Solubility and Ks

A saturated solution is an equilibrium

• A sparingly soluble ionic solid dissolves only slightly. In a saturated solution, the undissolved solid sits in equilibrium with its ions:

$$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$$

• Solid is dissolving and ions are re-forming solid at equal rates, so the concentrations stay constant.

The solubility product

• $K_s$ is the equilibrium constant for that dissolving process. The solid is not included — it is a pure solid, and its concentration does not change.

$$K_s = [\text{Ag}^+][\text{Cl}^-]$$

• Each ion is raised to the power of its coefficient in the equation.
• A smaller $K_s$ means a less soluble solid.

Solubility versus $K_s$

• Solubility, $s$, is the number of moles of the solid that dissolve per litre of solution — units mol L⁻¹.
• $K_s$ is a constant for that solid at that temperature.
• They are related, but they are not the same thing, and the relationship depends on the type of solid.

The three permitted types

The standard restricts you to AB, $\mathbf{A_2B}$ and $\mathbf{AB_2}$ types where neither ion reacts further with water.

AB type — for example $\text{AgCl}$:

$$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$$

If $s$ mol dissolve, each gives one of each ion, so $[\text{Ag}^+] = [\text{Cl}^-] = s$:

$$K_s = s \times s = s^2$$

$\text{A}_2\text{B}$ type — for example $\text{Ag}_2\text{CrO}_4$:

$$\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$$

Each formula unit gives two silver ions and one chromate, so $[\text{Ag}^+] = 2s$ and $[\text{CrO}_4^{2-}] = s$:

$$K_s = (2s)^2 \times s = 4s^3$$

$\text{AB}_2$ type — for example $\text{PbI}_2$:

$$\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+} + 2\text{I}^-$$

$$K_s = s \times (2s)^2 = 4s^3$$

• Both $K_s = s^2$ and $K_s = 4s^3$ are given in the resource booklet, so you do not have to derive them — but you do have to choose the right one.

Why the coefficient appears twice

• A coefficient of 2 does two jobs, and forgetting either is the usual error:
  • it multiplies the concentration, because two ions are released per formula unit → $2s$
  • it becomes the power in the $K_s$ expression → $(2s)^2$
• That is where the factor of 4 comes from: $(2s)^2 = 4s^2$.

Converting units

• $K_s$ questions often quote solubility in g L⁻¹ rather than mol L⁻¹.
• Convert with $n = m/M$ before substituting, and convert back at the end if the question asks for grams.