Solubility and Solubility Product of Sparingly Soluble Salts

Lessons

Senior 6 · Chemistry

Solubility and Solubility Product of Sparingly Soluble Salts

4 lessons

Lesson 4 of 4

Common Ions, pH, Complex Ions and Applications

25 min

Why vinegar removes limescale

Boil hard water in a kettle for a few months and a pale, rock-hard crust of calcium carbonate builds up inside. Scraping barely touches it, yet leaving vinegar or lemon juice in the kettle overnight loosens it with a fizz of carbon dioxide. Nothing about $K_{sp}$ for calcium carbonate has changed; what changed is the chemistry of the ions in solution.

This lesson shows the three ways to push a sparingly soluble salt's equilibrium: add one of its own ions, change the pH, or tie up one ion in a complex. Each has uses, from purifying chemicals to medicine.

Definition

Common-ion effect: the lowering of the solubility of a salt by the presence of one of its own ions from another source. Example: silver chloride is about 7500 times less soluble in 0.10 mol dm⁻³ NaCl than in pure water.

Complex ion: a central metal ion bonded to ligands by dative bonds (Unit 1). Example: the diamminesilver(I) ion, $\mathrm{[Ag(NH_3)_2]^{+}}$ .

Amphoteric hydroxide: a hydroxide that dissolves in both acids and excess strong alkali. Example: zinc hydroxide, which dissolves in excess NaOH to give $\mathrm{[Zn(OH)_4]^{2-}}$ .

1. Adding a common ion

For $\mathrm{AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)}$ , adding chloride ions (as NaCl) makes $Q > K_{sp}$ . By Le Chatelier's principle the equilibrium shifts left, AgCl precipitates, and less of it can dissolve. The same $K_{sp}$ now has to be shared with a large, fixed $[\mathrm{Cl^{-}}]$ , so $[\mathrm{Ag^{+}}]$ must be tiny. Chemists use this:

In gravimetric analysis, a slight excess of the precipitating reagent is added so that the precipitate is as complete as possible.
In salting out, soap is precipitated from the boiled mixture by adding a lot of sodium chloride: the extra $\mathrm{Na^{+}}$ reduces the solubility of the sodium salt of the fatty acid.

2. Changing the pH

If the anion of a salt is the conjugate base of a weak acid ( $\mathrm{OH^{-}}$ , $\mathrm{CO_3^{2-}}$ , $\mathrm{F^{-}}$ , $\mathrm{PO_4^{3-}}$ , $\mathrm{C_2O_4^{2-}}$ ), adding acid removes that anion and the salt dissolves:

$\mathrm{CaCO_3(s) + 2H^{+}(aq) \rightarrow Ca^{2+}(aq) + H_2O(l) + CO_2(g)}$

$\mathrm{Mg(OH)_2(s) + 2H^{+}(aq) \rightarrow Mg^{2+}(aq) + 2H_2O(l)}$

For a hydroxide the effect is easy to calculate: $s = [\mathrm{Mg^{2+}}] = \frac{K_{sp}}{[\mathrm{OH^{-}}]^2}$ , so each unit fall in pH makes magnesium hydroxide 100 times more soluble (Figure 4.1). A salt such as AgCl, whose anion comes from a strong acid, is unaffected by pH: $\mathrm{Cl^{-}}$ does not react with $\mathrm{H^{+}}$ .

Figure 4.1: The solubility of a hydroxide depends strongly on pH; that of a chloride does not.

The same chemistry decides the health of your teeth. Tooth enamel is mainly hydroxyapatite, $\mathrm{Ca_5(PO_4)_3OH}$ . Acids made by mouth bacteria from sugar remove its $\mathrm{OH^{-}}$ and $\mathrm{PO_4^{3-}}$ ions, so it dissolves. Fluoride helps by forming fluorapatite, which resists acid better.

3. Forming a complex ion

Silver chloride dissolves in aqueous ammonia because ammonia ties up the silver ions:

$\mathrm{AgCl(s) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^{+}(aq) + Cl^{-}(aq)}$

Removing free $\mathrm{Ag^{+}}$ keeps $Q$ below $K_{sp}$ , so more solid dissolves. AgBr (smaller $K_{sp}$ ) needs concentrated ammonia, and AgI (smaller still) does not dissolve; this is why ammonia helps tell the three halides apart. In the same way, the pale blue precipitate of copper(II) hydroxide dissolves in excess ammonia to give the deep blue tetraammine complex (photo below), and amphoteric hydroxides dissolve in excess alkali: $\mathrm{Zn(OH)_2(s) + 2OH^{-}(aq) \rightarrow [Zn(OH)_4]^{2-}(aq)}$ .

