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

• saturated solution, [ ]max

• equilibrium between:

solid crystals dissolved ions

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

1-(aq)

1st - H2O needs

to separate- H-bonding

needs to be overcome

- dissolving

- NRG input

1st

3rd

3rd

2nd

2nd - H2O removes

ions from the lattice- dissociation

- NRG input

3rd - hydration of ions

- Ion – dipole force

- with H2O

- NRG output

•solubility is a contest between:

• lattice energies vs hydration energies

• ionic bond strengths vs ion-dipole forces

• usually exothermic with increased disorder

• if undissolved solid is present, then it is a mixture with a saturated sol’n phase

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

(aq)

Then:ym+ x n-[A ] [B ]

Ksp = [A B ]x y

and [A B ] is constant (solid)x y

ym+ x n- Ksp = [A ] [B ]

If: Q = Ksp then it is a saturated solution

Q > Ksp there is noticeable precipitate

Q < Ksp unsaturated

• refers to the amount of solid that can dissolve not how much is in solution

• solubility units are g/100g or g/100 mL or g/L

To find Concentration from Solubility:

[ ]max = solubility Msolute and V

adjustment

0.591 mol/ L=

eg. Solubility of Mg(OH)2 is 3.45 g/100 mL.

0.591molsol Mg(OH) = x 2 L

58.33g x

mol

100mL

0.1L

= molar solubility

To find Solubility from Concentration:

Solubility = molar solubility x Msolute and V change

3.45g[Mg(OH) ] x 2 100mL

mol x

58.33g

0.1L

100mL

= 3.45 g/ 100 mL

1) Ksp from solubility

eg. Calculate the Ksp of Ag2CO3 given its

solubility of 0.0014 g/100g.

2 30.0014 g

[Ag CO ] = x100 g

1st - calculate the [ ] from solubility:100 g

x 0.1 L

1 mol

275.75 g

-5= 5.1 x 10 mol/ L

2nd - create ICE table.

- Since dealing with a solid, the equilibrium is simplified

+ 2-2 3 3Ag CO 2 Ag + CO ;

I

C

E

+ 2 2-sp 3K = [Ag ] [CO ]

0 0

-5and x = 5.1 x 10 mol/ L (from solubility)

+2 x + x+2 x + x

Q – not necessary

100 Rule – not necessary

+ 2 2-sp 3K = [Ag ] [CO ]

-5 2 -5= [2(5.1 x 10 )] (5.1 x 10 )

-13= 5.3 x 10

2) Solubility from Ksptypes: solubility, molar solubility, [ion]eq, amount

of solid that will dissolve

- all from calculating “x”

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

- when not given the Ksp and not asked to calculate it, find it in the reference table or the textbook

2+ 1-2PbI Pb + 2 I ; 2+ 1- 2

sp

-9

K = [Pb ][I ]

= 8.5 x 10

- as the solid is not included and the [ion]i = 0, we

can write ICE horizontally instead of vertically

ICE +x +2x

2 -9(x)(2x) = 8.5 x 10

2+ 1- 2 -9spK = [Pb ][I ] = 8.5 x 10

-3x = 1.3 x 10 mol/ L

3 -94 x = 8.5 x 102+

eq= [Pb ]

= the molar solubility

-3

21.3 x 10 mol

solubility of PbI = x L

461.00 g x

mol

= 0.060 g/ 100 mL

0.10 L

100 mL

Predicting Precipitation:

• Used to determine precipitation when mixing 2 sol’ns just like in double displacement rxns

Don’t by 100 - it is a unit 100 mL

1st – write down each reaction and calculate the V and C values

1+ 2-2 4 4Na CrO 2 Na + CrO

1 1V = 0.0050 L, C = 0.030 M

1+ 1-3 3AgNO Ag + NO

2 2V = 0.0010 L, C = 0.0050 M

+ 2-2 4 4Ag CrO 2 Ag + CrO ;

2nd – write down the equilibrium equation, calculate the new [ ] in the combined sol’n

+ 2 2-sp 4

-12

K = [Ag ] [CrO ]

= 1.1 x 10 1 1

total

C V

V

= 0.00083

2 2

total

C V

V

0.030M x 0.0050L

0.0060L

0.0050M x 0.0010L

0.0060L

= 0.025

3rd – calculate Q and compare with Ksp2Q = (0.025) (0.00083) -7= 5.2 x 10

a precipitate will form

-12Ksp (1.1 x 10 ) >