   Chapter 17, Problem 45PS

Chapter
Section
Textbook Problem

When 250 mg of SrF2, strontium fluoride, is added to 1.00 L of water, the salt dissolves to a very small extent. S r F 2 ( s ) ⇄ S r 2 + ( a q ) + 2 F − ( a q ) At equilibrium, the concentration of Sr2+ is found to be 1.03 − 10−3 M. What is the value of Ksp for SrF2?

Interpretation Introduction

Interpretation: Value of solubility product constant Ksp for SrF2 has to be calculated.

Concept introduction:

The solubility of a salt is defined as the maximium 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.

Expression for Ksp of a salt is given as,

AxBy(s)xAy+(aq)+yBx(aq)Ksp=[Ay+]x[Bx]y

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

Rearrange the expression for s.

(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

The solubility product constant Ksp of SrF2 is calculated below.

Given:

The equilibrium concentration of Sr2+ is 1.03×103M.

When SrF2 is dissolved in water, it dissociates as follows:

SrF2(s) Sr2+(aq)+ 2F1(aq)

[Sr2+]=1.03×103M[F1]=2[Sr2+]=2

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