Solubility and Solubility Product of Sparingly Soluble Salts

Lessons

Senior 6 · Chemistry

Solubility and Solubility Product of Sparingly Soluble Salts

4 lessons

Lesson 3 of 4

Predicting Precipitation and Fractional Precipitation

25 min

A white precipitate first, a red one at the end

Water laboratories measure the chloride in drinking water by titrating it with silver nitrate, with a little potassium chromate in the flask. As the silver nitrate runs in, white silver chloride forms. Only when almost all the chloride has gone does a red-brown tint of silver chromate appear and stay; that is the end point. This is Mohr's method.

Why does the silver ion "choose" chloride first, and why does chromate wait so precisely? The answer is a simple comparison between the concentrations actually present and the solubility product.

Definition

Ionic product, $Q$ : the product of the ion concentrations, each raised to the power of its coefficient, calculated for the solution as it is, whether or not it is at equilibrium. It has the same form as $K_{sp}$ . Example: for $\mathrm{PbI_2}$ , $Q = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2$ using the concentrations just after mixing.

Precipitation rule: if $Q > K_{sp}$ , the solution is supersaturated and the salt precipitates until $Q = K_{sp}$ ; if $Q = K_{sp}$ , the solution is just saturated; if $Q < K_{sp}$ , no precipitate forms.

Fractional (selective) precipitation: separating ions in a mixture by adding a reagent slowly, so that the least soluble salt precipitates first while the others stay in solution. Example: AgI precipitates before AgCl when silver nitrate is added to a mixture of iodide and chloride.

Will a precipitate form?

Work out each ion's concentration after mixing: mixing two solutions dilutes both.
Calculate $Q$ from the $K_{sp}$ expression.
Compare $Q$ with $K_{sp}$ .

Which precipitate forms first?

When a reagent is added slowly to a solution that contains two ions it can precipitate, calculate for each salt the concentration of reagent at which $Q$ just reaches $K_{sp}$ . The salt with the lower threshold precipitates first. Figure 3.1 shows the thresholds of silver ion concentration for a mixture of iodide and chloride: $8.5 \times 10^{-15}$ and $1.8 \times 10^{-8}\,\mathrm{mol\,dm^{-3}}$ . Between the two, silver iodide comes out of solution alone.

Figure 3.1: As the silver ion concentration rises, AgI reaches its $K_{sp}$ first; AgCl starts only when $[\mathrm{Ag^{+}}]$ is two million times larger.

Mohr's method explained

Suppose the flask contains $5.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ of chromate (an illustrative value). Silver chromate can form only once

$[\mathrm{Ag^{+}}] = \sqrt{\frac{K_{sp}(\mathrm{Ag_2CrO_4})}{[\mathrm{CrO_4^{2-}}]}} = \sqrt{\frac{1.12 \times 10^{-12}}{5.0 \times 10^{-3}}} = 1.5 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$

At that silver concentration, the chloride left is $\frac{1.77 \times 10^{-10}}{1.5 \times 10^{-5}} = 1.2 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ , almost exactly the $1.33 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ present at the true equivalence point, when the flask holds only saturated AgCl. So the red-brown colour appears within a drop of the equivalence point. The titration must be done near neutral pH: in acid, chromate is converted to hydrogenchromate and dichromate, which lowers $[\mathrm{CrO_4^{2-}}]$ and delays the end point; at high pH, silver hydroxide precipitates instead.

Separating metal ions by controlling pH

Metal hydroxides have very different $K_{sp}$ values, so raising the pH step by step precipitates them one after another. For a solution 0.010 mol dm⁻³ in both $\mathrm{Fe^{3+}}$ and $\mathrm{Mg^{2+}}$ :

Hydroxide	$[\mathrm{OH^{-}}]$ when precipitation starts	pH when it starts
$\mathrm{Fe(OH)_3}$	$\sqrt[3]{\frac{2.79 \times 10^{-39}}{0.010}} = 6.5 \times 10^{-13}$	1.8
$\mathrm{Mg(OH)_2}$	$\sqrt{\frac{5.61 \times 10^{-12}}{0.010}} = 2.4 \times 10^{-5}$	9.4

By pH 2.8, $[\mathrm{Fe^{3+}}]$ has already fallen to $1 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ , a thousandth of its starting value. Holding the pH between about 3 and 9 therefore removes iron as rust-brown $\mathrm{Fe(OH)_3}$ while leaving magnesium dissolved. Water treatment uses the same idea to remove dissolved iron.

