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NEET CHEMISTRYEquilibriumMedium

Question

Find out the solubility product KspK_{sp} of Ca(OH)2Ca(OH)_2 if the pH of its saturated solution is 9:

A

0.5×10100.5 \times 10^{-10}

B

0.5×10150.5 \times 10^{-15}

C

0.25×10100.25 \times 10^{-10}

D

0.125×10150.125 \times 10^{-15}

Step-by-Step Solution

Given, pH of the saturated solution = 9 We know that at 298K, pH+pOH=14pH + pOH = 14. pOH=149=5pOH = 14 - 9 = 5 Therefore, the concentration of hydroxyl ions is [OH]=105 M[OH^-] = 10^{-5} \text{ M}. The dissociation equilibrium of Ca(OH)2Ca(OH)_2 is: Ca(OH)2(s)Ca2+(aq)+2OH(aq)Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq) If the molar solubility of Ca(OH)2Ca(OH)_2 is SS, then [Ca2+]=S[Ca^{2+}] = S and [OH]=2S[OH^-] = 2S. Given [OH]=105 M[OH^-] = 10^{-5} \text{ M}, we have: 2S=105S=0.5×105 M2S = 10^{-5} \Rightarrow S = 0.5 \times 10^{-5} \text{ M} Now, the solubility product constant KspK_{sp} is given by: Ksp=[Ca2+][OH]2=(S)(2S)2K_{sp} = [Ca^{2+}][OH^-]^2 = (S)(2S)^2 Ksp=(0.5×105)(105)2K_{sp} = (0.5 \times 10^{-5})(10^{-5})^2 Ksp=0.5×105×1010=0.5×1015K_{sp} = 0.5 \times 10^{-5} \times 10^{-10} = 0.5 \times 10^{-15}

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.

CHEMISTRYEquilibriumsolubilityproductsaturatedsolution

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