Multiple Choice
The molar solubility of Ba3(PO4)2 is 8.89 × 10^-9 M in pure water. Calculate the Ksp for Ba3(PO4)2.
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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Given the molar solubility of Ag2SO3 in pure water is 1.55 x 10^-5 M, what is the solubility product constant (Ksp) for Ag2SO3?
A
Ksp = 3.72 x 10^-10
B
Ksp = 2.40 x 10^-10
C
Ksp = 5.00 x 10^-10
D
Ksp = 1.48 x 10^-10
Verified step by step guidance1
Identify the dissolution equation for Ag2SO3 in water: Ag2SO3(s) ⇌ 2Ag⁺(aq) + SO3²⁻(aq).
Define the molar solubility (s) of Ag2SO3 as 1.55 x 10^-5 M. This means that for every mole of Ag2SO3 that dissolves, 2 moles of Ag⁺ and 1 mole of SO3²⁻ are produced.
Express the concentrations of the ions in terms of s: [Ag⁺] = 2s and [SO3²⁻] = s.
Write the expression for the solubility product constant (Ksp) for Ag2SO3: Ksp = [Ag⁺]²[SO3²⁻].
Substitute the expressions for the ion concentrations into the Ksp expression: Ksp = (2s)²(s) = 4s³. Calculate Ksp using s = 1.55 x 10^-5 M.
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