Precipitation reactions chapter 17 part 2
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PRECIPITATION REACTIONS Chapter 17 Part 2. Insoluble Chlorides. All salts formed in this experiment are said to be INSOLUBLE and form precipitates when mixing moderately concentrated solutions of the metal ion with chloride ions. Insoluble Chlorides.

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PRECIPITATION REACTIONS Chapter 17 Part 2

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Precipitation reactions chapter 17 part 2

PRECIPITATION REACTIONSChapter 17 Part 2


Insoluble chlorides

InsolubleChlorides

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


Insoluble chlorides1

InsolubleChlorides

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

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

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


Insoluble chlorides2

InsolubleChlorides

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

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

This is equivalent to the SOLUBILITYof AgCl.

What is [Cl-]?

This is also equivalent to the AgCl solubility.


Precipitation reactions chapter 17 part 2

InsolubleChlorides

Make a chart.

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

some 00

- 1.34 x 10-5

1.34 x 10-5

1.34 x 10-5

some - 1.34 x 10-5

1.34 x 10-5

1.34 x 10-5


Insoluble chlorides3

InsolubleChlorides

Ksp = [Ag+] [Cl-]

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

= 1.80 x 10-10

Ksp = solubility product constant

See Table 18.2 and Appendix J 18A & 18B


Lead ii chloride

Lead(II) Chloride

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

Ksp = 1.9 x 10-5


Solubility of lead ii iodide

Solubility of Lead(II) Iodide

Consider PbI2 dissolving in water

PbI2(s) Pb2+(aq) + 2 I-(aq)

Calculate Ksp if solubility =0.00130 M

Solution

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

[I-] = _____________ ?

2(1.30 x 10-3 M)


Solubility of lead ii iodide1

Solubility of Lead(II) Iodide

Consider PbI2 dissolving in water

PbI2(s) Pb2+(aq) + 2 I-(aq)

Calculate Ksp if solubility =0.00130 M

Solution

1.Solubility = [Pb2+] = 1.30 x 10-3 M

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


Solubility of lead ii iodide2

Solubility of Lead(II) Iodide

Consider PbI2 dissolving in water

PbI2(s) Pb2+(aq) + 2 I-(aq)

Calculate Ksp if solubility = 0.00130 M

Solution

1.Solubility = [Pb2+] = 1.30 x 10-3 M

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

2.Ksp = [Pb2+] [I-]2

= [Pb2+] {2 • [Pb2+]}2

= 4 [Pb2+]3


Solubility of lead ii iodide3

Solubility of Lead(II) Iodide

Consider PbI2 dissolving in water

PbI2(s) Pb2+(aq) + 2 I-(aq)

Calculate Ksp if solubility = 0.00130 M

Solution

2.Ksp = 4 [Pb2+]3 = 4 (solubility)3

Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9

Sample Problems


Precipitating an insoluble salt

Precipitating an Insoluble Salt

Hg2Cl2(s) <--> Hg22+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-] 2

If [Hg22+] = 0.010 M, what [Cl-] is required to just begin the precipitation of Hg2Cl2?

What is the maximum [Cl-] that can be in solution with 0.010 M Hg22+ without forming Hg2Cl2?


Precipitating an insoluble salt1

Precipitating an Insoluble Salt

Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Recognize that

Ksp = product of maximum ion concentrations.

Precipitation begins when product of ion concentrations EXCEEDS the Ksp.


Precipitating an insoluble salt2

K

sp

-

-18

[

Cl

]

=

=

1.1 x 10

M

4(0.010)

Precipitating an Insoluble Salt

Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [2Cl-]2

Solution

[Cl-] that can exist when [Hg22+] = 0.010 M,

If this concentration of Cl- is just exceeded, Hg2Cl2 begins to precipitate.


Precipitating an insoluble salt3

Precipitating an Insoluble Salt

Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Now raise [Cl-] to 1.0 M.

What is the value of [Hg22+] at this point?

Solution

[Hg22+] = Ksp / [Cl-]2

= Ksp / (1.0)2 = 1.1 x 10-18 M

The concentration of Hg22+ has been reduced by 1016 !

