.7×10−4 M[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-4} \text{ M}

B

[Ag+]=1.8×10−7 M;[Pb2+]=1.7×10−6 M[\text{Ag}^+] = 1.8 \times 10^{-7} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-6} \text{ M}

C

[Ag+]=1.8×10−11 M;[Pb2+]=8.5×10−5 M[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 8.5 \times 10^{-5} \text{ M}

D

[Ag+]=1.8×10−9 M;[Pb2+]=1.7×10−3 M[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \text{ M}

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NEET CHEMISTRY: "In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts...." — Solved MCQ | TopperSquare