Solubility is an important factor of the measurements of the environmental pollution of salts. The s — Physical Chemistry — Kinetics Chemistry Question
Selective solubility
Solubility is an important factor of the measurements of the environmental pollution of salts. The solubility of a substance is defined as the amount that dissolves in a given quantity of solvent to form a saturated solution. The solubility varies greatly with the nature of a solute and the solvent, and the experimental conditions, such as temperature and pressure. The pH and the complex formation also may have influence on the solubility.
An aqueous solution contains BaCl2 and SrCl2 both in a concentration of 0.01 mol dm -3. The question is whether it will be possible to separate this mixture completely by adding a saturated solution of sodium sulphate. The criterion is that at least 99.9 % of the Ba 2+ has precipitated as BaSO4 and that SrSO4 may be contaminated with no more than 0.1 % BaSO4. The solubility product constants are as follows:
Ks(BaSO4) = 1×10–10 and Ks(SrSO4) = 3×10–7.
Complex formation may have a profound effect on the solubility. A complex is a charged species consisting of a central metal ion bonded to one of more ligands. For example Ag(NH3)2 + is a complex containing Ag + as the central ion and two NH3 molecules as ligands.
The solubility of AgCl in water is 1.3×10–5 mol dm -3.
The solubility product constant of AgCl is 1.7×10–10.
The equilibrium constant Kf for the formation of the complex has a value of 1.5×107.
(a) Give the relevant equations.
(b) Calculate the residual concentration of Ba 2+.
(c) Calculate the percentage of Ba 2+ and Sr 2+ in the separated substances.
Model Answer
The relevant equations are:
Ba 2+(aq) + SO4 2-(aq) BaSO4(s)
Sr 2+(aq) + SO4 2-(aq) SrSO4(s)
Precipitation of BaSO4 will start when
Ksp(BaSO4) = 1×10–10 = [Ba 2+][SO4 2–] ⇒ [SO4 2–] = 1×10–10 / 1×10–2 = 1×10–8
Precipitation of SrSO4 will start when
Ksp(SrSO4) = 3×10–7 = [Sr 2+][SO4 2–] ⇒ [SO4 2–] = 3×10–7 / 1×10–2 = 3×10–5
If there are no kinetic complications (for example when the formation of BaSO4 would be very slow) first BaSO4 will be formed. This results in a decrease of the concentration of Ba 2+ ions. If the concentration SO4 2– satisfies equation (2), the concentration of Ba 2+ can be calculated from the formula:
Ksp(BaSO4) = 1×10–10 = [Ba 2+] × 3×10–5
[Ba 2+] = 1×10–10 / 3×10–5 = 1/3 × 10–5
At the starting point the concentration of Ba 2+ was 1×10–2 mol dm–3. This means that the loss amounts to
(1/3 × 10–5 / 1×10–2) × 100 % = 0.033 %
The separation meets the criterion.
Show by calculation that the solubility of AgCl in aqueous ammonia (c = 1.0 mol dm -3) is higher than in pure water.
Model Answer
The following equilibrium reactions have to be considered:
AgCl(s) Ag+(aq) + Cl–(aq)
Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq)
Total: AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl–(aq)
Koverall = [Ag(NH3)2+][Cl–] / [NH3]2 = Ksp × Kf = 1.7×10–10 × 1.5×107 = 2.6×10–3
If x is the molar solubility of AgCl (mol dm-3) then the changes in concentration of AgCl as the result of the formation of the complex ion are:
AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl–(aq)
Concentration
Starting point: 1.0, 0.0, 0.0
Change: – 2 x, + x, + x
Equilibrium: (1.0 – 2 x), + x, + x
Kf is quite large so most of the Ag+ ions exist in the complexed form. In absence of NH3 at equilibrium holds [Ag+] = [Cl–].
Complex formation leads to: [Ag(NH3)2+] = [Cl–]
Koverall can be written as:
x^2 / (1.0 – 2x)^2 = 2.6×10–3 or x / (1.0 – 2x) = 0.051
and x = 0.046 mol dm–3
This result means that 4.6×10–2 mol of AgCl dissolves in 1 dm3 of NH3 aqueous solution with the concentration of 1.0 mol dm–3. Thus the formation of the complex ion Ag(NH3)2+ enhances the solubility of AgCl, because in pure water the solubility of AgCl amounts to only 1.3×10–5 mol dm–3.