Sample Problems


Review problems

REVIEW 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.


Review problems1

REVIEW 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 forCaF2, 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.


Separating metal ions cu 2 ag pb 2

Separating Metal Ions Cu2+, Ag+, Pb2+

Ksp Values

AgCl1.8 x 10-10

PbCl21.7 x 10-5

PbCrO4 1.8 x 10-14


Separating salts by differences in k sp

Separating Salts by Differences in Ksp

A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?

Ksp for Ag2CrO4 = 9.0 x 10-12

Ksp for PbCrO4 = 1.8 x 10-14

Solution

The substance whose Ksp is first exceeded will precipitate first.

The ion requiring the lesser amount of CrO42- precipitate first.

19


Separating salts by differences in k sp1

Separating Salts by Differences in Ksp

A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?

Ksp for Ag2CrO4 = 9.0 x 10-12

Ksp for PbCrO4 = 1.8 x 10-14

Solution

[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]

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

[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2

= 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M

PbCrO4 precipitates first.


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 precipitates first.

Ksp (Ag2CrO4)= 9.0 x 10-12

Ksp (PbCrO4) = 1.8 x 10-14

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

Solution

We know that [CrO42-] = 2.3 x 10-8 M to begin to precipitates Ag2CrO4.

What is the Pb2+ concentration at this point?


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 precipitates first.

Ksp (Ag2CrO4)= 9.0 x 10-12

Ksp (PbCrO4) = 1.8 x 10-14

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

Solution

[Pb2+] = Ksp / [CrO42-] = 1.8 x 10-14 / 2.3 x 10-8 M

= 7.8 x 10-7 M

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


Common ion effect adding an ion common to an equilibrium

Common Ion EffectAdding an Ion “Common” to an Equilibrium


The common ion effect

The Common Ion Effect

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

(b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

BaSO4(s) Ba2+(aq) + SO42-(aq)

Solution (a)

Solubility in pure water = [Ba2+] = [SO42-] = s

Ksp = [Ba2+] [SO42-] = s2

s = (Ksp)1/2 = 1.1 x 10-5 M


The common ion effect1

The Common Ion Effect

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

(b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

BaSO4(s) Ba2+(aq) + SO42-(aq)

Solution (b)

Now dissolve BaSO4 in water already containing 0.010 M Ba2+.

Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO4 be less than or greater than in pure water?___

Left

Less


The common ion effect2

The Common Ion Effect

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

(b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

BaSO4(s) Ba2+(aq) + SO42-(aq)

Solution (b)

[Ba2+] [SO42-]

initial

change

equilib.

0.010 0

+ s + s

0.010 + ss


The common ion effect3

The Common Ion Effect

Calculate the solubility of BaSO4 in:

(b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

BaSO4(s) Ba2+(aq) + SO42-(aq)

Solution

Ksp = [Ba2+] [SO42-] = (0.010 + s) (s)

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

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


The common ion effect4

The Common Ion Effect

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

(b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

BaSO4(s) Ba2+(aq) + SO42-(aq)

Solution

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

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

Le Chatelier’s Principle is followed!

Sample Problems


Review problems2

REVIEW PROBLEMS

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

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

    water.

    0.0025 M calcium nitrate.

    0.080 M sodium fluoride.

    Write appropriate net-ionic equations.


Solubility and ph

SOLUBILITY AND pH

  • We have discovered in Experiment 23 that salts of weak acids are generally soluble in acidic solutions. This principle is illustrated by combining the Ka equation with the Ksp equation. If we consider CaC2O4 in the presence of strong acid, the following is the net equilibrium equation:

    CaC2O4(s) + 2 H+ <======> H2C2O4(aq) + Ca+2

    Knet = Ksp. ( 1/Ka 1 ) . ( 1/Ka 2 )

  • Since Ka 1 and Ka 2 are both less than one, Knet > Ksp.

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


Solubility and complex ions

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.

    AgBr(s) + 2 NH3(aq) <=====> Ag(NH3)2+(aq) + Br-

    Knet = Ksp. Kf =

    ( 3.3 x 10-13 ) ( 1.6 x 107) = 5.3 x 10-6


Simultaneous equilibria

Simultaneous 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

1.7 x 10-5

1/1.8 x 10-14

9.4 x 108

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

Yes!


