If the solubility of \(\mathrm{BaF}_{2}\), is equal to \(x,\) which of the following expressions is equal to the solubility product, \(K_{s p}\) , for \(\operatorname{BaF}_{2} ?\) (A) \(x^{2}\) (B) 2\(x^{2}\) (C) 2\(x^{3}\) (D) 4\(x^{3}\)

Short Answer

Expert verified
The solubility product, Ksp, of BaF2 when the solubility is x, is equal to 4x^3. Hence, the correct answer is (D) 4x^3.

Step by step solution

01

Identify the dissociation

The first step is to understand how BaF2 dissociates in water. The equation is as follows: BaF2(s) ⇌ Ba2+(aq) + 2F-(aq)
02

Write the expression for the solubility product

The solubility product, Ksp, is the product of the concentrations of the ions produced when the compound dissolves, each raised to the power of its stoichiometric coefficient. So, for BaF2, the expression for Ksp is [Ba2+][F-]^2.
03

Substitute the concentrations in terms of x

Considering the fact that solubility of BaF2 is denoted as x, and since one formula unit of BaF2 produces one Ba2+ ion and two F- ions, the concentrations of Ba2+ and F- in the solution are x and 2x, respectively. Substitute these into the expression for Ksp from step 2: Ksp = (x)(2x)^2.
04

Simplify the expression

Solving the expression derived in step 3, we get Ksp = 4x^3.

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Most popular questions from this chapter

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