Solubility Product Constant, Ksp

Solubility:

• is the maximum amount of solute in a solvent at a

given temperature

• saturated solution, [ ]max

• equilibrium between: (see Dissolving salts OH)

solid crystals dissolved ions

eg. AgNO3(s) Ag1+(aq) + NO3

1-(aq)

• solubility is a contest between:

• ionic bond strengths vs ion-dipole forces

• lattice energies vs hydration energies

• usually exothermic with increased disorder

• if undissolved solid is present, then it is a mixture with

a saturated sol’n phase

Solubility Product Constant, Ksp: • equilibrium expression for slightly soluble salts

If: AxBy(s) xAm+(aq) + yBn-

(aq)

Then:

y]-n[B x]m[A Ksp

(solid)constant is ]yBx[A and

]yBx[A

y]-n[B x]m[A

Ksp

Saturation can be tested by Q:

If: Q = Ksp then: it is a saturation solution

Q > Ksp there is noticeable precipitate

Q < Ksp unsaturated

Terminology: • refers to the amount of solid that can dissolve not how

much is in solution • solubility in g/100g or g/100 mL or g/L

[ ]max = solubility x Msolute and volume adjustment

• molar solubility in mol/L (CONCENTRATION !)

= solubility / Msolute

Calculations: 1) Ksp from solubility

eg. Calculate the Ksp of Ag2CO3 given its solubility

of 0.0014 g/100g.

mol/L 10 x 5.1 and

+ 2 + E

+ 2 C

0 0 I

]CO[][Ag = K ; CO Ag 2 COAg

mol/L 10 x 5.1

g 275.75

mol 1 x

L 0.1

g 100 x

g 100

g 0.0014 ]CO[Ag

5-

-23

2sp

-2332

5-

32

x

xx

xx

Ksp = [Ag ]2[CO32- ]

(2(5.1 x 10 -5))2(5.1 x 10 -5 )

5.3 x 10 -13

2) Solubility from Ksp

types: solubility, molar solubility, [ion]eq, amount of

solid that will dissolve

eg. How much PbI2 at SATP will dissolve in 1.00 L of

water? Give the solute ion concentration, [Pb2+]eq,

solubility (g/100mL) and molar solubility.

(Ksp of PbI2 from textbook is 8.5 x 10-9)

xx

xx

2+ + E

2+ C

10 x 8.5 0 0 I

]I][[Pb = K ; I 2 Pb PbI

9-

2-12sp

-122

eq23-

9-3

9-2

9-2-12sp

][Pb mol/L 10 x 1.3

10 x 8.5 4

10 x 8.5 ))(2(

10 x 8.5 ]I][[Pb = K

x

x

xx

mL g/100 0.060

mL 100

L 0.10 x

mol

g 461.00 x

L

mol 10 x 1.3 PbI of solubility

mol/L 10 x 1.3 PbI of solubilitymolar

3-

2

3-2

Predicting Precipitation:

• Used to determine precipitation when mixing 2 sol’ns

as with double displacement rxns

eg. Will precipitation occur when 5.0 ml of 0.030 M

AgNO3 is mixed with 1.0 ml of 0.0050 M Na2CrO4?

(Ksp = 1.1 x 10-12 for Ag2CrO4).

AgNO3 Ag1 NO31- ; V1 0.0050 L, C1 0.030 M

Na2CrO4 2 Na 1 CrO42- ; V2 0.0010 L, C2 0.0050 M

Ag2CrO4 2 Ag CrO42- ; Ksp = [Ag ]2[CrO4

2- ] 1.1 x 10 -12

V1C1

Vtotal

V2C2

Vtotal

0.0050 x 0.030

0.0060

0.0010 x 0.0050

0.0060

= 0.025 0.00083

Q (0.025) 2(0.0083) 5.2 x 10 -7 > Ksp (1.1 x 10 -12 )

a precipitate will form

Common Ion Effect Application of Le Chatelier’s Principle to solubility

For:

AB2(s) A2+(aq) + 2 B1-

(aq)

can affect the equilibrium in:

i) forward direction by redissolving solid

ii) reverse direction by forming precipitate

Re-dissolving Precipitates

1. Add more solvent

then [A2+] & [B1-] shift

o AB A+ + B-

o A precipitate forms or exists if [A+][B-] >> ksp

o Adding solvent it lowers the ion concentrations

and equilibrium shifts to the right

2. Addition of an Acid

a) CaCO3(s) Ca2+ + CO32-

CO32- + 2H+ CO2 + H2O

As CO32- falls below ksp

The equilibrium shifts

To replace CO32- and CaCO3 dissolves

b) Cu(OH)2(s) Cu2+(aq) + 2OH1-

(aq)

H+ + OH- H2O

Equilibrium

Cu(OH)2 dissolves

3. Addition of a Complexing Agent

Cu(OH)2 Cu2+ (down) + 2OH-

Added:

Cu2+(aq) + NH3 Cu(NH3)4

2+

Equilibrium

More Cu(OH)2 dissolves

eg. Calculate the molar solubility of AgI in:

a) pure water

b) 0.10M NaI

a) Given that Ksp(AgI) = 4.5 x 10-17

AgI(s) Ag1+ + I1- ; Ksp = [Ag1+][I1-]

ICE +x +x = 4.5 x 10-17

Ksp = 4.5x10-17 = [x][x]

x = [Ag1+] = [I1-] = 6.7 x 10-9 M

b) NaI Na1+ + I-1 Ksp = infinite

0.1 0.1

AgI OH2 Ag1+ + I1-

Ksp = 4.5 x 10-17

ICE 0 0.1

Ksp = 4.5 x 10-17 = (x)(0.1)

x = 4.5 x 10-16

[Ag1+] = 4.5 x 10-16 = molar solubility