Simultaneous equilibria1

Simultaneous Equilibria

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+

1.8 x 10-10

1.6 x 107

2.9 x 10-3

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

No, unless very high [NH3]


Simultaneous equilibria2

Simultaneous Equilibria

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

2.3 x 10-9

1/6.4 x 10-5

1/5.9 x 10-2

6.1 x 10-4

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

No, unless very high [H+]


Review problems3

REVIEW PROBLEMS

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

  • Which precipitates first when I- is added?

    Ksp AgI = 1.5 x 10 -16 Ksp PbI2 = 8.7 x 10 -9.

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


Precipitation reactions chapter 17 part 2

Practice Problems

1. 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)


Precipitation reactions chapter 17 part 2

Practice Problems

4. 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


Precipitation reactions chapter 17 part 2

Practice 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


Mercury i chloride

Mercury(I) Chloride

Hg2Cl2(s) <--> Hg2+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18

Lead(II) Chloride

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

Ksp = 1.9 x 10-5

Silver Chloride

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

Ksp = 1.8 x 10-10


K sp from solubility

Ksp from Solubility

1. 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

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

Ksp = [Cu+] [Cl-]

= (0.000570)(0.000570)

= 3.25 x 10-7

Solid

- 0.000570

0.000570

0.000570

0.000570

0.000570

Solid


K sp from solubility1

Ksp from Solubility

2. A saturated solution of PbBr2 has [Pb2+] = 1.05 x 10-1 M. Calculate the Ksp.

PbBr2(s) <--> Pb2+(aq) + 2 Cl-(aq)

Ksp = [Pb2+] [Cl-]2

= (0.0105)(0.0210)2

= 4.63 x 10-3

Solid

- 0.0105

0.0105

0.0210

Solid

0.0105

0.0210


K sp from solubility2

Ksp from Solubility

3. A saturated solution of Ag2CrO4 has [Ag+] = 1.6 x 10-4 M. Calculate the Ksp.

Ag2CrO4(s) <--> 2 Ag+(aq) + CrO42-(aq)

Ksp = [Ag+]2 [CrO42-]

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

= 2.0 x 10-12

Solid

- 8.0 x 10-5

1.6 x 10-4

8.0 x 10-5

Solid

1.6 x 10-4

8.0 x 10-5


Solubility from k sp

Solubility from Ksp

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

SrCO3(s) <--> Sr2+(aq) + CO32-(aq)

Ksp = [Sr2+] [CO32-]

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

s = 2.6 x 10-5 M

Solid

- s

s

s

Solid

s

s


Solubility from k sp1

Solubility from Ksp

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

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

Ksp = [Ca2+] [OH-]2

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

s = 1.3 x 10-2 M

Solid

- s

s

2s

Solid

s

2s


Solubility from k sp2

Solubility from Ksp

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

Ksp = [Al3+] [OH-]3

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

s = 2.9 x 10-9 M

Solid

- s

s

3s

Solid

s

3s


Precipitating an insoluble salt4

Precipitating an Insoluble Salt

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? Hg2Cl2(s) <--> Hg22+(aq) + 2 Cl-(aq)

Q = [Hg22+] [Cl-] 2

[Hg22+] = 5.0 x 10-6 (200./300.) = 3.3 x 10-6 M

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

Q = (3.3 x 10-6)(1.7 x 10-8) 2 = 9.5 x 10-22

Q < Ksp No ppt


Precipitating an insoluble salt5

Precipitating an Insoluble Salt

Will mixing 100. mL 0.20 M magnesium nitrate and 300. mL 0.40 M sodium oxalate cause a precipitate to form?

