Equilibrium & Solubility:The Solubility Product Constant, Ksp

1 © Prof. Zvi C. Koren 20.07.2010

Pb2+ + 2NO3- + 2K+ + 2I- PbI2(s) + 2K+ + 2NO3

-

Pb2+ + 2I- PbI2(s)

Gross rxn.: Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)

(recall double-replacement rxns)

Net-Ionic rxn.:

Precipitation Reactions

2 © Prof. Zvi C. Koren 20.07.2010

saturated

solution of

PbI2

PbI2(s) Pb2+ + 2I–

Solubility of a Salt = S, units of “moles/L”

Equilibrium

between:

dissolved

and

undissolved

(precipitate)PbI2 (s)

Pb2+ 2I–

Solubility of a Salt

3 © Prof. Zvi C. Koren 20.07.2010

Al2S3(s) 2Al3+ + 3S2-

K = [Al3+]2 [S2-]3sp

solubility product constant

Let x = solubility of salt: x moles salt dissolved / L

2x 3x

= (2x)2 (3x)3 = 108x5

For example,

two types of problems: Solubility Ksp

Note:

Don’t confuse “solubility” with “solubility product constant”

Solubility Product Constant, Ksp

4 © Prof. Zvi C. Koren 20.07.2010

5 © Prof. Zvi C. Koren 20.07.2010

Problem:

Calculate the solubility of magnesium fluoride in g/L.

Ksp = 6.4x10–9 @250C.

Solution:

Case 1: Ksp Solubility

1. Write the equilibrium dissolution rxn: MgF2(s) Mg2+ + 2F–

2. Write the Ksp expression: Ksp = [Mg2+][F–]2

3. Do the stoichometric “ICE-box” right on the equilibrium rxn.

Let x = solubility of the salt in moles/L.

x 2x

4. Calculate using Ksp expression: Ksp = (x)(2x)2

6.4x10–9 = 4x3

Solubility of MgF2 = x = 1.17x10–3 mole of MgF2 dissolves/L @250C.

in g/L: (1.17x10–3 mole of MgF2 /L)(62.3018 g/mole) = 0.073 g MgF2 /L

6 © Prof. Zvi C. Koren 20.07.2010

Problem:

Calculate the solubility product constant, Ksp, of magnesium fluoride.

Solubility = 1.17x10–3 mole/L @250C = [Mg2+]

Solution:1. Write the equilibrium dissolution rxn: MgF2(s) Mg2+ + 2F–

2. Write the Ksp expression: Ksp = [Mg2+][F–]2

Case 2: Solubility Ksp

3. Do the stoichometric “ICE-box” right on the equilibrium rxn.

Let x = solubility of the salt in moles/L = 1.17x10–3 mole/L

x 2x

4. Calculate using Ksp expression:

Ksp = (x)(2x)2 = (1.17x10–3)(2.34x10–3)2 = 6.41x10–9

Experimental

Determination of

Ion Concentrations

Atomic Absorption (AA)

Spectrometer

7 © Prof. Zvi C. Koren 20.07.2010

Ksp in your Kishkes

8 © Prof. Zvi C. Koren 20.07.2010

The Meaning of the Ion-Product in Ksp

Consider MX(s) M+ + X–

Ksp = [M+][X–]

This equation pertains to:

• just before precipitation, or

• after precipitation

Maximum Concentrations BEFORE Precipitation:

[M+] and [X–] are the maximum concentrations that can “live with each

other” (“peaceful coexistence”) without “attacking each other” and

creating a ppt.

Maximum Concentrations AFTER Precipitation:

[M+] and [X–] are the maximum concentrations that can “live with each

other” after a precipitate of MX(s) is formed, no matter how much ppt is

present: Saturated solution.9 © Prof. Zvi C. Koren 20.07.2010

Al2S3(s) 2Al3+ + 3S2-

K = [Al3+]2 [S2-]3sp

Recall:

eq eq

Q = [Al3+]2 [S2-]3no ppt ppt no ppt;

equil.:

max concs.

