Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 12, Problem 12.34P
Interpretation Introduction

Interpretation:

Moles of PbO that will dissolve in 1 L solution with fixed pH of 10.5 has to be calculated.

Concept Introduction:

Systematic treatment of any chemical equilibrium involves steps indicated as follows:

  • Pertinent reactions possible for the system are written.
  • Charge balance and mas balance equation are written.
  • Equilibrium constant corresponding to equation written are listed.
  • Equations are solved to find unknown from available equations.

Expert Solution & Answer
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Explanation of Solution

Formula to calculate [H+] is given as follows:

  [H+]=10pH        (1)

Substitute 10.50 for pH in equation (1).

  [H+]=1010.50

In aqueous solution self-ionization of water occurs as follows:

  H2OH++OH

Corresponding ionic product expression is written as follows:

  Kw=[H+][OH]        (2)

Rearrange equation (2) to obtain [OH].

  [OH]=Kw[H+]        (3)

Equilibrium of PbO occurs as follows:

  PbO(s)Pb2++2OH

Corresponding solubility product expression is written as follows:

  Ksp=[Pb2+][OH]2        (4)

Substitute equation (3) in equation (4).

  Ksp=[Pb2+](Kw[H+])2        (5)

Substitute 1014 for Kw , 1010.50 for [H+] and 5×1016 for Ksp in equation (5) and rearrange.

  [Pb2+]=5×1016(1010.501014)2=5×109

Hence, 5×109 mol of PbO can dissolve in 1 L of solution.

Equilibrium reaction for PbOH+ occurs as follows:

  Pb2++H2OPbOH++H+

Dissociation constant expression is written as follows:

  Kb=[HCN][OH][CN]        (6)

Substitute 2.5×108 for Ka , 1010.50 for [H+] and 5×109 for [Pb2+] in equation (6) and rearrange.

  [PbOH+]=(2.5×108)(5×1091010.50)=3.95×1064.0×106

Thus, 4.0×106 mol of PbOH+ can dissolve in 1 L of solution.

If activities are included formula to calculate [OH]γOH is as follows:

  [OH]γOH=Kw10pH        (7)

Substitute 1014 for Kw and 10.50 for pH  in equation (7).

  [OH]γOH=10141010.50=103.50

Ionic product expression with activities included is written as follows:

  Kw=[H+](γH+)[OH](γOH)        (2)

Thus, formula to compute [OH]γOH is as follows:

  [OH]γOH=Kw10pH        (7)

Substitute 1014 for Kw and 9 for pH  in equation (7).

  [OH]γOH=1014109=105

Dissociation constant expression is written as follows:

  Kb=[HCN](γHCN)[OH](γOH)[CN](γCN)        (6)

Substitute 1.6×105 for Kb and 0.755 for γCN, 1 for  γHCN, 105 for [OH]γOH  in equation (6) and rearrange.

  [HCN]=(1.6×105)(0.755)[CN]1.0×105=1.208[CN]

Substitute 1.208[CN] for [HCN] in equation ().

  [Ag2+]=[CN]+1.208[CN][CN]=[Ag2+]2.208

Substitute  [Ag2+]/20208 in equation ().

Solubility product expression with activities included is written as follows:

  Ksp=[Ag2+](γAg2+)[CN](γCN)        (8)

Substitute 0.455 for γPb2+, 103.50 for [OH]γOH and 5×1016 for Ksp in equation (5).

  [Pb2+]=5×1016(0.455)(103.50)2=1.098×1081.1×108 M

Thus, 1.1×108 mol of Pb2+ can dissolve in 1 L of solution.

Charge balance equation that shows net postive charge equal to net negative charge is formulated as follows:

  [H+]+2[Mg2+]=[OH]        (3)

Mass balance equation is formulated as follows:

  [Ag2+]=[CN]+[HCN]        (4)

Substitute equation (4) and equation (2) in equation (3).

  [H+]+2(4.0×108 M)=Kw[H+]        (5)

Substitute 1014 for Kw in equation (5) and rearrange.

  [H+]2+(8×108)[H+]1014=0

Simplify to obtain value of [H+] as 6.8×108.

Substitute 6.8×108 for [H+] in equation (2) to obtain value of [OH].

  [OH]=10146.8×108=1.4×107 M

Hence, values of [Mg2+], [OH] and [H+] are 4.0×108 M, 1.4×107 M and 6.8×108 M respectively.

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