MgC2O4(s) <--> Mg2+(aq) + C2O42-(aq)

Q = [Mg2+][C2O42-]

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

[C2O42-] = 0.40 (300./400.) = 0.30 M

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

Ksp = 8.6 x 10-5 Q > Ksp ppt


Precipitating an insoluble salt6

Precipitating an Insoluble Salt

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

Q = [Ag+][Cl-]

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

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

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

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


Precipitating an insoluble salt7

Precipitating 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) <--> Sr2+(aq) + SO42-(aq)

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

x

0.20


Precipitating an insoluble salt8

Precipitating an Insoluble Salt

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

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

0.10

x


Precipitating an insoluble salt9

Precipitating an Insoluble Salt

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

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

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

0.100

x


Precipitating an insoluble salt10

Precipitating an Insoluble Salt

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

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

Ksp = [Pb2+] [Cl-]2

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

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

x

0.10


Precipitating an insoluble salt11

10.0

200.

Precipitating an Insoluble Salt

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

20.0 10.0

-10.0

-10.0

10.0

0

[Ag+] =

= 0.0500 M


Precipitation reactions chapter 17 part 2

Precipitating an Insoluble Salt

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

Ksp = [Ag+] [Cl-]

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

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

Solid

0.0500

- x

x

x

Solid

0.0500

x


Common ions

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

Common Ions

1. The Ksp of SrCO3 is 7.0 x10-10. Calculate the molar solubility of SrCO3 in 0.10 M Na2CO3.

SrCO3(s) <--> Sr2+(aq) + CO32-(aq)

Ksp = [Sr2+] [CO32-]

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

s = 7.0 x 10-9 M

0.10

Solid

- s

s

s

Solid

s

0.10


Common ions1

Remember in H2O: s= 1.3 x 10-2M

Common Ions

2. 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) <--> Ca2+(aq) + 2 OH-(aq)

Ksp = [Ca2+] [OH-]2

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

s = 3.2 x 10-5 M

Solid

0.50

- s

s

2s

Solid

s

0.50


Common ions2

Remember in H2O: s= 2.9 x 10-9 M

Common Ions

3. 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) <--> Al3+(aq) + 3 OH-(aq)

Ksp = [Al3+] [OH-]3

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

s = 2.0 x 10-33 M

1.0

Solid

- s

s

3s

Solid

s

1.0


Common ions3

Common Ions

4. Calculate the solubility of calcium chromate in 0.0050 M calcium chloride. (Ksp = 7.1 x 10-4)

CaCrO4(s) <--> Ca2+(aq) + CrO42-(aq)

Ksp = [Ca2+] [CrO42-]

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

s = 2.4 x 10-2 M

Solid

0.0050

- s

s

s

Solid

0.0050 + s

s


Separating salts by differences in k sp2

Separating Salts by Differences in Ksp

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

AgBrKsp=3.3 x 10-13

PbBr2Ksp=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 Ksp is first exceeded will precipitate first.

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


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

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

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

.10

x


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

PbBr2(s) <--> Pb2+(aq) + 2 Br-(aq)

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

.10

x


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

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?


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

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

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

x

7.9 x 10-3


Separating salts by differences in k sp3

Separating Salts by Differences in Ksp

2. Separation of .10 M CO32- and .10 M C2O42-

BaCO3Ksp=8.1 x 10-9

BaC2O4Ksp=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 Ksp is first exceeded will precipitate first.

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


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

BaCO3(s) <--> Ba2+(aq) + CO32-(aq)

Ksp = [Ba2+][CO32-]

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

x = 8.1 x 10-8 = [Ba2+]

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

x

.10


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

BaC2O4(s) <--> Ba2+(aq) + C2O42-(aq)

Ksp = [Ba2+] [C2O42-]

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

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

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

x

.10


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

For ppt BaCO3 [Ba2+] > 8.1 x 10-8

For ppt BaC2O4 [Ba2+] > 1.1 x 10-6

Therefore;

Start ppt of BaCO3 [Ba2+] > 8.1 x 10-8

Max. ppt of BaCO3 [Ba2+] > 1.1 x 10-6

At max. ppt of BaCO3, what is the

[CO32-] left in solution?


Precipitation reactions chapter 17 part 2

Separating Salts by Differences in Ksp

BaCO3(s) <--> Ba2+(aq) + CO32-(aq)

Ksp = [Ba2+][CO32-]

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

x = 7.4 x 10-3 = [CO32-]

Poor Separation: [CO32-] > 10-5

1.1 x 10-6

x


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