Worked example

Problem. Does lead(II) iodide ( $K_{sp} = 9.8 \times 10^{-9}$ ) precipitate when (a) 50.0 cm³ of $1.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ $\mathrm{Pb(NO_3)_2}$ is mixed with 50.0 cm³ of $2.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ KI; (b) the same volumes of 0.010 and 0.020 mol dm⁻³ solutions are mixed?

Solution. Mixing equal volumes halves each concentration.

(a) $[\mathrm{Pb^{2+}}] = 5.0 \times 10^{-4}$ , $[\mathrm{I^{-}}] = 1.0 \times 10^{-3}$ :

$Q = (5.0 \times 10^{-4})(1.0 \times 10^{-3})^2 = 5.0 \times 10^{-10} < K_{sp}$

(b) $[\mathrm{Pb^{2+}}] = 5.0 \times 10^{-3}$ , $[\mathrm{I^{-}}] = 1.0 \times 10^{-2}$ :

$Q = (5.0 \times 10^{-3})(1.0 \times 10^{-2})^2 = 5.0 \times 10^{-7} > K_{sp}$

Answer: (a) no precipitate; the solution stays clear and unsaturated. (b) Lead(II) iodide precipitates, until the remaining ion concentrations give $Q = K_{sp}$ . Ten times the concentrations raises $Q$ a thousand-fold, because $[\mathrm{I^{-}}]$ is squared.

Worked example

Problem. A solution is 0.010 mol dm⁻³ in both $\mathrm{I^{-}}$ and $\mathrm{Cl^{-}}$ . Solid silver nitrate is added a little at a time (ignore any volume change). (a) Which salt precipitates first? (b) What is $[\mathrm{I^{-}}]$ when AgCl begins to precipitate, and what fraction of the iodide is still in solution?

Solution.

(a) AgI starts when $[\mathrm{Ag^{+}}] = \frac{8.52 \times 10^{-17}}{0.010} = 8.5 \times 10^{-15}$ ; AgCl starts when $[\mathrm{Ag^{+}}] = \frac{1.77 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8}$ . AgI needs far less silver, so it precipitates first.

(b) When AgCl starts, $[\mathrm{Ag^{+}}] = 1.77 \times 10^{-8}$ , and the solution is still saturated with AgI:

$[\mathrm{I^{-}}] = \frac{8.52 \times 10^{-17}}{1.77 \times 10^{-8}} = 4.8 \times 10^{-9}\,\mathrm{mol\,dm^{-3}}$

Fraction left $= \frac{4.8 \times 10^{-9}}{0.010} = 4.8 \times 10^{-7}$ .

Answer: (a) silver iodide; (b) $4.8 \times 10^{-9}\,\mathrm{mol\,dm^{-3}}$ , only 0.000048% of the iodide, so the separation is essentially complete before any AgCl forms.

Warning

Common mistakes

Using the concentrations before mixing. Always allow for dilution: mixing 25 cm³ with 25 cm³ halves both concentrations.
Forgetting the powers in $Q$ : for $\mathrm{PbI_2}$ , $[\mathrm{I^{-}}]$ is squared in $Q$ just as in $K_{sp}$ .
Deciding the order of precipitation from $K_{sp}$ alone when the salts have different formulas (such as AgCl and $\mathrm{Ag_2CrO_4}$ ). Calculate the reagent concentration needed for each.
Thinking a precipitate removes an ion completely. Some always remains, set by $K_{sp}$ .

Tip

Remember "Q over K, out it comes": if $Q > K_{sp}$ , solid comes out of solution. For a mixture, find the threshold for each salt; the lowest threshold wins.

Khan Academy: Worked example, predicting whether a precipitate forms by comparing Q and Ksp. Watch how the concentrations after mixing are found first, before Q is calculated.