Applications of the solubility product

Application	The $K_{sp}$ idea behind it
Qualitative analysis: identifying halide and metal ions	Each ion is precipitated by a reagent when $Q > K_{sp}$ ; complexing (e.g. with ammonia) then tells similar precipitates apart.
Gravimetric analysis: sulfate weighed as $\mathrm{BaSO_4}$	Very small $K_{sp}$ plus a common ion (excess $\mathrm{Ba^{2+}}$ ) makes the precipitation essentially complete.
Water softening with washing soda, $\mathrm{Na_2CO_3}$	Excess $\mathrm{CO_3^{2-}}$ pushes $Q$ for $\mathrm{CaCO_3}$ far above $K_{sp}$ , removing hardness as a precipitate.
Removing iron and heavy metals from water	Raising the pH precipitates metal hydroxides with tiny $K_{sp}$ values.
Barium meal for X-ray imaging of the gut	Soluble barium compounds are toxic, but $\mathrm{BaSO_4}$ is so sparingly soluble ( $K_{sp} = 1.08 \times 10^{-10}$ ) that the patient absorbs no harmful amount.
Kidney stones	Stones of calcium oxalate ( $K_{sp} \approx 2.7 \times 10^{-9}$ ) can form when urine becomes supersaturated; drinking more water lowers the ion concentrations, and so $Q$ .

Worked example

Problem. Calculate the molar solubility of silver chloride (a) in pure water and (b) in 0.10 mol dm⁻³ sodium chloride, at 25 °C. By what factor does the common ion reduce the solubility?

Solution.

(a) $s = \sqrt{1.77 \times 10^{-10}} = 1.33 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ .

(b) Let the solubility be $s$ . Then $[\mathrm{Ag^{+}}] = s$ and $[\mathrm{Cl^{-}}] = 0.10 + s \approx 0.10$ , because $s$ is tiny:

$K_{sp} = s(0.10) \quad\Rightarrow\quad s = \frac{1.77 \times 10^{-10}}{0.10} = 1.77 \times 10^{-9}\,\mathrm{mol\,dm^{-3}}$

The approximation is valid: $1.77 \times 10^{-9}$ is negligible beside 0.10.

Answer: (a) $1.33 \times 10^{-5}$ ; (b) $1.77 \times 10^{-9}\,\mathrm{mol\,dm^{-3}}$ ; about 7500 times less soluble.

Worked example

Problem. Hard water contains $2.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ of $\mathrm{Ca^{2+}}$ . Enough sodium carbonate is added to precipitate calcium carbonate and leave $[\mathrm{CO_3^{2-}}] = 1.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ in excess. Calculate the calcium ion concentration left and the percentage of calcium removed.

Solution.

$[\mathrm{Ca^{2+}}] = \frac{K_{sp}}{[\mathrm{CO_3^{2-}}]} = \frac{3.36 \times 10^{-9}}{1.0 \times 10^{-3}} = 3.4 \times 10^{-6}\,\mathrm{mol\,dm^{-3}}$

$\text{removed} = \left(1 - \frac{3.4 \times 10^{-6}}{2.0 \times 10^{-3}}\right) \times 100 = 99.8\%$

Answer: $3.4 \times 10^{-6}\,\mathrm{mol\,dm^{-3}}$ remains; 99.8% of the calcium is removed, thanks to the common ion $\mathrm{CO_3^{2-}}$ .

Warning

Common mistakes

Forgetting to include the common ion's concentration: in 0.10 mol dm⁻³ NaCl, $[\mathrm{Cl^{-}}]$ is 0.10, not $s$ .
Forgetting to raise the common ion to its power: for $\mathrm{CaF_2}$ in NaF, $K_{sp} = s[\mathrm{F^{-}}]^2$ .
Thinking the common ion changes $K_{sp}$ . It changes the solubility; $K_{sp}$ depends only on temperature.
Expecting all salts to dissolve in acid. Only those whose anion is the conjugate base of a weak acid do; AgCl and $\mathrm{BaSO_4}$ barely respond.

Safety. Soluble barium and lead compounds are toxic, and concentrated ammonia gives off a choking vapour: use small quantities, work in a fume cupboard where possible, and wash your hands. If any reagent splashes into an eye, rinse it with clean running water for at least 15 minutes and get medical help.

Tip

Three levers, one idea: anything that adds one of the salt's ions lowers its solubility; anything that removes one of its ions (acid removing a weak-acid anion, a ligand removing a metal ion) raises it. Ask: "Does this change add to, or take away from, the right-hand side of the equilibrium?"

