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NEET CHEMISTRYEquilibriumMedium

Question

The concentration of Ag+Ag^+ ions in a saturated solution of Ag2C2O4Ag_2C_2O_4 is 2.2×104 mol L12.2 \times 10^{-4} \text{ mol L}^{-1}. The solubility product of Ag2C2O4Ag_2C_2O_4 is:

A

2.66×10122.66 \times 10^{-12}

B

4.5×10114.5 \times 10^{-11}

C

5.3×10125.3 \times 10^{-12}

D

2.42×1082.42 \times 10^{-8}

Step-by-Step Solution

The dissociation of silver oxalate is given by: Ag2C2O4(s)2Ag+(aq)+C2O42(aq)Ag_2C_2O_4(s) \rightleftharpoons 2Ag^+(aq) + C_2O_4^{2-}(aq) Given the concentration of Ag+Ag^+ ions, [Ag+]=2.2×104 M[Ag^+] = 2.2 \times 10^{-4} \text{ M}. From the stoichiometry of the dissociation reaction, the concentration of oxalate ions [C2O42][C_2O_4^{2-}] is half the concentration of Ag+Ag^+ ions: [C2O42]=[Ag+]2=2.2×1042=1.1×104 M[C_2O_4^{2-}] = \frac{[Ag^+]}{2} = \frac{2.2 \times 10^{-4}}{2} = 1.1 \times 10^{-4} \text{ M}. The solubility product expression is: Ksp=[Ag+]2[C2O42]K_{sp} = [Ag^+]^2[C_2O_4^{2-}] Substitute the values into the expression: Ksp=(2.2×104)2×(1.1×104)=(4.84×108)×(1.1×104)=5.324×10125.3×1012K_{sp} = (2.2 \times 10^{-4})^2 \times (1.1 \times 10^{-4}) = (4.84 \times 10^{-8}) \times (1.1 \times 10^{-4}) = 5.324 \times 10^{-12} \approx 5.3 \times 10^{-12}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumconcentrationsaturatedsolutionsolubilityproduct

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