The Solubility Product, Ksp – Solubility and Solubility Product of Sparingly Soluble Salts (Senior 6 Chemistry) | Ganzaa

Solubility and Solubility Product of Sparingly Soluble Salts

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Senior 6 · Chemistry

Solubility and Solubility Product of Sparingly Soluble Salts

4 lessons

Lesson 2 of 4

The Solubility Product, Ksp

25 min

How much of a "precipitate" stays dissolved?

Pour colourless potassium iodide solution into colourless lead(II) nitrate solution and a bright yellow solid, lead(II) iodide, swirls out at once. You may have written $\mathrm{Pb^{2+}(aq) + 2I^{-}(aq) \rightarrow PbI_2(s)}$ as if the reaction went to completion. It does not quite. A little lead(II) iodide always stays dissolved, about 0.6 g per dm³ at 25 °C, and the solid and its ions settle into an equilibrium.

In Unit 7 you wrote equilibrium constants for reactions in solution. Here you will write one for a dissolving salt, the solubility product, and use it to calculate how much of a "precipitate" dissolves.

Definition

Solubility product, $K_{sp}$ : the equilibrium constant for a sparingly soluble salt in equilibrium with its saturated solution. For $\mathrm{A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)}$ :

$K_{sp} = [\mathrm{A^{y+}}]^{x}\,[\mathrm{B^{x-}}]^{y}$

The concentrations are those in the saturated solution at equilibrium; the solid does not appear. $K_{sp}$ has units, which depend on the formula, and it changes only with temperature. Example: for $\mathrm{PbI_2}$ , $K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2 = 9.8 \times 10^{-9}\,\mathrm{mol^3\,dm^{-9}}$ at 25 °C.

Writing $K_{sp}$ expressions

Salt	Equilibrium	$K_{sp}$ expression	Units
silver chloride	$\mathrm{AgCl(s) \rightleftharpoons Ag^{+} + Cl^{-}}$	$[\mathrm{Ag^{+}}][\mathrm{Cl^{-}}]$	$\mathrm{mol^2\,dm^{-6}}$
calcium fluoride	$\mathrm{CaF_2(s) \rightleftharpoons Ca^{2+} + 2F^{-}}$	$[\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2$	$\mathrm{mol^3\,dm^{-9}}$
silver chromate	$\mathrm{Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+} + CrO_4^{2-}}$	$[\mathrm{Ag^{+}}]^2[\mathrm{CrO_4^{2-}}]$	$\mathrm{mol^3\,dm^{-9}}$
iron(III) hydroxide	$\mathrm{Fe(OH)_3(s) \rightleftharpoons Fe^{3+} + 3OH^{-}}$	$[\mathrm{Fe^{3+}}][\mathrm{OH^{-}}]^3$	$\mathrm{mol^4\,dm^{-12}}$
calcium phosphate	$\mathrm{Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}}$	$[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2$	$\mathrm{mol^5\,dm^{-15}}$

Linking $K_{sp}$ and molar solubility

If $s$ mol of salt dissolves per dm³ of pure water, each ion's concentration is $s$ times its number in the formula. Substituting gives:

1 : 1 salts (AgCl, $\mathrm{BaSO_4}$ ): $K_{sp} = s \times s = s^2$ , so $s = \sqrt{K_{sp}}$ .
1 : 2 or 2 : 1 salts ( $\mathrm{CaF_2}$ , $\mathrm{PbI_2}$ , $\mathrm{Ag_2CrO_4}$ ): $K_{sp} = s(2s)^2 = 4s^3$ , so $s = \sqrt[3]{\frac{K_{sp}}{4}}$ .
1 : 3 salts ( $\mathrm{Fe(OH)_3}$ ): $K_{sp} = s(3s)^3 = 27s^4$ .
3 : 2 salts ( $\mathrm{Ca_3(PO_4)_2}$ ): $K_{sp} = (3s)^3(2s)^2 = 108s^5$ .

Solubility products at 25 °C

Salt	$K_{sp}$	Salt	$K_{sp}$
AgCl	$1.77 \times 10^{-10}$	$\mathrm{CaCO_3}$ (calcite)	$3.36 \times 10^{-9}$
AgBr	$5.35 \times 10^{-13}$	$\mathrm{CaF_2}$	$3.45 \times 10^{-11}$
AgI	$8.52 \times 10^{-17}$	$\mathrm{CaSO_4}$	$4.93 \times 10^{-5}$
$\mathrm{Ag_2CrO_4}$	$1.12 \times 10^{-12}$	$\mathrm{Ca(OH)_2}$	$5.02 \times 10^{-6}$
$\mathrm{BaSO_4}$	$1.08 \times 10^{-10}$	$\mathrm{Mg(OH)_2}$	$5.61 \times 10^{-12}$
$\mathrm{BaCO_3}$	$2.58 \times 10^{-9}$	$\mathrm{PbI_2}$	$9.8 \times 10^{-9}$
$\mathrm{PbCl_2}$	$1.70 \times 10^{-5}$	$\mathrm{Fe(OH)_3}$	$2.79 \times 10^{-39}$

Units are omitted in the table; work them out from each formula. Figure 2.1 converts some of these values to molar solubilities.

