Answer:
(C) im pretty sure is the answer
Explanation:
Answer: I think that the answer was False.
Answer:
a. in pure water Solubility (x) = 1.26 x 10⁻⁴M
b. in 0.202M M⁺² Solubility (x) = 9.963 x 10⁻¹²M
The large drop in solubility is consistent with the common ion effect.
Explanation:
a. Solubility in pure water
Given: M(OH)₂ ⇄ M⁺² + 2OH⁻
I --- 0 0
C --- x 2x
E --- x 2x
Ksp = [M⁺²][OH⁻]² = (x)(2x)² = 4x³ => x = CubeRt(Ksp/4)
solubility in pure water = x = CubeRt(8.05 x 10⁻¹²/4) = 1.26 x 10⁻⁴M
b. Solubility in presence of 0.202M M⁺² as common ion.
Given: M(OH)₂ ⇄ M⁺² + 2OH⁻
I --- 0.202M 0
C --- +x +2x
E --- 0.202M + x 2x
≈ 0.202M
Ksp = [M⁺²][2x]² = (0.202)(2x)² = (0.202)(4x²) = 8.05 x 10⁻¹²
=> x = (8.05 x 10⁻¹²)/(0.202)(4) = 9.963 x 10⁻¹²M
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