Equilibrium and Solubility

This lesson covers: 

1. What the solubility product (K_sp) is
2. Calculating solubility from K_sp and vice versa
3. Predicting precipitation using K_sp
4. The common ion effect

Solubility product (K_sp)

Even 'insoluble' ionic compounds may dissolve to a very small extent in water.

An equilibrium is established between the undissolved solid and its ions in a saturated solution:

M_aX_b(s) ⇌ aMⁿ⁺_(aq) + bX^m-_(aq)

The equilibrium constant expression for this process is:

Kc​=[Ma​Xb(s)​][M(aq)n+​]a[X(aq)m−​]b​

As the concentration of a solid is constant, it can be combined with K_c to give the solubility product (K_sp):

K_sp = [Mⁿ⁺_(aq)]^a [X^m-_(aq)]^b

Where:

• a = number of Mⁿ⁺ cations in one formula unit
• b = number of X^m- anions in one formula unit

For example, the K_sp expression for Fe₂S₃ is:

K_sp = [Fe³⁺_(aq)]² [S^2-_(aq)]³

The units of K_sp depend on the number of each type of ion in the formula, for example, the units of K_sp for Fe₂S₃ are mol⁵ dm^-15.

The lower the K_sp value, the lower the solubility of the compound. K_sp is only applicable to sparingly soluble salts.

Worked example 1 - Calculating K_sp from solubility

Calculate the K_sp of silver bromide (AgBr) if its solubility is 5.71×10−7 mol dm−3 at 298 K. 

Step 1: Write the equilibrium equation

AgBr_(s) ⇌ Ag⁺_(aq) + Br^-_(aq)

Step 2: Calculate the concentration of each ion

[Ag+]=[Br−]=5.71×10−7 mol dm−3

Step 3: Write the K_sp expression

K_sp = [Ag⁺][Br^-]

Step 4: Substitution and correct evaluation

Ksp​=(5.71×10−7)×(5.71×10−7)=3.26×10−13

Step 5: Determine the units 

K_sp will have units of (mol dm^-3)² = mol² dm^-6. 

Therefore, the K_sp of silver bromide is 3.26 x 10^-13 mol² dm^-6.

Worked example 2 - Calculating solubility from K_sp

Calculate the solubility of lead chloride (PbCl₂) given that K_sp for PbCl₂ is 1.7×10−5 mol³ dm^-9 at 298 K.

Step 1: Write the equilibrium equation

PbCl_2(s) ⇌ Pb²⁺_(aq) + 2Cl^-_(aq)

Step 2: Write the K_sp expression

Let the solubility of PbCl₂ = x mol dm−3

Ksp​=[Pb2+][Cl−]2=x×2x2=4x3

Step 4: Substitution and correct evaluation

1.7×10−5=4x3

x=3√4(1.7×10−5)​​=1.6×10−2 mol dm−3

The solubility of PbCl2​ is 1.6×10−2 mol dm−3

Predicting precipitation using K_sp

K_sp can be used to predict if a precipitate will form when two solutions are mixed.

• If the product of [Mⁿ⁺] and [X^m-] exceeds K_sp, precipitation occurs.
• If the product is less than K_sp, no precipitate forms.

Worked example 3 - Predicting precipitation using K_sp

Determine if a precipitate of CaF₂ will form when 100 cm³ of 1.00 x 10^-3 mol dm^-3 Ca(NO₃)₂ is mixed with 200 cm³ of 1.50 x 10^-3 mol dm^-3 NaF. 

K_sp of CaF₂ = 3.9 x 10^-11 mol³ dm^-9. 

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

100 cm³ = 0.100 dm³

200 cm³ = 0.200 dm³

Step 2: Calculate number of moles of Ca²⁺ and F^- before mixing

Ca²⁺:  n = c× V=(1.00×10−3)×0.100=1.00×10−4 mol

F^-:  n = c× V=(1.50×10−3)×0.200=3.00×10−4 mol

Step 3: Calculate [Ca²⁺] and [F^-] after mixing

Total volume after mixing = 0.300 dm³

[Ca2+]=Vn​=0.300(1.00×10−4)​=3.33×10−4 mol dm−3

[F−]=Vn​=0.300(3.00×10−4)​=1.00×10−3 mol dm−3

Step 2: Calculate [Ca²⁺][F^-]²

[Ca2+][F−]2=(3.33×10−4)×(1.00×10−3)2=3.33×10−10 mol3 dm−9

Step 3: Compare [Ca²⁺][F^-]² with K_sp to determine precipitation

3.3 x 10^-10 > 3.9 x 10^-11

As [Ca²⁺][F^-]² > K_sp, a precipitate of CaF₂ will form when the two solutions are mixed.

The common ion effect

The common ion effect refers to the reduced solubility of an ionic compound when a soluble salt containing one of the same ions is added to the solution. This can lead to precipitation.

For example, consider the equilibrium for AgCl:

AgCl_(s) ⇌ Ag⁺_(aq) + Cl^-_(aq)

Adding NaCl introduces the common ion Cl^-, shifting the equilibrium position to the left and precipitating AgCl. This occurs when [Ag⁺][Cl^-] exceeds the K_sp for AgCl.

The solubility of an ionic compound in a solution with a common ion is always lower than its solubility in pure water.

For instance, the solubility of silver chloride (AgCl) is:

• 1.3×10−5 mol dm−3 in water
• 1.8×10−9 mol dm−3 in 0.10 mol dm−3 NaCl(aq)​

We can explain this using the K_sp expression for AgCl:

Ksp​=[Ag+][Cl−]=1.8×10−10 mol2 dm−6

Ignoring the small [Cl^-] from AgCl, the value of [Ag⁺] in 0.10 mol dm^-3 NaCl_(aq) is:

[Ag+]=[Cl−]Ksp​​=0.10(1.8×10−10)​=1.8×10−9 mol dm−3

1.8 x 10^-9 < 1.3 x 10^-5 

Therefore, the common chloride ion from NaCl lowers the solubility of AgCl in NaCl_(aq) compared to water.