Predict the solubility of an ionic substance by applying solubility equilibrium principles, and perform calculations involving the solubility product constant (Ksp) and the ionic product

What this dot point is asking

NESA wants you to write Ksp expressions for sparingly soluble salts, calculate molar solubility, predict whether a precipitate will form by comparing the ionic product Q to Ksp, and understand the common ion effect. This dot point is examined as a 3-5 mark calculation question and as part of extended response questions on environmental chemistry.

The answer

The solubility product Ksp

For a sparingly soluble ionic compound, dissolution is an equilibrium between the solid and its ions in solution:

The solubility product constant Ksp is the equilibrium constant for this dissolution. The pure solid is excluded, so:

Examples:

Ksp values are small (typically $10^{-5}$ to $10^{-50}$), reflecting the fact that the compound is only slightly soluble.

The ionic product Q

For any solution (not necessarily at equilibrium), the same expression evaluated using current concentrations is the ionic product Q.

Compare Q to Ksp to predict precipitation:

• If $Q < K_{sp}$, the solution is unsaturated. More solid can dissolve. No precipitate forms.
• If $Q = K_{sp}$, the solution is saturated. The system is at equilibrium.
• If $Q > K_{sp}$, the solution is supersaturated. A precipitate forms until Q falls back to Ksp.

The common ion effect

Adding a common ion to a saturated solution decreases solubility. For example, adding NaCl to a saturated solution of AgCl increases $[Cl^-]$, so $[Ag^+]$ must decrease to keep $Q = K_{sp}$. Some AgCl precipitates.

This is a direct application of Le Chatelier's principle. The added ion shifts the dissolution equilibrium to the left (toward the solid).

Worked example 2: the common ion effect

Calculate the molar solubility of $AgCl$ ($K_{sp} = 1.8 \times 10^{-10}$) in 0.10 mol/L $NaCl$, and compare with its solubility in pure water.

Step 1: Account for the common ion. $NaCl$ fully dissociates, so $[Cl^-]$ from the salt is 0.10 mol/L before any AgCl dissolves. Let $s'$ be the molar solubility of AgCl in this solution.

Step 2: Substitute into Ksp.

Step 3: Approximate. Because Ksp is tiny and $[Cl^-]$ from NaCl is much larger than $s'$, $0.10 + s' \approx 0.10$.

Step 4: Compare. In pure water $s = 1.34 \times 10^{-5}$ mol/L; in 0.10 mol/L NaCl, $s' = 1.8 \times 10^{-9}$ mol/L. The solubility falls by a factor of ~7400. Adding the common ion drives the equilibrium to the left (toward solid), exactly as Le Chatelier's principle predicts.

Check the approximation. $s' / 0.10 = 1.8 \times 10^{-8}$, far below 5%. Valid.

Examples in context

Example 1. Cadia gold mine cyanide leach circuit. At Newmont's Cadia operation near Orange, gold is dissolved as the soluble dicyanoaurate complex, but trace metals must be precipitated out before tailings discharge. Engineers control the silver content of the spent solution by manipulating the equilibrium $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$, with $K_{sp} = 1.8 \times 10^{-10}$. Adding excess chloride to the leach residue drives the common-ion effect, dropping $[Ag^+]$ to well below regulatory limits. The same Ksp framework students apply to lead chloride precipitation problems in HSC Section II underpins the real-time process control at the tailings dam.

Example 2. Hard-water scaling in Hunter Valley irrigation pipes. Bore water along the Hunter Valley vineyard belt contains calcium and carbonate above the saturation threshold for $CaCO_3$, with $K_{sp} = 3.4 \times 10^{-9}$. Pipe inspections after summer reveal a hard white scale on the inside of irrigation lines. Calculating the ionic product $Q = [Ca^{2+}][CO_3^{2-}]$ from a typical bore reading of $[Ca^{2+}] = 2.0 \times 10^{-3}$ and $[CO_3^{2-}] = 3.0 \times 10^{-5}$ mol L$^{-1}$ gives $Q = 6.0 \times 10^{-8}$, well above Ksp, so calcite precipitates. The same calculation tells maintenance crews when to dose with acid or replace pipe runs.

Try this

Q1. Write the Ksp expression for silver chromate, Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO_4^{2-}_{(aq)}, and state the units. [2 marks]

• Cue. $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$, with units mol$^3$ L$^{-3}$ for the unsymmetrical 2:1 salt.

Q2. The Ksp of $PbI_2$ is $7.1 \times 10^{-9}$. Calculate the molar solubility of $PbI_2$ in pure water. [3 marks]

• Cue. Let $s$ = molar solubility, then $[Pb^{2+}] = s$ and $[I^-] = 2s$, so $K_{sp} = s(2s)^2 = 4s^3$; $s = \sqrt[3]{7.1 \times 10^{-9} / 4} = 1.21 \times 10^{-3}$ mol L$^{-1}$.

Q3. A solution contains $[Ba^{2+}] = 1.0 \times 10^{-4}$ mol L$^{-1}$ and $[SO_4^{2-}] = 5.0 \times 10^{-5}$ mol L$^{-1}$. Given that $K_{sp}(BaSO_4) = 1.1 \times 10^{-10}$, (a) calculate $Q$, (b) predict whether a precipitate forms, (c) explain how adding sodium sulfate would change the answer. [2+1+2 marks]

• Cue. (a) $Q = 1.0 \times 10^{-4} \times 5.0 \times 10^{-5} = 5.0 \times 10^{-9}$. (b) $Q > K_{sp}$, so $BaSO_4$ precipitates. (c) Common-ion effect raises $[SO_4^{2-}]$, $Q$ increases further, more precipitate forms.