PRECIPITATION REACTIONS Chapter 17 Part 2

PRECIPITATION PRECIPITATION REACTIONSREACTIONS

Chapter 17 Part 2Chapter 17 Part 2

2

InsolubleChloridesInsolubleChlorides

All salts formed in All salts formed in this experiment are this experiment are said to be said to be INSOLUBLEINSOLUBLE and and form precipitates form precipitates when mixing when mixing moderately moderately concentrated concentrated solutions of the solutions of the metal ion with metal ion with chloride ions.chloride ions.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

3


Although all salts formed in this Although all salts formed in this experiment are said to be insoluble, they experiment are said to be insoluble, they do dissolve to some SLIGHT extent.do dissolve to some SLIGHT extent.

AgCl(s) AgCl(s) AgAg++(aq) + Cl(aq) + Cl--(aq)(aq)

When equilibrium has been established, When equilibrium has been established, no more AgCl dissolves and the solution no more AgCl dissolves and the solution is is SATURATEDSATURATED..

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

4


AgCl(s) <--> AgAgCl(s) <--> Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

When the solution is When the solution is SATURATEDSATURATED, , experiment shows that [Agexperiment shows that [Ag++] = 1.34 x 10] = 1.34 x 10-5-5 M. M.

This is equivalent to the This is equivalent to the SOLUBILITYSOLUBILITY of AgCl.of AgCl.

What is [ClWhat is [Cl--]? ]?

This is also equivalent to the AgCl solubility.This is also equivalent to the AgCl solubility.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

5

Make a chart. Make a chart.

AgCl(s) <--> AgAgCl(s) <--> Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

somesome 0 0 00

- - 1.34 x 101.34 x 10-5-5

1.34 x 101.34 x 10-5-51.34 x 101.34 x 10-5-5

1.34 x 101.34 x 10-5-51.34 x 101.34 x 10-5-5

somesome - - 1.34 x 101.34 x 10-5-5


Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

6


Ksp = [Ag+] [Cl-]

= (1.34 x 10-5)(1.34 x 10-5)

= 1.80 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

KKspsp = solubility product constant = solubility product constant

See Table 18.2 and Appendix J 18A & 18BSee Table 18.2 and Appendix J 18A & 18B

7

Lead(II) ChlorideLead(II) ChlorideLead(II) ChlorideLead(II) ChloridePbClPbCl22(s) <--> Pb(s) <--> Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)

KKspsp = 1.9 x 10 = 1.9 x 10-5-5

8

Consider PbIConsider PbI22 dissolving in water dissolving in water

PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2 I(aq) + 2 I--(aq)(aq)

Calculate KCalculate Kspsp if solubility =0.00130 M if solubility =0.00130 M

SolutionSolution

Solubility = [PbSolubility = [Pb2+2+] = 1.30 x 10] = 1.30 x 10-3-3 M M

[I[I--] = _____________ ?] = _____________ ?

Solubility of Lead(II) IodideSolubility of Lead(II) Iodide

2(1.30 x 102(1.30 x 10-3-3 M) M)

9



Calculate KCalculate Kspsp if solubility =0.00130 M if solubility =0.00130 M


SolutionSolution

1.1. Solubility = Solubility =

[Pb[Pb2+2+] ]

= 1.30 x 10= 1.30 x 10-3-3 M M

[I[I--] = 2 x [Pb] = 2 x [Pb2+2+] ]

= 2.60 x 10= 2.60 x 10-3-3 M M

10



Calculate KCalculate Kspsp if solubility = 0.00130 M if solubility = 0.00130 M

SolutionSolution

1.1. Solubility = [PbSolubility = [Pb2+2+] = 1.30 x 10] = 1.30 x 10-3-3 M M

[I[I--] = 2 x [Pb] = 2 x [Pb2+2+] = 2.60 x 10] = 2.60 x 10-3-3 M M

2.2. KKspsp = [Pb = [Pb2+2+] [I] [I--]]22

= [Pb= [Pb2+2+] {2 • [Pb] {2 • [Pb2+2+]}]}22

= 4 [Pb= 4 [Pb2+2+]]33


11



Calculate KCalculate Kspsp if solubility = 0.00130 M if solubility = 0.00130 M

SolutionSolution

2.2. KKspsp = 4 [Pb = 4 [Pb2+2+]]33 = 4 (solubility) = 4 (solubility)33

KKspsp = 4 (1.30 x 10 = 4 (1.30 x 10-3-3))33 = = 8.8 x 108.8 x 10-9-9


Sample Problems

12Precipitating an Insoluble SaltPrecipitating an Insoluble Salt

HgHg22ClCl22(s) (s) <--> <--> Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--] ] 22

If [HgIf [Hg222+2+] = 0.010 M, what [Cl] = 0.010 M, what [Cl--] is required to ] is required to

just begin the precipitation of Hgjust begin the precipitation of Hg22ClCl22??

