Solution Equilibria

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

How ionic compounds dissolve

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

When an ionic solid such as $NaCl$ goes into water:

Water molecules surround the ions at the surface of the crystal.
The $δ -$ oxygen ends are attracted to positive ions ( $N a^{+}$ ); the $δ +$ hydrogen ends to negative ions ( $C l^{-}$ ). These are ion-dipole forces.
If these attractions are strong enough, ions are pulled out of the lattice.
Each ion ends up surrounded by water molecules. It is hydrated.

$NaCl (s) \to N a^{+} (aq) + C l^{-} (aq)$

Solubility rules

Usually soluble	Exceptions
All $N a^{+}$ , $K^{+}$ , $N H_{4}^{+}$ compounds	none
All nitrates $N O_{3}^{-}$ and acetates $C H_{3} CO O^{-}$	none
Chlorides, bromides, iodides	$A g^{+}$ , $P b^{2 +}$
Sulfates $S O_{4}^{2 -}$	$B a^{2 +}$ , $P b^{2 +}$ , $C a^{2 +}$ (slightly), $A g^{+}$ (slightly)

Usually insoluble	Exceptions
Carbonates, phosphates	with $N a^{+}$ , $K^{+}$ , $N H_{4}^{+}$
Hydroxides	with $N a^{+}$ , $K^{+}$ , $N H_{4}^{+}$ ; $B a^{2 +}$ and $C a^{2 +}$ slightly

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

The solubility product, Ksp

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

$PbC l_{2} (s) ⇌ P b^{2 +} (aq) + 2C l^{-} (aq)$

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

$K_{s p} = [P b^{2 +}] [C l^{-}]^{2}$

This is the solubility product. A smaller $K_{s p}$ (for salts of the same type) means less soluble. The HSC data sheet lists $K_{s p}$ 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	$K_{s p}$ in terms of $s$
$AgCl$ (1 : 1)	$s$ and $s$	$s^{2}$
$PbC l_{2}$ (1 : 2)	$s$ and $2 s$	$4 s^{3}$
$A g_{3} P O_{4}$ (3 : 1)	$3 s$ and $s$	$27 s^{4}$

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 $K_{s p}$ .

Will a precipitate form?

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

$Q > K_{s p}$ : too many ions. A precipitate forms.
$Q \leq K_{s p}$ : no precipitate.

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

The common ion effect

Dissolve $AgCl$ in a solution that already contains $C l^{-}$ (such as $NaCl$ ). By Le Chatelier, the extra $C l^{-}$ shifts $AgCl (s) ⇌ A g^{+} + C l^{-}$ to the left. Less $AgCl$ 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.

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 12Harder Ksp problems

Solubility with a common ion. $K_{s p} (AgCl) = 1.77 \times 1 0^{- 10}$ . In pure water, $s = 1.77 \times 1 0^{- 10} = 1.33 \times 1 0^{- 5}$ mol/L.

In 0.10 mol/L $NaCl$ , $[C l^{-}] \approx 0.10$ , so $s = \frac{1.77 \times 1 0 ^{- 10}}{0.10} = 1.77 \times 1 0^{- 9}$ mol/L. That is about 7500 times less soluble.

Hydroxides and pH. $K_{s p} (Mg (OH)_{2}) = 5.61 \times 1 0^{- 12}$ . $4 s^{3} = 5.61 \times 1 0^{- 12}$ gives $s = 1.12 \times 1 0^{- 4}$ , so $[O H^{-}] = 2.24 \times 1 0^{- 4}$ mol/L. Then $pOH = 3.65$ and $pH = 14 - 3.65 = 10.35$ .

Which precipitates first? Add $A g^{+}$ slowly to a mixture of $C l^{-}$ , $B r^{-}$ and $I^{-}$ (same concentration). $AgI$ has the smallest $K_{s p}$ , so its $Q$ passes $K_{s p}$ first: $AgI$ , then $AgBr$ , then $AgCl$ .

From pH back to Ksp. A saturated $Ca (OH)_{2}$ solution at pH 12.30 has pOH 1.70, so $[O H^{-}] = 0.0200$ and $s = 0.0100$ mol/L. $K_{s p} = s (2 s)^{2} = 4.0 \times 1 0^{- 6}$ .

Common ion with a 2 : 1 salt. For $A g_{2} Cr O_{4}$ in 0.010 mol/L $AgN O_{3}$ : $K_{s p} = [A g^{+}]^{2} [Cr O_{4}^{2 -}]$ , so $s = \frac{1.1 \times 1 0 ^{- 12}}{0.01 0 ^{2}} = 1.1 \times 1 0^{- 8}$ mol/L.

How much water to dissolve a solid? Work out $s$ , change it to g/L with the molar mass, then divide. $PbS O_{4}$ ( $K_{s p} = 2.53 \times 1 0^{- 8}$ ) has $s = 1.59 \times 1 0^{- 4}$ 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 $P b^{2 +}$ (0.00200 mol) and 40.0 mL of 0.100 mol/L $C l^{-}$ (0.00400 mol) react 1 : 2, so at most 0.00200 mol of $PbC l_{2}$ (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.
$K_{s p}$ is the product of ion concentrations in a saturated solution, each to the power of its coefficient.
1 : 1 salt: $K_{s p} = s^{2}$ . 1 : 2 salt: $K_{s p} = 4 s^{3}$ .
After mixing (and diluting), if $Q > K_{s p}$ a precipitate forms.
A common ion makes a salt less soluble.
Leaching in running water removes soluble toxins from foods.

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When an ionic solid dissolves in water, the ions are attracted to water by:
1. A Covalent bonds
2. B Ion-dipole forces
3. C Metallic bonds
4. D Dispersion forces only
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B: Ion-dipole forces
The charged ions attract the partly charged ends of polar water molecules.
Which compound is insoluble in water?
1. A Sodium chloride
2. B Potassium nitrate
3. C Silver chloride
4. D Ammonium sulfate
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C: Silver chloride
Silver and lead halides are the exceptions to "chlorides are soluble".
Which compound is soluble in water?
1. A Barium sulfate
2. B Calcium carbonate
3. C Sodium carbonate
4. D Lead(II) iodide
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C: Sodium carbonate
All sodium compounds are soluble.

How ionic compounds dissolve

Solubility rules

The solubility product, Ksp

Molar solubility

Will a precipitate form?

The common ion effect

Removing toxins from food

Quick summary

Quiz

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