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

Answer: (D)

Verified step by step guidance

1. Identify the dissolution equation for Ag2SO3 in water: Ag2SO3(s) ⇌ 2Ag⁺(aq) + SO3²⁻(aq).
2. 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.
3. Express the concentrations of the ions in terms of s: [Ag⁺] = 2s and [SO3²⁻] = s.
4. Write the expression for the solubility product constant (Ksp) for Ag2SO3: Ksp = [Ag⁺]²[SO3²⁻].
5. 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.