Multiple Choice
Determine the molar solubility of CaCO3 in a solution of 0.500 M Na2CO3. Ksp for CaCO3 is 4.96 × 10−9.
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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What is the solubility of CN− ions in a solution of 5.5 M Hg2(CN)2, with a Ksp of 5.0 × 10−40?
A
4.5× 10−20 M
B
5.0 × 10−14 M
C
1.1 × 10−20 M
D
1.0 × 10−13 M
Verified step by step guidance1
Understand the dissociation of Hg2(CN)2 in water. The compound dissociates into Hg2^2+ and CN^- ions. The balanced equation is: Hg2(CN)2 ⇌ Hg2^2+ + 2CN^-.
Identify the expression for the solubility product constant (Ksp). For the dissociation of Hg2(CN)2, the Ksp expression is: Ksp = [Hg2^2+][CN^-]^2.
Let 's' be the solubility of Hg2(CN)2 in mol/L. Since each formula unit of Hg2(CN)2 produces one Hg2^2+ ion and two CN^- ions, [Hg2^2+] = s and [CN^-] = 2s.
Substitute the concentrations into the Ksp expression: Ksp = (s)(2s)^2 = 4s^3.
Solve for 's' using the given Ksp value: 5.0 × 10^-40 = 4s^3. Rearrange to find s, which represents the solubility of CN^- ions.
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