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NEET CHEMISTRYEquilibriumMedium

Question

The solubility product of BaSO4\text{BaSO}_4 in water is 1.5×1091.5 \times 10^{-9}. The molar solubility of BaSO4\text{BaSO}_4 in 0.1 M0.1\text{ M} solution of Ba(NO3)2\text{Ba(NO}_3\text{)}_2 is:

A

2.0×108 M2.0 \times 10^{-8}\text{ M}

B

0.5×108 M0.5 \times 10^{-8}\text{ M}

C

1.5×108 M1.5 \times 10^{-8}\text{ M}

D

1.0×108 M1.0 \times 10^{-8}\text{ M}

Step-by-Step Solution

Let the molar solubility of BaSO4\text{BaSO}_4 in 0.1 M Ba(NO3)20.1\text{ M Ba(NO}_3\text{)}_2 solution be SS'. The dissociation of BaSO4\text{BaSO}_4 is: BaSO4(s)Ba2+(aq)+SO42(aq)\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)

In the solution, Ba(NO3)2\text{Ba(NO}_3\text{)}_2 dissociates completely to give 0.1 M Ba2+0.1\text{ M Ba}^{2+} ions. Due to the common ion effect, the total concentration of Ba2+\text{Ba}^{2+} ions will be: [Ba2+]=S+0.10.1 M[\text{Ba}^{2+}] = S' + 0.1 \approx 0.1\text{ M} (since SS' is very small compared to 0.1 M0.1\text{ M})

The concentration of SO42\text{SO}_4^{2-} ions is: [SO42]=S[\text{SO}_4^{2-}] = S'

The solubility product expression (KspK_{sp}) is: Ksp=[Ba2+][SO42]K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] 1.5×109=(0.1)(S)1.5 \times 10^{-9} = (0.1)(S') S=1.5×1090.1=1.5×108 MS' = \frac{1.5 \times 10^{-9}}{0.1} = 1.5 \times 10^{-8}\text{ M}

Thus, the molar solubility of BaSO4\text{BaSO}_4 in the given solution is 1.5×108 M1.5 \times 10^{-8}\text{ M}.

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.

CHEMISTRYEquilibriumsolubilityproducttextbasosolubilitytextbaso

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