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NEET CHEMISTRYEquilibriumMedium

Question

MY and NY3NY_3, two nearly insoluble salts, have the same KspK_{sp} values of 6.2×10−136.2 \times 10^{-13} at room temperature. Which statement would be true in regard to MY and NY3NY_3?

A

The molar solubility of MY in water is less than that of NY3NY_3

B

The salts MY and NY3NY_3 are more soluble in 0.5 M KY than in pure water

C

The addition of the salt of KY to solution of MY and NY3NY_3 will have no effect on their solubilities

D

The molar solubilities of MY and NY3NY_3 in water are identical

Step-by-Step Solution

Let the molar solubility of MY be s1s_1 and that of NY3NY_3 be s2s_2. For the salt MY: MY⇌M++Y−MY \rightleftharpoons M^+ + Y^- Ksp=[M+][Y−]=s1×s1=s12K_{sp} = [M^+][Y^-] = s_1 \times s_1 = s_1^2 s1=Ksp=6.2×10−13=62×10−14≈7.87×10−7 Ms_1 = \sqrt{K_{sp}} = \sqrt{6.2 \times 10^{-13}} = \sqrt{62 \times 10^{-14}} \approx 7.87 \times 10^{-7}\text{ M}.

For the salt NY3NY_3: NY3⇌N3++3Y−NY_3 \rightleftharpoons N^{3+} + 3Y^- Ksp=[N3+][Y−]3=s2×(3s2)3=27s24K_{sp} = [N^{3+}][Y^-]^3 = s_2 \times (3s_2)^3 = 27s_2^4 s2=(Ksp27)1/4=(6.2×10−1327)1/4≈3.89×10−4 Ms_2 = \left(\frac{K_{sp}}{27}\right)^{1/4} = \left(\frac{6.2 \times 10^{-13}}{27}\right)^{1/4} \approx 3.89 \times 10^{-4}\text{ M}.

Comparing the two solubilities, s1<s2s_1 < s_2. Therefore, the molar solubility of MY in water is less than that of NY3NY_3. Additionally, the addition of KY to the solutions introduces a common ion (Y−Y^-), which will decrease the solubility of both salts due to the common ion effect (Le Chatelier's principle), making options B and C incorrect.

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.

CHEMISTRYEquilibriumnearlyinsolublevaluestemperaturestatement

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