Figure 2.1: Molar solubilities calculated from $K_{sp}$ . Silver chromate (orange) has a smaller $K_{sp}$ than silver chloride, yet it is about five times more soluble.

Figure 2.1 shows why you may compare $K_{sp}$ values directly only for salts of the same formula type. AgCl, AgBr and AgI are all 1 : 1, so their order of $K_{sp}$ is their order of solubility. Silver chromate is 2 : 1, so its $K_{sp}$ involves $s^{3}$ , not $s^{2}$ , and the comparison breaks down.

Worked example

Problem. At 25 °C a saturated solution of lead(II) iodide contains 0.621 g of $\mathrm{PbI_2}$ per dm³ ( $M = 461.0\,\mathrm{g\,mol^{-1}}$ ). Calculate $K_{sp}$ .

Solution.

$s = \frac{0.621}{461.0} = 1.347 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$

$\mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)}$ , so $[\mathrm{Pb^{2+}}] = s$ and $[\mathrm{I^{-}}] = 2s = 2.694 \times 10^{-3}$ .

$K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2 = (1.347 \times 10^{-3})(2.694 \times 10^{-3})^2 = 9.78 \times 10^{-9}\,\mathrm{mol^3\,dm^{-9}}$

Answer: $K_{sp} = 9.8 \times 10^{-9}\,\mathrm{mol^3\,dm^{-9}}$ , matching the table.

Worked example

Problem. Calculate (a) the molar solubility of barium sulfate and its mass solubility in mg dm⁻³ ( $M = 233.39\,\mathrm{g\,mol^{-1}}$ ), and (b) the pH of a saturated solution of calcium hydroxide at 25 °C. Use the table.

Solution.

(a) $s = \sqrt{K_{sp}} = \sqrt{1.08 \times 10^{-10}} = 1.04 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ ; mass: $1.04 \times 10^{-5} \times 233.39 = 2.43 \times 10^{-3}\,\mathrm{g\,dm^{-3}} = 2.43\,\mathrm{mg\,dm^{-3}}$ .

(b) $\mathrm{Ca(OH)_2(s) \rightleftharpoons Ca^{2+} + 2OH^{-}}$ , so $K_{sp} = 4s^3$ :

$s = \sqrt[3]{\frac{5.02 \times 10^{-6}}{4}} = 1.08 \times 10^{-2}\,\mathrm{mol\,dm^{-3}}, \qquad [\mathrm{OH^{-}}] = 2s = 2.16 \times 10^{-2}$

$\mathrm{pOH} = -\log_{10}(2.16 \times 10^{-2}) = 1.67, \qquad \mathrm{pH} = 14.00 - 1.67 = 12.33$

Answer: (a) $1.04 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ , 2.43 mg dm⁻³; (b) pH 12.33. Limewater is strongly alkaline even though very little calcium hydroxide dissolves.

Worked example

Problem. Which is more soluble in water at 25 °C, silver chloride ( $K_{sp} = 1.77 \times 10^{-10}$ ) or silver chromate ( $K_{sp} = 1.12 \times 10^{-12}$ )? Show your working.

Solution. AgCl: $s = \sqrt{1.77 \times 10^{-10}} = 1.33 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ .

$\mathrm{Ag_2CrO_4}$ : $K_{sp} = (2s)^2 s = 4s^3$ , so $s = \sqrt[3]{\frac{1.12 \times 10^{-12}}{4}} = 6.54 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ .

Answer: silver chromate is about five times more soluble, although its $K_{sp}$ is about 160 times smaller. Never compare $K_{sp}$ values of salts with different formula types; compare their molar solubilities.

Warning

Common mistakes with $K_{sp}$

Forgetting to double (or triple) the concentration of an ion that appears twice (or three times) in the formula, or forgetting to square (or cube) it. For $\mathrm{PbI_2}$ , $[\mathrm{I^{-}}] = 2s$ and it is squared: $4 s^{3}$ , not $2 s^{2}$ or $s^{3}$ .
Putting the solid into the expression. Its "concentration" is constant and is already part of $K_{sp}$ .
Comparing $K_{sp}$ values of salts with different formula types to rank their solubilities.
Applying $K_{sp}$ to soluble salts such as NaCl. The simple expression works well only when the ion concentrations are very small; even for the more soluble $\mathrm{Ca(OH)_2}$ and $\mathrm{PbCl_2}$ , values calculated this way are only approximate.

