# Solution Equilibria

> How ionic compounds dissolve, solubility rules, the solubility product Ksp, molar solubility, predicting precipitates with Q, the common ion effect, and how Aboriginal peoples removed toxins from food.

From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/solution-equilibria/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Bread from poisonous seeds
Cycad seeds grow across northern Australia. Eaten raw, they are poisonous. Yet Aboriginal peoples made them into a safe, nourishing bread for thousands of years.

The trick was water. The seeds were cut up and left for days in running streams or soaked in water many times over. The poison slowly dissolved and washed away, until the seeds were safe. This is chemistry: a **dissolving equilibrium** pushed along by constantly removing what had dissolved.

**In short:** When an ionic solid dissolves, its saturated solution is an equilibrium between the solid and its ions in water. The solubility product $K_{sp}$ measures how much dissolves, and a precipitate forms when mixing two solutions gives an ion product greater than $K_{sp}$.

## How ionic compounds dissolve

Water molecules are **polar**: the oxygen end is slightly negative ($\delta-$) and the hydrogen ends slightly positive ($\delta+$).

When an ionic solid such as $\mathrm{NaCl}$ goes into water:

1. Water molecules surround the ions at the surface of the crystal.
2. The $\delta-$ oxygen ends are attracted to positive ions ($\mathrm{Na^+}$); the $\delta+$ hydrogen ends to negative ions ($\mathrm{Cl^-}$). These are **ion-dipole forces**.
3. If these attractions are strong enough, ions are pulled out of the lattice.
4. Each ion ends up surrounded by water molecules. It is **hydrated**.

$$\mathrm{NaCl(s)} \rightarrow \mathrm{Na^+(aq)} + \mathrm{Cl^-(aq)}$$

## Solubility rules

| Usually **soluble** | Exceptions |
|---|---|
| All $\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{NH_4^+}$ compounds | none |
| All nitrates $\mathrm{NO_3^-}$ and acetates $\mathrm{CH_3COO^-}$ | none |
| Chlorides, bromides, iodides | $\mathrm{Ag^+}$, $\mathrm{Pb^{2+}}$ |
| Sulfates $\mathrm{SO_4^{2-}}$ | $\mathrm{Ba^{2+}}$, $\mathrm{Pb^{2+}}$, $\mathrm{Ca^{2+}}$ (slightly), $\mathrm{Ag^+}$ (slightly) |

| Usually **insoluble** | Exceptions |
|---|---|
| Carbonates, phosphates | with $\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{NH_4^+}$ |
| Hydroxides | with $\mathrm{Na^+}$, $\mathrm{K^+}$, $\mathrm{NH_4^+}$; $\mathrm{Ba^{2+}}$ and $\mathrm{Ca^{2+}}$ slightly |

When two solutions are mixed and an insoluble combination forms, a solid **precipitate** appears.

### Worked example: Net ionic equations
Silver nitrate mixed with sodium chloride:

Full equation: $\mathrm{AgNO_3(aq)} + \mathrm{NaCl(aq)} \rightarrow \mathrm{AgCl(s)} + \mathrm{NaNO_3(aq)}$

The $\mathrm{Na^+}$ and $\mathrm{NO_3^-}$ ions do nothing. They are **spectator ions**. Leave them out:

Net ionic equation: $\mathrm{Ag^+(aq)} + \mathrm{Cl^-(aq)} \rightarrow \mathrm{AgCl(s)}$

## The solubility product, Ksp

Even "insoluble" salts dissolve a tiny bit. In a **saturated** solution, the solid and its ions are in dynamic equilibrium:

$$\mathrm{PbCl_2(s)} \rightleftharpoons \mathrm{Pb^{2+}(aq)} + \mathrm{2Cl^-(aq)}$$

The equilibrium constant leaves out the solid, so it is just the product of the ion concentrations:

$$K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2$$

This is the **solubility product**. A smaller $K_{sp}$ (for salts of the same type) means less soluble. The HSC data sheet lists $K_{sp}$ values at 25 °C.