“ppt” = precipitate

Unsaturatedsolution

Saturatedsolution

Reaction Quotient, Q (Again):Can a Precipitation Reaction Occur?

Q < Ksp Q > Ksp Q = Ksp

(continued)10 © Prof. Zvi C. Koren 20.07.2010

Question:

50.0 mL of 0.34 M Pb(NO3)2 solution is mixed with 25.0 mL of

0.10 M NaCl. Will a precipitate be formed?

Answer:

Overall rxn.: Pb(NO3)2 (aq) + NaCl(aq) PbCl2(s) + NaNO3(aq)2 2

Net ionic rxn.: Pb2+ + 2Cl– PbCl2(s)

?

?

Ksp = [Pb2+] [Cl–]2 = 1.7x10–5 Qsp = [Pb2+][Cl–]2

Calculate concentrations after mixing, but before rxn.:

[Pb2+] = (50.0/75.0)(3.4x10-1 M) = 2.27x10–1 M

[Cl–] = (25.0/75.0)(1.0x10-1 M) = 3.33x10–2 M

Q = (2.27x10-1)(3.33x10-2)2 = 2.5x10–4 > Ksp = 1.7x10–5 ppt!

What are the concentrations of all ions and how many moles of

precipitate are obtained? (see next slide)

eq eq

Can a Precipitation Reaction Occur?

11 © Prof. Zvi C. Koren 20.07.2010

Net ionic rxn.: Pb2+ + 2Cl- PbCl2(s),

Ksp

What are the concentrations of all ions and how many moles of

precipitate are obtained?

0.227 0.0333I

-0.0333 +0.0167-0.0167CE,I 0.210 - 0.0167

C +x +2x -x

E 0.210+x 2x 0.0167-x

1.7x10-5 = = [Pb2+][Cl-]2

= (0.210+x)(2x)2 (0.210)(4x2), x = 4.5x10-3 M

Concentrations of ions:

[Cl-] = 2x = 9.0x10-3 M

[Pb2+] = 0.210+x 0.21 M

Moles of ppt formed:

PbCl2(s): 0.0167-0.0045 (0.0122 mole/L)x(0.075 L) = 9.15x10-4 mole

K >> 1

Check: Ksp = [Pb2+][Cl-]2 1.7x10-5

(continued)

K >> 1:

12 © Prof. Zvi C. Koren 20.07.2010

MX(s) M+ + X–sparingly soluble salt:+ MA(aq) M+ + A- or LX(aq) L+ + X–soluble salt:

What happens to the solubility of a salt when there is a common ion?

according to Le Chatelier’s Principle, the solubility of the sparingly

soluble salt will decrease in the presence of a common ion.

The Common Ion Effect (Again)

13 © Prof. Zvi C. Koren 20.07.2010

Question:

The solubility of AgCl is 0.0019 g/L.

If some AgCl is placed in 500.0 mL of a 0.50 M solution of NaCl, how

many grams of AgCl dissolve?

Answer:

Note the presence of 2 salts, one sparingly soluble and one soluble.

0.50 M

2. Write the equilibrium dissolution rxn: AgCl(s) Ag+ + Cl–

1. Write the total dissolution rxn: NaCl(aq) Na+ + Cl–

4. Do the stoichometries (and ICE-box for sp. sol. salt) right on the rxns.

Let x = solubility of the salt in moles/L

3. Write the Ksp expression for the sparingly soluble salt:

Ksp = [Ag+][Cl–] = (x)(x+0.50) (x)(0.50)

0.50 M

x x+0.50

(continued)

The Common Ion Effect - Calculations

14 © Prof. Zvi C. Koren 20.07.2010

(continued)

Need to calculate Ksp from solubility:

Given: The solubility of AgCl is 0.0019 g/L.

From before: Ksp = [Ag+][Cl–] = (x)(x+0.50) (x)(0.50)

x = Solubility = 0.0019 g/L (1 mole/143.321 g) 1.33x10-5 M

[Ag+] = x = 1.33x10-5 M, [Cl-] = x = 1.33x10-5 M.

