Chemistry 112

## The common ion effect in solubility.

The presence of a common ion reduces the solubility of a sparingly soluble salt. This is a direct consequence of LeChatelier's Principle.

Suppose some lead(II) chloride is dissolved in water and equilibrium is established:

PbCl2(s)Pb2+(aq) + 2 Cl(aq)

NaCl(s)Na+(aq) + Cl(aq)

The added chloride ion pushes the equilibrium reaction towards reactants, i.e. creating more solid lead(II) chloride and removing some of the lead ion from solution.

This can be an effective way to remove ions from solution.

#### Example

How much is the solubility of lead(II) chloride changed in the presence of 0.85 M NaCl?

Recall that lead(II) chloride has a solubility of 1.6×10–2 M in pure water.

In the presence of NaCl, we need to redo the calculation, accounting for the common ion.

 NaCl(s) Na+(aq) + Cl–(aq) PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Ksp = [Pb2+]e[Cl–]e2 = 1.6×10–5 Initial 0 0.85 Change + x + 2x Equilibrium x 0.85 + 2x

Since x < 1.6×10–2, 0.85 + 2x ~ 0.85

1.6×10–5 = [x][0.85]2

x = 2.2×10–5

or [Pb2+]e = 2.2×10–5 M

The concentration is reduced by about 1000.

## Quantitative prediction of precipitation:

Since salts are sparingly soluble, whether or not a precipitation occurs depends upon the concentrations of ions in the solution.

We use the reaction quotient to make the decision.

For a proposed precipitation reaction, we calculate the reaction quotient for the mass action expression of the solubilization of the sparingly soluble salt. This allows direct comparison to Ksp to decide which direction the reaction will proceed.

if Qc > Ksp

Reaction favors reactants and precipitation occurs.

if Qc < Ksp

Reaction favors products and no precipitation occurs.

if Qc = Ksp

Reaction is at equilibrium and no precipitation occurs

#### Example

50.0 mL of a 0.0010 M solution of calcium nitrate is mixed with 25.0 mL of a 0.0025 M solution of potassium sulfate at 25 °C. Will precipitation occur?

First, determine the possible reaction:

Solubilization reactions:

Ca(NO3)2(aq) Ca2+(aq) +2 NO3(aq)

K2SO4(aq) 2 K+(aq) + SO42–(aq)

Precipitation reaction:

Ca2+(aq) + SO42–(aq) CaSO4(s)

Net reaction:

Ca(NO3)2(aq) + K2SO4(aq) ? CaSO4(s) +2 K+(aq) +2 NO3(aq)

The solubilization reaction of the sparingly soluble salt is:

CaSO4(s)Ca2+(aq) + SO42–(aq)

Ksp = [Ca2+]e[SO42–]e = 9.1×10–6

Qc = [Ca2+][SO42–]

The initial concentrations of Ca2+ and SO42– must be found by doing a dilution problem:

The reaction quotient is Qc = [6.7×10–4][8.3×10–4] = 5.6×10–7

5.6×10–7 < 9.1×10–6 so the equilibrium reaction tends towards products, i.e. dissolved ions, so no precipitation occurs.

Ca(NO3)2(aq) + K2SO4(aq) NR (under these conditions)

Suppose you added solid potassium sulfate to the aforementioned calcium nitrate solution (50.0 mL, 0.0010 M). How many grams of potassium sulfate would be required to just start precipitation?

First, find the required concentration of sulfate. Then convert this to g of K2SO4.

 Ca(NO3)2(aq) Ca2+(aq) + 2 NO3–(aq) K2SO4(aq) 2 K+(aq) + SO42–(aq) CaSO4(s) Ca2+(aq) + SO42–(aq) Ksp = [Ca2+]e[SO42–]e = 9.1×10–6 Initial 0.0010 0 Change +0 + x Equilibrium 0.0010 x

9.1×10–6 = [0.0010][x]

x = 9.1×10–3

The required concentration of SO42– is 9.1×10–3 M.

To find the number of grams of K2SO4:

x = 0.079 g

How many g of potassium sulfate are needed to precipitate 99% of the calcium in the above solution?

After 99% of the Ca2+ has precipitated, there will be only 1% remaining. The concentration of Ca2+ remaining in solution will be:

(0.01)×(0.0010) = 1.0×10–5 M

 Ca(NO3)2(aq) Ca2+(aq) + 2 NO3–(aq) K2SO4(aq) 2 K+(aq) + SO42–(aq) CaSO4(s) Ca2+(aq) + SO42–(aq) Ksp = [Ca2+]e[SO42–]e = 9.1×10–6 Initial 0.0010 0 Change –0.99×0.0010 + x Equilibrium 1.0×10–5 x

9.1×10–6 = [1.0×10–5][x]

x = 0.91

The required concentration of SO42– is 0.91 M.

To find the number of grams of K2SO4:

x = 7.9 g K2SO4