## Molar solubility

The **molar solubility** $s$ is how many moles of the salt dissolve in one litre of saturated solution.

| Salt type | Ion concentrations | $K_{sp}$ in terms of $s$ |
|---|---|---|
| $\mathrm{AgCl}$ (1 : 1) | $s$ and $s$ | $s^2$ |
| $\mathrm{PbCl_2}$ (1 : 2) | $s$ and $2s$ | $4s^3$ |
| $\mathrm{Ag_3PO_4}$ (3 : 1) | $3s$ and $s$ | $27s^4$ |

### Worked example: Lead(II) chloride
$K_{sp}(\mathrm{PbCl_2}) = 1.70 \times 10^{-5}$.

$4s^3 = 1.70 \times 10^{-5}$, so $s^3 = 4.25 \times 10^{-6}$ and $s = 0.0162$ mol/L.

So $[\mathrm{Pb^{2+}}] = 0.0162$ and $[\mathrm{Cl^-}] = 0.0324$ mol/L.

**Going the other way.** If you know the solubility in g/L, divide by the molar mass to get $s$ in mol/L, then work out $K_{sp}$.

## Will a precipitate form?

Work out the **ionic product** $Q$ using the concentrations **after mixing**, and compare it to $K_{sp}$:

- $Q > K_{sp}$: too many ions. A **precipitate forms**.
- $Q \le K_{sp}$: no precipitate.

**Mixing dilutes both solutions.** Mixing 50 mL with 50 mL halves each concentration.

### Worked example: Mixing silver and chloride
10.0 mL of $1.0 \times 10^{-4}$ mol/L $\mathrm{AgNO_3}$ is mixed with 30.0 mL of $2.0 \times 10^{-4}$ mol/L $\mathrm{NaCl}$. $K_{sp}(\mathrm{AgCl}) = 1.77 \times 10^{-10}$.

Total volume = 40.0 mL.

$[\mathrm{Ag^+}] = 1.0 \times 10^{-4} \times \dfrac{10.0}{40.0} = 2.5 \times 10^{-5}$

$[\mathrm{Cl^-}] = 2.0 \times 10^{-4} \times \dfrac{30.0}{40.0} = 1.5 \times 10^{-4}$

$Q = 2.5 \times 10^{-5} \times 1.5 \times 10^{-4} = 3.75 \times 10^{-9}$

$Q > K_{sp}$, so a precipitate of $\mathrm{AgCl}$ forms.

## The common ion effect

Dissolve $\mathrm{AgCl}$ in a solution that **already contains** $\mathrm{Cl^-}$ (such as $\mathrm{NaCl}$). By Le Chatelier, the extra $\mathrm{Cl^-}$ shifts $\mathrm{AgCl(s)} \rightleftharpoons \mathrm{Ag^+} + \mathrm{Cl^-}$ to the **left**. Less $\mathrm{AgCl}$ dissolves.

A salt is **less soluble** in a solution containing one of its own ions.

## Removing toxins from food

Aboriginal and Torres Strait Islander peoples used solubility to make toxic plants safe:

- **Leaching**: soaking cut or crushed cycad seeds, or yams, in running water. The water-soluble toxin dissolves, and the flowing water carries it away. Its concentration in the water stays low, so more keeps dissolving.
- **Cutting and grinding** gives a larger surface area, so the toxin dissolves faster.
- **Fermenting** (leaving seeds in containers for months) lets microbes break the toxin down.

**Mind map**

```text
Solution equilibria
  Dissolving
    Polar water
    Ion-dipole forces
    Hydrated ions
  Solubility rules
    Na, K, NH4, nitrates soluble
    Ag, Pb halides insoluble
    Net ionic equations
  Ksp
    Product of ion concentrations
    Leave out the solid
    s from Ksp
  Precipitate
    Q > Ksp forms
    Dilute on mixing
  Common ion
    Less soluble
  Toxins
    Leaching in running water
```

### Year 11: Molar mass and concentration
Ksp questions often give amounts in grams. Two Year 11 formulas do the work:

$$n = \frac{m}{MM} \qquad c = \frac{n}{V}$$

**Example.** 0.016 g of calcium fluoride ($\mathrm{CaF_2}$, 78.08 g/mol) dissolves in 1.0 L of water.

$s = 0.016 \div 78.08 = 2.05 \times 10^{-4}$ mol/L.

$K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2 = s \times (2s)^2 = 4s^3 = 3.4 \times 10^{-11}$.

### Year 12: Harder Ksp problems
**Solubility with a common ion.** $K_{sp}(\mathrm{AgCl}) = 1.77 \times 10^{-10}$. In pure water, $s = \sqrt{1.77 \times 10^{-10}} = 1.33 \times 10^{-5}$ mol/L.