What is the maximum [ClWhat is the maximum [Cl--] that can be in ] that can be in

solution with 0.010 M Hgsolution with 0.010 M Hg222+2+ without forming without forming

HgHg22ClCl22??

13

Recognize thatRecognize that

KKspsp = product of maximum ion = product of maximum ion

concentrations.concentrations.

Precipitation begins when product of Precipitation begins when product of

ion concentrations EXCEEDS the Kion concentrations EXCEEDS the Kspsp..

Precipitating an Insoluble SaltPrecipitating an Insoluble Salt

HgHg22ClCl22(s) (s) Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]] 2

2

14

SolutionSolution

[Cl[Cl--] that can exist when [Hg] that can exist when [Hg222+2+] = 0.010 M,] = 0.010 M,



KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [2Cl] [2Cl--]] 2

2

[Cl ] = Ksp

4(0.010) = 1.1 x 10 -18 M

If this concentration of ClIf this concentration of Cl-- is just exceeded, is just exceeded,

HgHg22ClCl22 begins to precipitate. begins to precipitate.

15

Now raise [ClNow raise [Cl--] to 1.0 M. ] to 1.0 M.

What is the value of [HgWhat is the value of [Hg222+2+] at this point?] at this point?

SolutionSolution

[Hg[Hg222+2+] = K] = Ksp sp / [Cl/ [Cl--]]22

= K= Kspsp / (1.0) / (1.0)22 = 1.1 x 10 = 1.1 x 10-18-18 M M

The concentration of HgThe concentration of Hg222+2+ has been reduced by has been reduced by

10101616 ! !



KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]] 2

2

Sample Problems

16REVIEW PROBLEMS

• Write the equilibrium equation and the equilibrium constant expression for saturated solutions of: Ag2S and PbI2.

• The molar solubility of barium carbonate is 9.0 x 10-5 M. Calculate the solubility product constant.

• The molar solubility of barium fluoride is 7.5 x 10-3 M. Calculate the solubility product constant.

17REVIEW PROBLEMS

• Calculate the molar solubility of galena, PbS, given Ksp= 8.4 x 10-28.

• Calculate the molar solubility of calcium fluoride given Ksp = 3.9 x 10-11.

• Compare the molar solubilities for CaF2, PbCl2, and Ag2CrO4.

• A solution is found to be 0.0060 M in barium ion and 0.019 M in fluoride ion. Is the system in equilibrium? If not what will occur as equilibrium is reached. Ksp = 1.7 x 10 -6.

18

Separating Metal Ions Separating Metal Ions CuCu2+2+, Ag, Ag++, Pb, Pb2+2+

Ksp Values

AgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14

Ksp Values

AgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14

19

Separating Salts by Differences in KSeparating Salts by Differences in KspspSeparating Salts by Differences in KSeparating Salts by Differences in Kspsp

A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add CrOCrO44

2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. Which precipitates first?. Which precipitates first?

KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12

KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14

SolutionSolution

The substance whose KThe substance whose Kspsp is first is first exceeded will precipitate first. exceeded will precipitate first.

The ion requiring the lesser amount The ion requiring the lesser amount of CrOof CrO44

2-2- precipitate first. precipitate first. 1919

20

[CrO[CrO442-2-] to ppt. PbCrO] to ppt. PbCrO4 4 = K= Ksp sp / [Pb/ [Pb2+2+] ]

= 1.8 x 10= 1.8 x 10-14-14 / 0.020 = 9.0 x 10 / 0.020 = 9.0 x 10-13-13 M M

[CrO[CrO442-2-] to ppt. Ag] to ppt. Ag22CrOCrO4 4 = K= Ksp sp / [Ag/ [Ag++]]22

= 9.0 x 10= 9.0 x 10-12-12 / (0.020) / (0.020)22 = 2.3 x 10 = 2.3 x 10-8-8 M M

PbCrOPbCrO44 precipitates first. precipitates first.