✦

Practice

State what happens when the ionic product of a salt is (a) less than, (b) equal to, (c) greater than its $K_{sp}$ .
10.0 cm³ of $1.0 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ $\mathrm{BaCl_2}$ is mixed with 90.0 cm³ of $1.0 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ $\mathrm{Na_2SO_4}$ . Will barium sulfate precipitate?
25.0 cm³ of 0.020 mol dm⁻³ $\mathrm{CaCl_2}$ is mixed with 25.0 cm³ of 0.020 mol dm⁻³ $\mathrm{Na_2SO_4}$ . Will calcium sulfate precipitate?
Water contains $2.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ of $\mathrm{Ca^{2+}}$ . What is the minimum carbonate ion concentration needed to start precipitating calcium carbonate?
A solution is 0.010 mol dm⁻³ in both $\mathrm{Cl^{-}}$ and $\mathrm{CrO_4^{2-}}$ . Silver nitrate is added slowly. Which precipitate forms first? Show your working.
A solution is 0.010 mol dm⁻³ in both $\mathrm{Ba^{2+}}$ and $\mathrm{Ca^{2+}}$ . Sodium sulfate is added slowly. Which sulfate precipitates first, and what is $[\mathrm{Ba^{2+}}]$ when the second begins to precipitate? Is this a good separation?
Explain why Mohr's titration gives a poor result in acidic solution, and evaluate what would happen to the end point if far too much potassium chromate (say 0.10 mol dm⁻³) were used.

Show answersHide answers

(a) Unsaturated: no precipitate, more could dissolve. (b) Just saturated. (c) Supersaturated: the salt precipitates until $Q = K_{sp}$ .
After mixing: $[\mathrm{Ba^{2+}}] = 1.0 \times 10^{-5}$ , $[\mathrm{SO_4^{2-}}] = 9.0 \times 10^{-5}$ ; $Q = 9.0 \times 10^{-10} > 1.08 \times 10^{-10}$ . Yes, a little $\mathrm{BaSO_4}$ precipitates.
Each becomes 0.010 mol dm⁻³; $Q = 1.0 \times 10^{-4} > 4.93 \times 10^{-5}$ . Yes, $\mathrm{CaSO_4}$ precipitates.
$[\mathrm{CO_3^{2-}}] = \frac{3.36 \times 10^{-9}}{2.0 \times 10^{-3}} = 1.7 \times 10^{-6}\,\mathrm{mol\,dm^{-3}}$ .
AgCl needs $[\mathrm{Ag^{+}}] = \frac{1.77 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8}$ ; $\mathrm{Ag_2CrO_4}$ needs $\sqrt{\frac{1.12 \times 10^{-12}}{0.010}} = 1.1 \times 10^{-5}$ . AgCl precipitates first, although its $K_{sp}$ is larger.
$\mathrm{BaSO_4}$ first: it needs $[\mathrm{SO_4^{2-}}] = 1.1 \times 10^{-8}$ , against $4.9 \times 10^{-3}$ for $\mathrm{CaSO_4}$ . When $\mathrm{CaSO_4}$ starts, $[\mathrm{Ba^{2+}}] = \frac{1.08 \times 10^{-10}}{4.93 \times 10^{-3}} = 2.2 \times 10^{-8}\,\mathrm{mol\,dm^{-3}}$ , about 0.0002% of the barium. Yes, an excellent separation.
In acid, $\mathrm{2CrO_4^{2-} + 2H^{+} \rightleftharpoons Cr_2O_7^{2-} + H_2O}$ lowers the chromate concentration, so more $\mathrm{Ag^{+}}$ is needed before $\mathrm{Ag_2CrO_4}$ forms and the end point comes late. With 0.10 mol dm⁻³ chromate, $\mathrm{Ag_2CrO_4}$ would form at $[\mathrm{Ag^{+}}] = \sqrt{\frac{1.12 \times 10^{-12}}{0.10}} = 3.3 \times 10^{-6}$ , before the equivalence point ( $1.33 \times 10^{-5}$ ), so the end point would come early; the intense yellow of so much chromate would also hide the colour change.

Sources and credits

$K_{sp}$ values at 25 °C: CRC Handbook of Chemistry and Physics, 84th edition (2004), as tabulated in E3: Solubility Constants for Compounds at 25°C, Chemistry LibreTexts (numerical data only; no text adapted).
Mohr's method and its pH conditions: Argentometry, Wikipedia contributors, Wikipedia, licensed under CC BY-SA 4.0, adapted by Ganzaa.
Image: Определување на хлориди во вода според Моровиот метод (1).jpg by Deamit, via Wikimedia Commons, licensed under CC BY 4.0.
Video: Worked example: Predicting whether a precipitate forms by comparing Q and Ksp, Khan Academy, embedded from YouTube.
The text, scale diagram, table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1A white precipitate first, a red one at the end

2Will a precipitate form?

3Which precipitate forms first?

4Mohr's method explained

5Separating metal ions by controlling pH

A white precipitate first, a red one at the end

Will a precipitate form?

Which precipitate forms first?

Mohr's method explained

Separating metal ions by controlling pH