   # A solution is 1 × 10 −4 M in NaF, Na 2 S, and Na 3 PO 4 . What would be the order of precipitation as a source of Pb 2+ is added gradually to the solution? The relevant K sp values are K sp (PbF 2 ) = 4 × 10 −8 , K sp (PbS) = 7 × 10 −29 , and K sp [Pb 3 (PO 4 ) 2 ] = 1 × 10 −54 . ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 59E
Textbook Problem
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## A solution is 1 × 10−4 M in NaF, Na2S, and Na3PO4. What would be the order of precipitation as a source of Pb2+ is added gradually to the solution? The relevant Ksp values are Ksp(PbF2) = 4 × 10−8, Ksp(PbS) = 7 × 10−29, and Ksp[Pb3(PO4)2] = 1 × 10−54.

Interpretation Introduction

Interpretation: The solubility product of PbF2,PbS,Pb3(PO4)2 and the concentration of NaF,Na2S and Na3PO4 is given. The order of precipitation of Pb2+ is to be stated.

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

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 solubility of salt is inversely proportional to the order of precipitation. It means that, the least soluble salt will precipitate out first.

### Explanation of Solution

Explanation

To determine: The order of precipitation of Pb2+ when it is added gradually into the solution.

The concentration of Pb2+ in PbF2 is 4.0M_ .

Given

Concentration of NaF is 1×104M .

Solubility product of PbF2 is 4×108 .

The PbF2 will only precipitate out during mixing of Pb2+ and F if,

Ksp<Q

The dissociation reaction of PbF2 is,

PbF2(s)Pb2+(aq)+2F(aq)

It is assumed that the concentration of F is same as the concentration of NaF . Hence, concentration of F is,

[F]=1×104M

The initial concentration of Pb2+ will be same as the minimum concentration of PbF2 . If the concentration of Pb2+ exceeds the Ksp=Q condition, then precipitate will form.

Formula

The ion product of PbF2 at Ksp=Q is calculated as,

Q=[Pb2+]0[F]02

Where,

• Q is ion product.
• [Pb2+]0 is initial concentration of Pb2+ .
• [F]0 is initial concentration of F .

Substitute the value of [F]0 and Ksp in the above expression.

Q=[Pb2+]0[F]024×108=[Pb2+]0[1×104M]2[Pb2+]0=4.0M_

Hence, PbF2 will precipitate out if the concentration of Pb2+ exceeds 4.0M_ .

The concentration of Pb2+ in PbS is 7.0×1025M_ .

Given

Concentration of Na2S is 1×104M .

Solubility product of PbS is 7×1029 .

The PbS will only precipitate out during mixing of Pb2+ and S2 if,

Ksp<Q

The dissociation reaction of PbS is,

PbS(s)Pb2+(aq)+S2(aq)

It is assumed that the concentration of S2 is same as the concentration of PbS . Hence, concentration of S2 is,

[S2]=1×104M

The initial concentration of Pb2+ will be same as the minimum concentration of PbS . If the concentration of Pb2+ exceeds the Ksp=Q condition, then precipitate will form.

Formula

The ion product of PbS at Ksp=Q is calculated as,

Q=[Pb2+]0[S2]0

Where,

• Q is ion product.
• [Pb2+]0 is initial concentration of Pb2+ .
• [S2]0 is initial concentration of S2 .

Substitute the value of S2 and Ksp in the above expression.

Q=[Pb2+]0[S2]07×1029=[Pb2+]0[1×104M][Pb2+]0=7.0×1025M_

Hence, PbS will precipitate out if the concentration of Pb2+ exceeds 7

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