Separating Salts by Differences in Ksp

A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. . Add CrOAdd CrO44

2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and and yellow PbCrOyellow PbCrO44. Which precipitates first?. Which precipitates first?

KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12

KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14

SolutionSolution

21


2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. . PbCrOPbCrO44 precipitates first. precipitates first.

KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12

KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14


How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?

SolutionSolution

We know that [CrOWe know that [CrO442-2-] = 2.3 x 10] = 2.3 x 10-8-8 M to begin to M to begin to

precipitates Agprecipitates Ag22CrOCrO44. .

What is the PbWhat is the Pb2+2+ concentration at this point? concentration at this point?

22


2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. . PbCrOPbCrO44 precipitates first. precipitates first.

KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12

KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14


How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?

SolutionSolution

[Pb[Pb2+2+] = K] = Kspsp / [CrO / [CrO442-2-] = 1.8 x 10] = 1.8 x 10-14-14 / 2.3 x 10 / 2.3 x 10-8-8 M M

= 7.8 x 10= 7.8 x 10-7-7 M M

Lead ion has dropped from 0.020 M to < 10Lead ion has dropped from 0.020 M to < 10-6-6 M M

23

Common Ion EffectCommon Ion EffectAdding an Ion “Common” to an Adding an Ion “Common” to an

EquilibriumEquilibrium

Common Ion EffectCommon Ion EffectAdding an Ion “Common” to an Adding an Ion “Common” to an

EquilibriumEquilibrium

24

Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in:in:(a) pure water and (a) pure water and

(b) in 0.010 M Ba(NO(b) in 0.010 M Ba(NO33))22..

KKspsp for BaSO for BaSO4 4 = 1.1 x 10= 1.1 x 10-10-10

BaSOBaSO44(s) Ba(s) Ba2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)

Solution (a)Solution (a)

Solubility in pure water = [BaSolubility in pure water = [Ba2+2+] = [SO] = [SO442-2-] = s] = s

KKspsp = [Ba = [Ba2+2+] [SO] [SO442-2-] = s] = s22

s = (Ks = (Kspsp))1/21/2 = 1.1 x 10 = 1.1 x 10-5-5 M M

The Common Ion EffectThe Common Ion EffectThe Common Ion EffectThe Common Ion Effect

25





Solution (b)Solution (b)


Now dissolve BaSONow dissolve BaSO44 in water already in water already containing 0.010 M Bacontaining 0.010 M Ba2+2+. .

Which way will the “common ion” shift the Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSOequilibrium? ___ Will solubility of BaSO44 be be less than or greater than in pure water?___less than or greater than in pure water?___

LeftLeftLessLess

26

[Ba[Ba2+2+]] [SO [SO442-2-]]

initialinitial

changechange

equilib.equilib.

The Common Ion EffectThe Common Ion EffectThe Common Ion EffectThe Common Ion Effect Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in:in:

(a) pure water and (a) pure water and




Solution (b)Solution (b)

0.0100.010 0 0

+ + ss + + ss

0.010 + 0.010 + ss ss

27

The Common Ion EffectThe Common Ion EffectThe Common Ion EffectThe Common Ion Effect Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in:in:




SolutionSolution KKspsp = [Ba = [Ba2+2+] [SO] [SO44

2-2-] = (0.010 + ] = (0.010 + ss) () (s)s)

ss < 1.1 x 10 < 1.1 x 10-5-5 M (solubility in pure water), this M (solubility in pure water), this means 0.010 + means 0.010 + ss is about equal to 0.010. is about equal to 0.010. Therefore,Therefore, KKsp sp = 1.1 x 10 = 1.1 x 10-10-10 = (0.010)( = (0.010)(ss))

s s = 1.1 x 10= 1.1 x 10-8-8 M = solubility in presence of M = solubility in presence of added Baadded Ba2+2+ ion. ion.