Ksp = 1.77x10-10

1.77x10-10 = Ksp = [Ag+][Cl–] = (x)(x+0.50) (x)(0.50)

x = 3.54x10-10 mole/L 5.07x10-8 g AgCl dissolve /L (0.500 L)

2.5x10-8 g AgCl dissolve!

15 © Prof. Zvi C. Koren 20.07.2010

Selective Separation of Ions & Solubility

Ag+

Pb2+

Cu2+

Cl–

AgCl(s)

PbCl2(s)

Cu2+

Δ

AgCl(s)

Pb2+ + CrO42– PbCrO4(s)

16 © Prof. Zvi C. Koren 20.07.2010

Separation of Two Ions by Difference in Solubility: Cl– and CrO42–

Which salt is more insoluble, AgCl or Ag2CrO4?

Ksp: 1.8x10–10 9.0x10–12

Check Solubilities!!! Let x = solubility of a salt in moles/L:

x x

Ksp = [Ag+][Cl-] = x2 x = 1.3x10–5 mol/L

Ag2CrO4(s) 2Ag+ + CrO42– ,

2x x

Ksp = [Ag+]2 [CrO42–] = (2x)2(x) x = 1.3x10–4 mol/L

Surprise, surprise!

(continued)

Selective Separation of Ions & Solubility - Calculations

AgCl(s) Ag+ + Cl– ,

17 © Prof. Zvi C. Koren 20.07.2010

Question:

A solution contains 0.010 M NaCl and 0.0010 M K2CrO4.

Solid AgNO3 is added slowly to the solution.

a) Which precipitates first?

b) What is [Cl-] when the second precipitate begins to form?

Answer:

First calculate [Ag+] required to just begin the precipitation of each salt:

To just begin precipitation of AgCl when [Cl–] = 0.010 M:

Ksp = [Ag+][Cl-] [Ag+] = 1.8x10-8 M.

To just begin precipitation of Ag2CrO4 when [CrO42–] = 0.0010 M:

Ksp = [Ag+]2 [CrO42–] [Ag+] = 9.5x10-5 M.

AgCl precipitates first!

When does Ag2CrO4 begin to precipitate?

When [Ag+] = 9.5x10-5 M Ksp = [Ag+][Cl-] [Cl-] = 1.9x10-6 M

% Cl- remaining in solution = 100x(1.9x10-6 M)/(0.010 M)= 0.019%

Separation of Two Ions by Difference in Solubility: Cl– and CrO42–

Selective Separation of Ions & Solubility (continued)

18 © Prof. Zvi C. Koren 20.07.2010

PbCl2(s) + CrO42– PbCrO4(s) + 2Cl–, K = ?

PbCl2(s) Pb2+ + 2Cl–, K1 = Ksp(PbCl2)

Pb2+ + CrO42– PbCrO4(s), K2 = 1/Ksp(PbCrO4)

?

K = K1•K2 = 9.4x108

Simultaneous Equilibria:

Two or more equilibria rxns occurring at the same time

When a salt solution is added to a salt precipitate … Check Krxn!!!

PbCl2(s) is only sparingly soluble in water (Ksp = 1.7x10–5).

What is its solubility in certain other ionic solutions, e.g., CrO42– ?

PbCl2 is “soluble” in a solution of CrO42–, but it becomes another ppt.

Simultaneous Equilibria

19 © Prof. Zvi C. Koren 20.07.2010

Some sparingly soluble salts are more soluble in water than normally

expected. Why???

Answer: Because the resultant anion can undergo hydrolysis as a base.

