Unit 11

Solubility Equilibrium

• Substances are considered insoluble if a saturated

solution in water at 25ºC in less than 0.1 M

• Ksp has to do with solid substances usually

considered insoluble in water.

1) The substance is dissociated when dissolved.

2) There is an equilibrium between the undissolved solid and

the dissolved ions.

• Let us consider silver chloride, AgCl, as an example.

When it dissolves, it dissociates like this:

AgCl(𝑠) ⇌ Ag(𝑎𝑞)+ + Cl(𝑎𝑞)

−

• An equilibrium expression can be written:

𝐾𝑐 =Ag+ Cl−

AgCl

• But…

𝐾𝑐 AgCl = Ag+ Cl−

• Recognizing that Kc[AgCl] is itself a constant we can define

Ksp = Kc[AgCl]

• Which makes our equilibrium expression become

Ksp = [Ag+][Cl−]

• As always, coefficients in the balanced

equation become exponents in the equilibrium

expression.

Sn(OH)2(𝑠) ⇌ Sn(𝑎𝑞)2+ + 2OH(𝑎𝑞)

−

Ksp = [Sn2+][OH–]2

• Using Ksp we can find the molar solubility of a

compound.

• Molar solubility is the moles of a substance

that will dissolve in 1 L of water at 25ºC

• Calculate the molar solubility (in mol/L) of a saturated solution of silver chloride (Ksp = 1.77 × 10−10).

AgCl(𝑠) ⇌ Ag(𝑎𝑞)+ + Cl(𝑎𝑞)

−

• The Ksp expression is as before:

Ksp = [Ag+][Cl−]

• Therefore

1.77 × 10−10 = [Ag+][Cl−]

• Noticing that the ratio of Ag+ to Cl− in the equation is 1:1, we can conclude that

[Ag+] = [Cl−]

• If we let x be the concentration of both ions we have

1.77 × 10−10 = (x)(x) = x2

• Taking the square root of both sides we get

x = 1.33 × 10−5

• Thus the molar solubility of AgCl is 1.33 × 10−5 M

• Determine the molar solubility for tin(II) hydroxide

(Ksp = 5.45 × 10−27).

Sn(OH)2(𝑠) ⇌ Sn(𝑎𝑞)2+ + 2OH(𝑎𝑞)

−

• The Ksp expression is

Ksp = [Sn2+][OH−]2

5.45 × 10−27 = [Sn2+][OH−]2

• Examining the balanced equation we note that [OH−]

will be twice that of [Sn2+]

• If we let x be [Sn2+] then [OH−] will be 2x

• Or if we let x be [OH−] the [Sn2+] will be ½x

• Since we want the moles of Sn(OH)2, the wise choice

is for x to be [Sn2+] since they exist in a 1:1 ratio

5.45 × 10−27 = (x)(2x)2

5.45 × 10−27 = 4x3

• Don’t forget to square the coefficient also!

• Solving for x

x = 1.11 × 10−9

• Therefore the molar solubility of tin(II) hydroxide is 1.11 × 10−9 M

• Determine the molar solubility of calcium phosphate

(Ksp = 1.0 × 10−26).

Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43−

1.0 × 10−26 = [Ca2+]3[PO43−]2

• If we let x = [Ca2+] then the molar solubility will be ⅓x

• Or if we let x = [PO43−] then the molar solubility will be

½x.

• Since we are looking for the molar solubility, and not

one of the actual concentrations, the best choice is to let

x equal the molar solubility, then [Ca2+] = 3x, and

[PO43−] = 2x.

1.0 × 10−26 = (3x)3(2x)2

1.0 × 10−26 = 108x5

x = 2.5 × 10−6

• Therefore the molar solubility of calcium

phosphate is 2.5 × 10−6 M

• When one of the ions in a compound is already

present in a solution the compound will be less

soluble.

• This is because the product of the

concentrations in the equilibrium expression

will reach the value of Ksp much sooner.

• This is known as the common ion effect

• What is the molar solubility of silver chloride (Ksp = 1.77 × 10−10) in a 2.5 × 10−2 M solution of sodium chloride?