Professor Dave Explains: The common ion effect. Watch how adding an ion already present shifts the dissolving equilibrium to the left, and how the calculation simplifies because s is tiny.

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Practice

Explain, using Le Chatelier's principle, why barium sulfate is less soluble in sodium sulfate solution than in pure water.
Calculate the solubility of barium sulfate in 0.050 mol dm⁻³ $\mathrm{Na_2SO_4}$ , in mol dm⁻³ and mg dm⁻³, and compare it with its solubility in pure water (2.43 mg dm⁻³).
Calculate the molar solubility of $\mathrm{PbI_2}$ (a) in 0.10 mol dm⁻³ KI; (b) in 0.10 mol dm⁻³ $\mathrm{Pb(NO_3)_2}$ .
Calculate the molar solubility of calcium fluoride in 0.010 mol dm⁻³ NaF.
Calculate the solubility of magnesium hydroxide in a solution buffered at pH 9.0, and explain why antacid tablets containing $\mathrm{Mg(OH)_2}$ dissolve readily in the stomach.
Explain why silver chloride dissolves in dilute aqueous ammonia but silver iodide does not.
A patient asks why swallowing barium sulfate is safe when barium chloride is a poison. Write a short explanation using $K_{sp}$ , and evaluate whether the barium sulfate would still be safe if it dissolved in stomach acid as calcium carbonate does.

Show answersHide answers

$\mathrm{BaSO_4(s) \rightleftharpoons Ba^{2+} + SO_4^{2-}}$ . Adding $\mathrm{SO_4^{2-}}$ raises $Q$ above $K_{sp}$ , so the equilibrium shifts left and $\mathrm{BaSO_4}$ precipitates; less can dissolve.
$s = \frac{1.08 \times 10^{-10}}{0.050} = 2.16 \times 10^{-9}\,\mathrm{mol\,dm^{-3}}$ ; $\times 233.39 \times 1000 = 5.0 \times 10^{-4}\,\mathrm{mg\,dm^{-3}}$ , about 4800 times less than in pure water.
(a) $[\mathrm{I^{-}}] = 0.10$ : $s = \frac{9.8 \times 10^{-9}}{(0.10)^2} = 9.8 \times 10^{-7}\,\mathrm{mol\,dm^{-3}}$ . (b) $[\mathrm{Pb^{2+}}] = 0.10$ , $[\mathrm{I^{-}}] = 2s$ : $0.10 \times 4s^2 = 9.8 \times 10^{-9}$ , $s = 1.6 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ . The common ion that appears squared (iodide) has the bigger effect.
$s = \frac{3.45 \times 10^{-11}}{(0.010)^2} = 3.45 \times 10^{-7}\,\mathrm{mol\,dm^{-3}}$ , about 600 times less than in water.
At pH 9.0, $[\mathrm{OH^{-}}] = 1.0 \times 10^{-5}$ ; $s = \frac{5.61 \times 10^{-12}}{(1.0 \times 10^{-5})^2} = 0.056\,\mathrm{mol\,dm^{-3}}$ . Stomach acid (pH about 1–2) removes $\mathrm{OH^{-}}$ almost completely, so $Q$ stays far below $K_{sp}$ and the hydroxide dissolves as it neutralises the acid.
Ammonia forms $\mathrm{[Ag(NH_3)_2]^{+}}$ , lowering $[\mathrm{Ag^{+}}]$ . For AgCl ( $K_{sp} = 1.77 \times 10^{-10}$ ) this lowers $Q$ below $K_{sp}$ , so it dissolves. AgI's $K_{sp}$ ( $8.52 \times 10^{-17}$ ) is so small that dilute ammonia cannot lower $[\mathrm{Ag^{+}}]$ enough.
Barium chloride is soluble, so swallowing it releases large amounts of toxic $\mathrm{Ba^{2+}}$ . Barium sulfate's $K_{sp}$ is $1.08 \times 10^{-10}$ : a saturated solution holds only about $1 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ of $\mathrm{Ba^{2+}}$ , too little to harm. It stays safe in the stomach because sulfate is only a very weak base, so stomach acid raises its solubility only slightly; a barium salt of a weak acid, such as barium carbonate, would dissolve in stomach acid and be poisonous.

1Why vinegar removes limescale

21. Adding a common ion

32. Changing the pH

43. Forming a complex ion

5Applications of the solubility product

Why vinegar removes limescale

1. Adding a common ion

2. Changing the pH

3. Forming a complex ion

Applications of the solubility product