   Chapter 17, Problem 84GQ

Chapter
Section
Textbook Problem

A sample of hard water contains about 2.0 × 10−3 M Ca2+. A soluble fluoride-containing salt such as NaF is added to “fluoridate” the water (to aid in the prevention of dental cavities). What is the maximum concentration of F− that can be present without precipitating CaF2?

Interpretation Introduction

Interpretation:

Maximum concentration of fluoride ions in that present in the hard water sample containing 2.0×103M Ca2+ ions without precipitating out CaF2 has to be calculated.

Concept introduction:

Solubility product constant,Ksp is 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,

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

Here,

• [Ay+] and [Bx] are equilibrium concentration.

Reaction quotient, Q, for a reaction is defined as the product of the concentration of the ions at any time of the reaction (other than equilibrium time ) of the salt raised to the power of their coefficients in the balanced chemical equation.

The expression for Q of a salt is given as,

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

Here,

• [Ay+] and [Bx] are the concentration at any time except equilibrium.
1. 1. If Q=Ksp, this implies that the solution is saturated solution and the concentration of the ions have reached their maximum limit.
2. 2. If Q<Ksp, this implies that the solution is not saturated and more salt can be added to the solution or the salt present in the solution already will dissolve more until the precipitation starts.
3. 3. If Q>Ksp, this implies that the solution is oversaturated and precipitation of salt will occur.
Explanation

Maximum concentration of fluoride ions that can be present in the hard water sample containing 2.0×103M Ca2+ ions without precipitating out CaF2 is calculated below.

Given:

Refer to the Appendix J in the textbook for the value of Ksp.

The value of solubility product constant,Ksp, for CaF2 is 5.3×1011.

The concentration of Ca2+ ions present in the solution is 2.0×103M.

When CaF2 is dissolved in water it undergoes dissociation in as follows,

CaF2(s)Ca2+(aq)+2F(aq)

The Ksp expression for CaF2 is,

Ksp=[Ca2+][F]2

Rearrange for [F].

[F]=Ksp[Ca2+]

Substitute 2

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