28





SolutionSolution


Solubility in pure water = s = 1.1 x 10Solubility in pure water = s = 1.1 x 10 -5-5 M M

Solubility in presence of added BaSolubility in presence of added Ba2+2+ = 1.1 x 10= 1.1 x 10-8-8 M M

Le Chatelier’s Principle is followed!Le Chatelier’s Principle is followed!Sample Problems

29REVIEW PROBLEMS

• Will a precipitate of lead (II) sulfatelead (II) sulfate form when 150 ml of 0.030 M sodium sulfate is mixed with 120 mL of 0.020 M lead (II) nitrate. Ksp = 1.8 x 10 -8.

• Calculate the molar solubilitymolar solubility for calcium calcium fluoridefluoride, Ksp = 3.9 x 10 -11, in:

water.

0.0025 M calcium nitrate.

0.080 M sodium fluoride.

Write appropriate net-ionic equationsWrite appropriate net-ionic equations.

30SOLUBILITY AND pH

• We have discovered in Experiment 23 that salts of weak We have discovered in Experiment 23 that salts of weak acids are generally soluble in acidic solutions. This acids are generally soluble in acidic solutions. This principle is illustrated by combining the Kprinciple is illustrated by combining the Kaa equation with equation with the Kthe Kspsp equation. If we consider CaC equation. If we consider CaC22OO44 in the presence in the presence of strong acid, the following is the net equilibrium of strong acid, the following is the net equilibrium equation:equation:

CaCCaC22OO4(s)4(s) + 2 H + 2 H++ <======> H <======> H22CC22OO4(aq)4(aq) + Ca + Ca+2+2

KKnetnet = K = Kspsp.. ( 1/K ( 1/Ka 1a 1 ) ) .. ( 1/K ( 1/Ka 2a 2 ) )

• Since KSince Ka 1a 1 and K and Ka 2a 2 are both less than one, K are both less than one, Knetnet > K > Kspsp. .

• If the acid is weak enough, KIf the acid is weak enough, Knet net may be greater than one may be greater than one and products be favored. If the anion is the conjugate and products be favored. If the anion is the conjugate base of a strong acid, the Kbase of a strong acid, the Kspsp equation is the only equation is the only equilibrium equation.equilibrium equation.

31

SOLUBILITY AND COMPLEX IONS• If the metal cation can form a complex ion with

the other species present, a new net equilibrium will exist. The process is similar to that in the previous slide.

• If silver bromide is treated with ammonia solution, some of the solid dissolves and the complex ion is formed.

AgBrAgBr(s)(s) + 2 NH + 2 NH3(aq)3(aq) <=====> Ag(NH <=====> Ag(NH33))22++

(aq)(aq) + Br + Br--

KKnetnet = K = Kspsp . . KKff = =

( 3.3 x 10( 3.3 x 10-13-13 ) ( 1.6 x 10 ) ( 1.6 x 1077) = 5.3 x 10) = 5.3 x 10-6-6

32Simultaneous Equilibria

1. If you add sufficient chromate ion to an aqueous suspension of PbCl2, can PbCl2 be converted to PbCrO4?

PbCl2 <--> Pb2+ + 2 Cl-

Pb2+ + CrO42- <--> PbCrO4

PbCl2 + CrO42- <--> PbCrO4 + 2 Cl-

1.7 x 10-5

1/1.8 x 10-14

9.4 x 108

Yes!


2. Can AgCl be dissolved by adding a solution of NH3?

Write the overall equation and determine the K value.

AgCl <--> Ag+ + Cl-

Ag+ + 2 NH3 <--> Ag(NH3)2+

AgCl + 2 NH3 <--> Ag(NH3)2+

+ Cl-

1.8 x 10-10

1.6 x 107

2.9 x 10-3

No, unless very high [NH3]


3. Can CaC2O4 be dissolved by adding a solution of HCl? Write the overall equation and determine the K value.

CaC2O4 <--> Ca2+ + C2O42-

H+ + C2O42- <--> HC2O4

-

H+ + HC2O4- <--> H2C2O4

CaC2O4 + 2 H+ <--> H2C2O4 + Ca2+

2.3 x 10-9

1/6.4 x 10-5

1/5.9 x 10-2

No, unless very high [H+]

6.1 x 10-4

35REVIEW PROBLEMS

• A solution contains 0.0035 M AgA solution contains 0.0035 M Ag++ and and 0.15 M Pb0.15 M Pb+2+2. .

• Which precipitates first when IWhich precipitates first when I-- is added? is added?