Tip

For any $K_{sp}$ problem, write the dissolving equation first, then write each ion's concentration in terms of $s$ directly under it, exactly as in an equilibrium table. The expression then writes itself: each concentration raised to the power of its coefficient.

Professor Dave Explains: Solubility product constant (Ksp). Watch how the dissolving equation for a salt is turned into a Ksp expression, and how the coefficients become powers.

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Practice

Write the $K_{sp}$ expression, with units, for (a) $\mathrm{BaCO_3}$ ; (b) $\mathrm{Mg(OH)_2}$ ; (c) $\mathrm{Ag_3PO_4}$ .
The molar solubility of silver bromide at 25 °C is $7.31 \times 10^{-7}\,\mathrm{mol\,dm^{-3}}$ . Calculate its $K_{sp}$ .
Calculate the molar solubility of calcium fluoride at 25 °C, and the concentration of fluoride ions in its saturated solution in $\mathrm{mg\,dm^{-3}}$ .
Calculate the pH of a saturated solution of magnesium hydroxide at 25 °C.
Show that for a salt of formula $\mathrm{AB_3}$ , $K_{sp} = 27s^4$ , and use it to find the molar solubility of iron(III) hydroxide.
Arrange AgCl, AgBr and AgI in order of increasing solubility, and explain why you may use their $K_{sp}$ values directly to do so.
A student concludes from the table that $\mathrm{CaF_2}$ ( $K_{sp} = 3.45 \times 10^{-11}$ ) is less soluble than $\mathrm{BaSO_4}$ ( $K_{sp} = 1.08 \times 10^{-10}$ ). Evaluate this conclusion with calculations.

Show answersHide answers

(a) $[\mathrm{Ba^{2+}}][\mathrm{CO_3^{2-}}]$ , $\mathrm{mol^2\,dm^{-6}}$ ; (b) $[\mathrm{Mg^{2+}}][\mathrm{OH^{-}}]^2$ , $\mathrm{mol^3\,dm^{-9}}$ ; (c) $[\mathrm{Ag^{+}}]^3[\mathrm{PO_4^{3-}}]$ , $\mathrm{mol^4\,dm^{-12}}$ .
$K_{sp} = s^2 = (7.31 \times 10^{-7})^2 = 5.34 \times 10^{-13}\,\mathrm{mol^2\,dm^{-6}}$ .
$s = \sqrt[3]{\frac{3.45 \times 10^{-11}}{4}} = 2.05 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ ; $[\mathrm{F^{-}}] = 2s = 4.10 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ , $\times 19.00 \times 1000 = 7.8\,\mathrm{mg\,dm^{-3}}$ .
$s = \sqrt[3]{\frac{5.61 \times 10^{-12}}{4}} = 1.12 \times 10^{-4}$ ; $[\mathrm{OH^{-}}] = 2.24 \times 10^{-4}$ , pOH = 3.65, pH = 10.35.
$K_{sp} = [\mathrm{Fe^{3+}}][\mathrm{OH^{-}}]^3 = s(3s)^3 = 27s^4$ . $s = \left(\frac{2.79 \times 10^{-39}}{27}\right)^{\frac{1}{4}} = 1.01 \times 10^{-10}\,\mathrm{mol\,dm^{-3}}$ .
AgI < AgBr < AgCl. All three are 1 : 1 salts, so $s = \sqrt{K_{sp}}$ in each case and the larger $K_{sp}$ always means the larger solubility.
The conclusion is wrong. $\mathrm{BaSO_4}$ : $s = \sqrt{1.08 \times 10^{-10}} = 1.04 \times 10^{-5}$ . $\mathrm{CaF_2}$ : $s = \sqrt[3]{\frac{3.45 \times 10^{-11}}{4}} = 2.05 \times 10^{-4}$ , about 20 times larger. The salts have different formula types, so their $K_{sp}$ values cannot be compared directly.

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).
Image: Lead (II) iodide precipitating out of solution.JPG by PRHaney, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Solubility Product Constant (Ksp), Professor Dave Explains, embedded from YouTube.
The text, bar chart, tables, 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.

1How much of a "precipitate" stays dissolved?

2Writing Ksp K_{sp} Ksp​ expressions

3Linking Ksp K_{sp} Ksp​ and molar solubility

4Solubility products at 25 °C

How much of a "precipitate" stays dissolved?

Writing $K_{sp}$ expressions

Linking $K_{sp}$ and molar solubility

Solubility products at 25 °C