Multiple Choice
Given the solubility product constant (Ksp) for PbBr₂ is 6.3 x 10⁻⁶, calculate the maximum concentration of Br⁻ ions that can exist in 0.5 L of 0.1 M PbBr₂ solution.
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
Calculate the molar solubility of barium fluoride (BaF2) in pure water, given that its solubility product constant (Ksp) is 2.45 x 10^-5.
A
4.9 x 10^-5 M
B
1.7 x 10^-2 M
C
3.1 x 10^-4 M
D
2.5 x 10^-3 M
Verified step by step guidance1
Start by writing the balanced chemical equation for the dissolution of barium fluoride (BaF2) in water: BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq).
Define the expression for the solubility product constant (Ksp) for BaF2. The Ksp expression is: Ksp = [Ba²⁺][F⁻]².
Let the molar solubility of BaF2 be 's'. When BaF2 dissolves, it produces 's' moles of Ba²⁺ and '2s' moles of F⁻. Substitute these into the Ksp expression: Ksp = (s)(2s)².
Simplify the expression: Ksp = 4s³. This equation relates the solubility product constant to the molar solubility.
Solve for 's' by substituting the given Ksp value (2.45 x 10^-5) into the equation: 4s³ = 2.45 x 10^-5. Rearrange to find s: s = (2.45 x 10^-5 / 4)^(1/3).
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