KKsp AgI sp AgI = 1.5 x 10 = 1.5 x 10 -16 -16 KKsp PbIsp PbI2 2

= 8.7 x 10 = 8.7 x 10 -9-9. .

• Calculate the concentration of the first Calculate the concentration of the first precipitated ion when the second ion begins precipitated ion when the second ion begins to precipitate.to precipitate.

• Write the equation for silver bromide changing to silver iodide with the addition of iodide ion. Calculate K for this reaction. Solubility product constants for silver bromide and silver iodide are 3.3 x 10 -13 and 1.5 x 10-16 respectively.

36

Practice ProblemsPractice Problems1. A saturated solution of lead chloride

contains 4.50 g of lead chloride per liter. Calculate the Ksp for lead chloride.

2. The Ksp for Al(OH)3 is 1.9 x 10-33. Calculate the molar solubility of Al(OH)3 and determine [Al3+] and [OH1-].

3. What is the molar solubility of BaSO4 in a solution that contains 0.100 M Na2SO4? (Ksp for BaSO4 = 1.1 x 10-10)

37

Practice ProblemsPractice Problems4. Will precipitation occur when 50.0 ml of

0.030 M Pb(NO3)2 is added to 50.0 ml of 0.0020 M KBr? (Ksp for lead bromide = 6.3 x 10-6)

5. Would it be possible to separate a solution containing 0.0020 M Pb2+ and 0.030 M Ag+ by adding drops of Na2CO3 solution? (Ksp for lead carbonate = 1.5 x 10-13 and

Ksp for silver carbonate = 8.2 x 10-12)

6. Can CuBr be dissolved by adding a solution of NaCl? Write the overall equation and determine the K value

38

Practice Problems AnswersPractice Problems Answers

1. 1.7 x 10-5

2. 2.9 x 10-9 M, 2.9 x 10-9 M, 8.7 x 10-9 M

3. 1.1 x 10-9 M

4. no

5. yes

6. No, unless [Cl-] is very large,

K = 5.3 x 10-4

The End

39

Mercury(I) ChlorideMercury(I) ChlorideMercury(I) ChlorideMercury(I) ChlorideHgHg22ClCl22(s) <--> Hg(s) <--> Hg22

++(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)

KKspsp = 1.1 x 10 = 1.1 x 10-18-18

Lead(II) ChlorideLead(II) ChlorideLead(II) ChlorideLead(II) ChloridePbClPbCl22(s) <--> Pb(s) <--> Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)

KKspsp = 1.9 x 10 = 1.9 x 10-5-5

Silver ChlorideSilver ChlorideSilver ChlorideSilver ChlorideAgCl(s) <--> AgAgCl(s) <--> Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

KKspsp = 1.8 x 10 = 1.8 x 10-10-10

40

Ksp from SolubilityKsp from Solubility1. A saturated solution of CuCl has a gram

solubility of 0.05643 g/L. Calculate the Ksp.

(0.05643g/L)(1 mole/99.0g) = 0.000570 M(0.05643g/L)(1 mole/99.0g) = 0.000570 M

CuCl(s) <--> CuCuCl(s) <--> Cu++(aq) + Cl(aq) + Cl--(aq)(aq)

- 0.000570

0.0005700.000570

0.0005700.000570

Ksp = [Cu+] [Cl-]

= (0.000570)(0.000570)

= 3.25 x 10-7

SolidSolid

SolidSolid

41

Ksp from SolubilityKsp from Solubility2. A saturated solution of PbBr2 has

[Pb2+] = 1.05 x 10-1 M. Calculate the Ksp.

PbBrPbBr22(s) <--> Pb(s) <--> Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

- 0.0105

0.02100.0105

0.02100.0105

Ksp = [Pb2+] [Cl-]2

= (0.0105)(0.0210)2

= 4.63 x 10-3

Solid

Solid

42

Ksp from SolubilityKsp from Solubility3. A saturated solution of Ag2CrO4 has [Ag+] =

1.6 x 10-4 M. Calculate the Ksp.