𝐴𝑔𝐶𝑙(𝑠) ⇌ 𝐴𝑔(𝑎𝑞)

+ + 𝐶𝑙(𝑎𝑞)−

Ksp = [Ag+][Cl−]

1.77 × 10−10 = [Ag+][Cl−]

• Next we need to make an ICE chart.

AgCl(s) <===> Ag+

(aq) + Cl−

(aq)

Initial − 0 0.025

Change − +x +x

Equilibrium − x 0.025 + x

• Our equilibrium expression becomes:

1.77 × 10−10 = (x)(0.025 + x)

• Since 0.025 >> 1.77 × 10−10 (by at least a factor of 100) 0.025 + x ≈ 0.025

• Which means our expression becomes:

1.77 × 10−10 = (x)(0.025)

x = 7.1 × 10−9

• Therefore the molar solubility is 7.1 × 10−9 M.

• Compare this to 1.33 × 10−5 M which we found for the solubility of AgCl in pure water!

• What is the molar solubility of calcium

phosphate (Ksp = 1.0 × 10−26) in a solution that

is 5.9 × 10−3 M in calcium nitrate?

Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43−

1.0 × 10−26 = [Ca2+]3[PO43−]2

• Next we need to make an ICE chart.

Ca3(PO4)2 <===> Ca2+

+ PO43−

Initial − 0.0059 0

Change − +3x +2x

Equilibrium − 0.0059 + 3x 2x

• Our equilibrium expression becomes:

1.0 × 10−26 = (0.0059 + 3x)3(2x)2

• Since 0.0059 >> 1.0 × 10−26 (by at least a

factor of 100) 0.0059 + 2x ≈ 0.0059

• Which means our expression becomes:

1.0 × 10−26 = (0.0059)3(2x)2

x = 1.1 × 10−10

• Therefore the molar solubility of calcium

phosphate is 1.1 × 10−10 M.

A solution is 1.97 × 10−5 M lead(II) ions and

8.45 × 10−7 M in iron(III) ions. In order to

separate them, hydroxide ions will be added.

For Fe(OH)3 Ksp = 2.79×10−39 and for Pb(OH)2

Ksp = 1.43×10−20.

(a) What will the concentration of hydroxide

ions be when the first precipitate begins to

form? What is that precipitate?

Fe(OH)3

2.79×10−39 = (8.45 × 10−7)[OH-]3

[OH-] = 1.49 × 10−11

Pb(OH)2

1.43×10−20 = (1.97 × 10−5) [OH-]2

[OH-] = 2.69 × 10−8

Precipitate is Fe(OH)3

[OH-] = 1.49 × 10−11

(b) What will the concentration of hydroxide

ions be when the second precipitate begins to

form? What is the precipitate?

[OH-] = 2.69 × 10−8

Pb(OH)2

(c) What will the concentration of the first ion be

when the second begins to precipitate?

2.79×10−39 = [Fe3+](2.69 × 10−8)3

[Fe3+] = 1.43 × 10−16 M

(d) If the solution has a volume of 56 mL how many moles of hydroxide ions will have been added just as the second precipitate begins to form?

The [Fe3+] changed from 8.45 × 10−7 M to 1.43 × 10−16 M, for a difference of 8.45 × 10−7 M

Fe(OH)3 ⇌ Fe3+ + 3OH−

3 Times as much OH- is needed as Fe3+

[OH-] = 3 × (8.45 × 10−7) = 2.54 × 10−6 M

Moles OH- = (2.54 × 10−6 M)(0.056 L) = 1.42 × 10−7 moles

• (e) When the second precipitate just begins to form,

what mass of the first precipitate has been formed?

1.42 × 10−7𝑚𝑜𝑙 𝑂𝐻−

1×

1 𝑚𝑜𝑙 𝐹𝑒(𝑂𝐻)3

3 𝑚𝑜𝑙 𝑂𝐻−×

106.8 𝑔 𝐹𝑒(𝑂𝐻)3

1 𝑚𝑜𝑙 𝐹𝑒(𝑂𝐻)3= 5.06