In 0.10 mol/L $\mathrm{NaCl}$, $[\mathrm{Cl^-}] \approx 0.10$, so $s = \dfrac{1.77 \times 10^{-10}}{0.10} = 1.77 \times 10^{-9}$ mol/L. That is about 7500 times less soluble.

**Hydroxides and pH.** $K_{sp}(\mathrm{Mg(OH)_2}) = 5.61 \times 10^{-12}$. $4s^3 = 5.61 \times 10^{-12}$ gives $s = 1.12 \times 10^{-4}$, so $[\mathrm{OH^-}] = 2.24 \times 10^{-4}$ mol/L. Then $\mathrm{pOH} = 3.65$ and $\mathrm{pH} = 14 - 3.65 = 10.35$.

**Which precipitates first?** Add $\mathrm{Ag^+}$ slowly to a mixture of $\mathrm{Cl^-}$, $\mathrm{Br^-}$ and $\mathrm{I^-}$ (same concentration). $\mathrm{AgI}$ has the smallest $K_{sp}$, so its $Q$ passes $K_{sp}$ first: $\mathrm{AgI}$, then $\mathrm{AgBr}$, then $\mathrm{AgCl}$.

**From pH back to Ksp.** A saturated $\mathrm{Ca(OH)_2}$ solution at pH 12.30 has pOH 1.70, so $[\mathrm{OH^-}] = 0.0200$ and $s = 0.0100$ mol/L. $K_{sp} = s(2s)^2 = 4.0 \times 10^{-6}$.

**Common ion with a 2 : 1 salt.** For $\mathrm{Ag_2CrO_4}$ in 0.010 mol/L $\mathrm{AgNO_3}$: $K_{sp} = [\mathrm{Ag^+}]^2[\mathrm{CrO_4^{2-}}]$, so $s = \dfrac{1.1 \times 10^{-12}}{0.010^2} = 1.1 \times 10^{-8}$ mol/L.

**How much water to dissolve a solid?** Work out $s$, change it to g/L with the molar mass, then divide. $\mathrm{PbSO_4}$ ($K_{sp} = 2.53 \times 10^{-8}$) has $s = 1.59 \times 10^{-4}$ mol/L, or 0.0482 g/L, so 1.0 g needs at least 21 L.

**Mass of precipitate.** Treat it as a limiting reagent problem. 20.0 mL of 0.100 mol/L $\mathrm{Pb^{2+}}$ (0.00200 mol) and 40.0 mL of 0.100 mol/L $\mathrm{Cl^-}$ (0.00400 mol) react 1 : 2, so at most 0.00200 mol of $\mathrm{PbCl_2}$ (0.556 g) forms. A little stays dissolved, so the real mass is slightly less.

## Quick summary

- Ionic solids dissolve when ion-dipole forces with water pull ions out of the lattice.
- Spectator ions are left out of net ionic equations.
- $K_{sp}$ is the product of ion concentrations in a saturated solution, each to the power of its coefficient.
- 1 : 1 salt: $K_{sp} = s^2$. 1 : 2 salt: $K_{sp} = 4s^3$.
- After mixing (and diluting), if $Q > K_{sp}$ a precipitate forms.
- A common ion makes a salt less soluble.
- Leaching in running water removes soluble toxins from foods.

## Quiz

Try these questions from the chapter quiz.

1. When an ionic solid dissolves in water, the ions are attracted to water by:
   - A. Covalent bonds
   - B. Ion-dipole forces
   - C. Metallic bonds
   - D. Dispersion forces only
   - Answer: B. Ion-dipole forces. The charged ions attract the partly charged ends of polar water molecules.

2. Which compound is insoluble in water?
   - A. Sodium chloride
   - B. Potassium nitrate
   - C. Silver chloride
   - D. Ammonium sulfate
   - Answer: C. Silver chloride. Silver and lead halides are the exceptions to "chlorides are soluble".

3. Which compound is soluble in water?
   - A. Barium sulfate
   - B. Calcium carbonate
   - C. Sodium carbonate
   - D. Lead(II) iodide
   - Answer: C. Sodium carbonate. All sodium compounds are soluble.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/solution-equilibria/#quiz