Ag2CrO4(s) <--> 2 Ag(s) <--> 2 Ag++(aq) + CrO(aq) + CrO442-2-(aq)(aq)

8.0 x 10-51.6 x 10-4

8.0 x 10-51.6 x 10-4

Ksp = [Ag+]2 [CrOCrO442-2-]]

= (1.6 x 10-4)2(8.0 x 10-5)

= 2.0 x 10-12

Solid

Solid

- 8.0 x 10-5

43

Solubility from KspSolubility from Ksp

1. The Ksp of SrCO3 is 7.0 x10-10. Calculate the molar solubility of SrCO3.

SrCO3(s) <--> Sr(s) <--> Sr2+2+(aq) + CO(aq) + CO332-2-(aq)(aq)

- s

ss

ss

Ksp = [SrSr2+2+]] [COCO332-2-]]

= (s)(s) = s2 = 7.0 x 10-10

s = 2.6 x 10-5 M

Solid

Solid

44


2.The Ksp of Ca(OH)2 is 7.9 x10-6. Calculate the molar solubility of Ca(OH)2.

Ca(OH)2(s) <--> Ca(s) <--> Ca2+2+(aq) + 2 OH(aq) + 2 OH--(aq)(aq)

- s

2ss

2ss

Ksp = [CaCa2+2+]] [OH--]]22

= (s)(2s)2 = 4s3 = 7.9 x 10-6

s = 1.3 x 10-2 M

Solid

Solid

45


3.The Ksp of Al(OH)3 is 2.0 x 10-33. Calculate the molar solubility of Al(OH)3.

Al(OH)3(s) <--> Al(s) <--> Al3+3+(aq) + 3 OH(aq) + 3 OH--(aq)(aq)

- s

3ss

3ss

Ksp = [AlAl3+3+]] [OH--]]33

= (s)(3s)3 = 27s4 = 2.0 x 10-33

s = 2.9 x 10-9 M

Solid

Solid


Will mixing 200. mL 5.0 x 10-6 M

mercury(I) nitrate and 100. mL 5.0 x 10-8 M

sodium chloride cause a precipitate to form?

HgHg22ClCl22(s) (s) <--> <--> Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

Q = [HgQ = [Hg222+2+] [Cl] [Cl--] ] 22

[Hg[Hg222+2+] = ] = 5.0 x 10-6 (200./300.) = 3.3 x 10-6 M

[Cl[Cl--] = ] = 5.0 x 10-8 (100./300.) = 1.7 x 10-8 M

Q = (Q = (3.3 x 10-6)()(1.7 x 10-8) ) 22 = 9.5 x 10 = 9.5 x 10-22-22

Q < KQ < Kspsp No ppt No ppt


Will mixing 100. mL 0.20 M magnesium

nitrate and 300. mL 0.40 M sodium oxalate

cause a precipitate to form?

MgCMgC22OO44(s) (s) <--> <--> Mg Mg2+2+(aq) + C(aq) + C22OO442-2-(aq)(aq)

Q = [MgQ = [Mg2+2+][C][C22OO442-2-]]

[Mg[Mg2+2+] = 0] = 0.20 (100./400.) = 0.050 M

[C[C22OO442-2-] = 0] = 0.40 (300./400.) = 0.30 M

Q = (0Q = (0.050)(0)(0.30) = 1.5 x 10) = 1.5 x 10-2-2

Ksp = 8.6 x 10-5 Q > K Q > Kspsp ppt ppt


Will mixing 1.0 L 0.00010 M sodium

chloride and 2.0 L 0.0090 M silver nitrate

cause a precipitate to form?

AgCl(s) AgCl(s) <--> <--> Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

Q = [AgQ = [Ag++][Cl][Cl--]]

[Ag[Ag++] = 0] = 0.0090 (2.0/3.0) = 0.0060 M

[Cl[Cl--] = 0] = 0.00010 (1.0/3.0) = 0.000033 M M

Q = (0Q = (0.0060)(0)(0.000033) = 2.0 x 10) = 2.0 x 10-7-7

(Ksp = 1.8 x 10-10) Q > K Q > Kspsp ppt ppt

49

Precipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble Salt

What [Sr2+] is required to ppt SrSO4 in a 0.20 M Na2SO4 solution? (Ksp = 2.8 x 10-7)

SrSO4(s) (s) <--> <--> Sr Sr2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)

0.20x

Ksp = [Sr2+] [SO42-]

2.8 x 10-7 = (x)(0.20)

x = 1.4 x 10-6 M = [Sr2+]