For example:

PbS(s) Pb2+ + S2–, Ksp(PbS) = 3.2x10–28

AND

S2– + H2O HS– + OH–, K = Kb(S2–) = Kw/Ka(HS–) =

*(Note: For H2S: Ka1 = 1x10–7, Ka2 = 1x10–19)

K2(H2S)*

1x105

PbS(s) + H2O Pb2+ + HS– + OH–,

K = 3x10–23 >> Ksp

(continued)

PbS is more soluble in water than expected

Solubility and pH

20 © Prof. Zvi C. Koren 20.07.2010

How can we determine the ability of a salt to be dissolved by an acid?(A powerful display of the wonders of chemical principles!!!)

Example: CaCO3(s) + H+? What is the value of K for this rxn?

The Trick: “Build” this rxn from other known salt solubility and acid-

base hydrolysis rxns.:

CaCO3(s) Ca2+ + CO32-, K = Ksp = 3.8x10-9

CO32- + H2O HCO3

- + OH-, K = Kb(CO32-) = Kw/Ka(HCO3

-) = 2.1x10-4

H+ + OH- H2O, K = 1/ Kw = 1.0x1014

CaCO3(s) + H+ Ca2+ + HCO3

-, K = 80.0 > 1 (but read on)

But Also:

HCO3- + H+

H2CO3, K = 1/Ka1 = 1/4.2x10-7 = 2.4x106

But Also:

H2CO3(aq) CO2(g) + H2O, K 105

In general: Must examine all possible subsequent rxns.

CaCO3(s) + 2H+ Ca2+ + CO2(g) + H2O, K 1013

(Note: CO2(g) bubbles out of system, so equil. moves even more to right!)

Solubility and pH – Salt in an Acidic Solution

21 © Prof. Zvi C. Koren 20.07.2010

Question:

The Ksp of Mg(OH)2 is 1.5x10-11.

If solid Mg(NO3)2 is added to a

solution with a pH of 9.00,

at what [Mg2+] does precipitation

begin?

Answer:

What happens to the solubility of

this hydroxide as the pH is changed,

up or down?

Mg(OH)2(s) Mg2+ + 2OH-,

Ksp = [Mg2+][OH-]2

pH = 9.00 [OH-] = 1.0x10-5 M.

[Mg2+] = 0.15 M.

Solubility and pH – Metal Hydroxides

22 © Prof. Zvi C. Koren 20.07.2010

Complex Metal surrounded by two or more groups

(neutral or cation) (anion or molecule )

Complex-Ion Formation and “Formation Constant”:

Ag+ + 2NH3(aq) Ag(NH3)2+,

Rxn in two steps:

Ag+ + NH3(aq) Ag(NH3)+, K1

Ag(NH3)+ + NH3(aq) Ag(NH3)2

+, K2

Kf = K1•K2 = 1.6x107

Dissolving a Precipitate by Complex-Ion Formation:

“ligands”

AgCl(s) + 2NH3(aq) Ag(NH3)2+ + Cl-. K = ?

If K > Ksp, then bingo!

We know that the solubility of AgCl(s) in water is only slight, Ksp =

1.8x10-10. Can another solvent, e.g., NH3(aq), better disssolve the solid?

Build this rxn from others. That’s the trick!

AgCl(s) Ag+ + Cl-, Ksp

Ag+ + 2NH3(aq) Ag(NH3)2+, Kf

K = Ksp•Kf

= 2.9x10-3. Yesh!

Solubility and Complex Ion Formation

23 © Prof. Zvi C. Koren 20.07.2010

Problem:

How many moles of ammonia must be added to dissolve 0.050 mol

of solid AgCl present in 1.0 L of water?

Answer:

AgCl(s) + 2NH3(aq) Ag(NH3)2+ + Cl-,

0.050 mol/LI

C -0.050 +0.050 +0.050

E - y-0.10 0.050 0.050

2.9x10-3 = K = [Ag(NH3)2+][Cl-] / [NH3]

2

y-0.10 = [NH3] = 0.93 M

y

-0.10

y = 1.03 mol/L

And

n = (1.03 mol/L) x (1 L) = 1.03 mol NH3 must be added initially.

K = 2.9x10-3.

Solubility and Complex Ion Formation - Calculations

24 © Prof. Zvi C. Koren 20.07.2010