   Chapter 17.4, Problem 17.8CYU

Chapter
Section
Textbook Problem

The barium ion concentration, [Ba2+], in a saturated solution of barium fluoride is 3.6 × 10−3M. Calculate the value of the Ksp for BaF2.BaF2(s) ⇄ Ba2+(aq) + 2 F−(aq)

Interpretation Introduction

Interpretation:

Value of solubility product constant Ksp for BaF2 has to be calculated.

Concept introduction:

The solubility of a salt is defined as the maximum amount of salt that can be dissolved in definite amount of solvent. It is expressed in moles per liter or grams per liter. Solubility in terms of moles per liter is called molar solubility and is defined as the number of moles of solute (salt) dissolved in per liter of solution.

Solubility product constant Ksp is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

The expression for Ksp of a salt is given as,

AxByxAy++yBxKsp=[Ay+]x[Bx]y

The relation between Ksp and s is derived as follows:

EquationAxByxAy++yBxInitial(M)00Change(M)+xs+ysEquilibrium(M)xsys

Ksp=[Ay+]x[Bx]yKsp=(xs)x(ys)y=xxyy(s)x+y(s)x+y=Kspxxyy=(Kspxxyy)1/(x+y)

Here,

• x is the coefficient of cation A+y
• y is the coefficient of anion Bx
• s is the molar solubility

Ksp is calculated by using molar solubility of the salt.

Explanation

Solubility product constant Ksp for BaF2 is to be calculated as follows,

Given:

BaF2 when dissolved in water dissociates as follows,

BaF2 Ba2++ 2F1

Ksp=[Ba2+][F1]2 (1)

Here,

• Ksp is solubility product constant

From the above reaction, it can be concluded that when BaF2 is dissolved in water for

each molecule of BaF2 dissolved, the concentration of F1 is twice that of Ba2+.

[Ba2+]= 3

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