For ppt [Sr2+] > 1.4 x 10-6 M

50


How many moles of HCl are required to ppt AgCl from 100. mL 0.10 M AgNO3? (Ksp = 1.8 x 10-10)

AgCl(s) (s) <--> <--> Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

x0.10

Ksp = [Ag+] [Cl-] 1.8 x 10-10 = (0.10)(x)

x = 1.8 x 10-9 M = [Cl-]

1.8 x 10-9 mole/L)(0.100 L) = 1.8 x 10-10 mole

For ppt mole HCl > 1.8 x 10-10

51


Calculate [Cl-] required to ppt PbCl2 from 0.100 M Pb(NO3)2. (Ksp = 1.7 x 10-5)

PbCl2(s) (s) <--> <--> Pb Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

x0.100

Ksp = [Pb2+] [Cl-] 2

1.7 x 10-5 = (0.100)(x) 2

x = 1.3 x 10-2 M = [Cl-]

For ppt [Cl-] > 1.3 x 10-2 M

52


If [Cl-] is raised to 0.10 M, calculate [Pb2+]

PbCl2(s) (s) <--> <--> Pb Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

0.10x

Ksp = [Pb2+] [Cl-]2

1.7 x 10-5 = (x)(0.10)2

x = 1.7 x 10-3 M = [Pb2+]

53


100. mL 0.200 M silver nitrate is mixed with 100. mL 0.100 M hydrochloric acid. Calculate [Ag+] and [Cl-]. (Ksp = 1.8 x 10-10)

AgCl(s) (s) <-- <-- Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

20.0 10.020.0 10.0

010.0

-10.0 -10.0

[Ag+] = = 0.0500 M10.0

200.

54

AgCl(s) (s) <--> <--> Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

- x

x0.0500

xx

Ksp = [AgAg++]] [ClCl--]]

1.8 x 10-10 = (0.0500)(x)

x = 3.6 x 10-9 M = [Cl-]

0.0500


Solid

Solid

55

Common IonsCommon Ions1. The Ksp of SrCO3 is 7.0 x10-10. Calculate the

molar solubility of SrCO3 in 0.10 M Na2CO3.

SrCO3(s) <--> Sr(s) <--> Sr2+2+(aq) + CO(aq) + CO332-2-(aq)(aq)

- s

0.10s

ss

Ksp = [SrSr2+2+]] [COCO332-2-]]

= (s)(.10) = s2 = 7.0 x 10-10

s = 7.0 x 10-9 M

0.10

Remember in H2O: s= 2.6 x 10-5 M

Solid

Solid

56

Common IonsCommon Ions2. The Ksp of Ca(OH)2 is 7.9 x10-6. Calculate the

molar solubility of Ca(OH)2 in 0.50 M NaOH.

Ca(OH)2(s) <--> Ca(s) <--> Ca2+2+(aq) + 2 OH(aq) + 2 OH--(aq)(aq)

- s

0.50s

2ss

Ksp = [CaCa2+2+]] [OH--]]22

7.9 x 10-6 = (s)(0.50)2

s = 3.2 x 10-5 M

0.50


Solid

Solid

57

Common IonsCommon Ions3. The Ksp of Al(OH)3 is 2.0 x 10-33. Calculate the

molar solubility of Al(OH)3 in 1.0 M KOH.

Al(OH)3(s) <--> Al(s) <--> Al3+3+(aq) + 3 OH(aq) + 3 OH--(aq)(aq)

- s

1.0s

3ss

Ksp = [AlAl3+3+]] [OH--]]33

2.0 x 10-33 = (s)(1.0)3

s = 2.0 x 10-33 M

1.0


Solid

Solid

58

Common IonsCommon Ions4. Calculate the solubility of calcium chromate

in 0.0050 M calcium chloride. (Ksp = 7.1 x 10-4)

CaCrO4(s) <--> Ca(s) <--> Ca2+2+(aq) + CrO(aq) + CrO442-2-(aq)(aq)

- s

s0.0050 + s

ss

Ksp = [CaCa2+2+]] [CrOCrO442-2-]]

7.1 x 10-4 = (0.0050 + s)(s)

s = 2.4 x 10-2 M

0.0050Solid

Solid

59Separating Salts by Differences in KSeparating Salts by Differences in KspspSeparating Salts by Differences in KSeparating Salts by Differences in Kspsp

1. Separation of .10 M Ag+ and .10 M Pb2+

AgBr Ksp=3.3 x 10-13

PbBr2 Ksp=6.3 x 10-6

Plan: Add Br- until all AgBr is ppt,

but no PbBr2 is ppt.

a. Calculate [Br-] required to ppt.

b. Calculate [Ag+] left in solution.

