   Chapter 16, Problem 51E

Chapter
Section
Textbook Problem

# A solution is prepared by mixing 100.0 mL of 1.0 × 10−2 M Pb(NO3)2 and 100.0 mL of 1.0 × 10−3 M NaF. Will PbF2(s) (Ksp = 4 × 10−8) precipitate?

Interpretation Introduction

Interpretation: The concentration and volume of each solution Pb(NO3)2 and NaF is given. It is to be checked that precipitate will form or not when given amount of Pb(NO3)2 is added to NaF solution.

Concept introduction: The formation of solid in a solution is known as precipitation.

Solubility product, Ksp is defined as the concentration of ions in a saturated solution where each ion is raised to the power of their coefficients. Ion product, Q is defined as the initial concentration of ions in any solution where each ion is raised to the power of their coefficients.

The relation between ion product, Q and solubility product, Ksp is as follows.

• If Ksp>Q , then no precipitate will form.
• If Ksp<Q , then precipitate will form.
• If Ksp=Q , then the solution will be in just saturation.
Explanation

Explanation

To determine: If precipitate will form when 100mL of Pb(NO3)2 is added to 100mL of NaF .

The initial concentration of Pb2+ is 5.0×103M_ and initial concentration of F is 5.0×104M_ .

Given

Concentration of Pb(NO3)2 is 1.0×102M .

Concentration of NaF is 1.0×103M .

Solubility product of PbF2 is 4.0×108 .

Volume of Pb(NO3)2 is 100mL .

Volume of NaF is 100mL .

The PbF2 will only precipitate out during mixing of Pb(NO3)2 and NaF if,

Ksp<Q

The reaction of Pb(NO3)2 and NaF is,

Pb(NO3)2+2NaFPbF2+2NaNO3

It is assumed that the concentration of Pb2+ and F is same as the concentration of Pb(NO3)2 and NaF . Hence, concentration of Pb2+ and F is,

[Pb2+]=1.0×102M[F]=1.0×103M

Formula

The initial concentration of compound is calculated using the formula,

Substitute the values of volume, given concentration of Pb2+ and volume of original solution in the above equation.

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