The substance whose KThe substance whose Kspsp is first exceeded will is first exceeded will precipitate first. precipitate first.

The ion requiring the lesser amount of BrThe ion requiring the lesser amount of Br-- precipitate first. precipitate first.

60

AgBr(s) AgBr(s) <--> <--> Ag Ag++(aq) + Br(aq) + Br--(aq)(aq)

x.10

Ksp = [Ag+] [Br-]

3.3 x 10-13 = (.10)(x)

x = 3.3 x 10-12 = [Br-]

For ppt [Br-] > 3.3 x 10-12


61

PbBrPbBr22(s) (s) <--> <--> Pb Pb2+2+(aq) + 2 Br(aq) + 2 Br--(aq)(aq)

x.10

Ksp = [Pb2+] [Br-]2

6.3 x 10-6 = (.10)(x)2

x = 7.9 x 10-3 = [Br-]

For ppt [Br-] > 7.9 x 10-3


62

For ppt AgBr [Br-] > 3.3 x 10-12

For ppt PbBr2 [Br-] > 7.9 x 10-3

Therefore;

Start ppt of AgBr [Br-] > 3.3 x 10-12

Max. ppt of AgBr [Br-] > 7.9 x 10-3

At max. ppt of AgBr, what is the [Ag+] left in solution?


63

AgBr(s) AgBr(s) <--> <--> Ag Ag++(aq) + Br(aq) + Br--(aq)(aq)

7.9 x 10-3x

Ksp = [Ag+] [Br-]

3.3 x 10-18 = (x)(7.9 x 10-3)

x = 4.2 x 10-11 = [Ag+]

Good Separation: [Ag+] < 10-5


64


2. Separation of .10 M CO32- and .10 M C2O4

2-

BaCO3 Ksp=8.1 x 10-9

BaC2O4 Ksp=1.1 x 10-7

Plan: Add Ba2+ until all BaCO3 is ppt,

but no BaC2O4 is ppt.

a. Calculate [Ba2+] required to ppt.

b. Calculate [CO32-] left in solution.

The substance whose KThe substance whose Kspsp is first exceeded will is first exceeded will precipitate first. precipitate first.

The ion requiring the lesser amount of BaThe ion requiring the lesser amount of Ba2+2+ precipitate first. precipitate first.

65

BaCOBaCO33(s) (s) <--> <--> Ba Ba2+2+(aq) + CO(aq) + CO332-2-(aq)(aq)

.10x

Ksp = [BaBa2+2+]][COCO332-2-]

8.1 x 10-9 = (x)(.10)

x = 8.1 x 10-8 = [BaBa2+2+]]

For ppt [BaBa2+2+] ] > 8.1 x 10-8


66

BaCBaC22OO44(s) (s) <--> <--> Ba Ba2+2+(aq) + C(aq) + C22OO442-2-(aq)(aq)

.10x

Ksp = [Ba2+] [CC22OO442-2-]

1.1 x 10-7 = (x)(.10)

x = 1.1 x 10-6 = [Ba2+]

For ppt [Ba2+] > 1.1 x 10-6


67

For ppt BaCOBaCO33 [Ba2+] > 8.1 x 10-8

For ppt BaCBaC22OO44 [Ba2+] > 1.1 x 10-6

Therefore;

Start ppt of BaCOBaCO33 [Ba2+] > 8.1 x 10-8

Max. ppt of BaCOBaCO33 [Ba2+] > 1.1 x 10-6

At max. ppt of BaCOBaCO33, what is the

[CO32-] left in solution?


68

BaCOBaCO33(s) (s) <--> <--> Ba Ba2+2+(aq) + CO(aq) + CO332-2-(aq)(aq)

1.1 x 10-6 x

Ksp = [BaBa2+2+]][COCO332-2-]

8.1 x 10-9 = (1.1 x 10-6)(x)

x = 7.4 x 10-3 = [COCO332-2-]

Poor Separation: [COCO332-2-] > > 10-5


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PRECIPITATION REACTIONS Chapter 17 Part 2