Solubility Calculations Revision Notes for HSC SSCE Chemistry

Solubility Calculations

When working with sparingly soluble or insoluble salts in water, there is a direct mathematical relationship between the solubility product constant ( $K_{sp}$ ) and the concentration of ions in solution. This connection is essential for understanding and predicting the behaviour of ionic compounds in aqueous solutions. In this note, you'll learn how to calculate $K_{sp}$ values from solubility data and, conversely, how to determine solubility from known $K_{sp}$ values.

Understanding molar solubility

Molar solubility refers to the concentration of a dissolved substance expressed in moles per litre (mol L⁻¹). This is different from solubility expressed as grams per volume, which is commonly seen on product labels.

When a salt dissolves in pure distilled water (with no other contaminants present), the molar solubility is directly connected to the solubility product constant. For sparingly soluble and insoluble salts, an equilibrium is established between the undissolved solid and the dissolved ions in solution. This equilibrium relationship allows us to perform calculations in both directions: from solubility to $K_{sp}$ , and from $K_{sp}$ to solubility.

infoNote

This direct relationship only applies when the salt is dissolved in pure water. The presence of other ions or substances can affect solubility through the common ion effect or other chemical interactions.

Calculating $K_{sp}$ from solubility data

When you know the solubility of a salt, you can determine its solubility product constant by following a systematic approach. Let's break this down using barium sulfate as an example.

Step-by-step process

Step 1: Write the balanced dissolution equation

Begin by writing the equation showing how the solid salt dissociates into ions in water. For barium sulfate:

$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

Step 2: Convert solubility to mol L⁻¹

Solubility data is often given as grams per 100 grams of water. You need to convert this to molar concentration (mol L⁻¹).

For barium sulfate with solubility of 0.00025 g/100 g water:

First, convert to g L⁻¹: $\text{Solubility} = 0.00025 \text{ g/100 g water} = 0.0025 \text{ g L}^{-1}$

Then, calculate the number of moles using the molar mass: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.0025}{233.4} = 1.07 \times 10^{-5} \text{ mol}$

Therefore, the solubility is $1.07 \times 10^{-5}$ mol L⁻¹.

Step 3: Determine ion concentrations

Use the stoichiometric ratios from your balanced equation. From the equation above, 1 mole of BaSO₄ produces 1 mole of Ba²⁺ ions and 1 mole of SO₄²⁻ ions.

This means:

[Ba²⁺] = $1.07 \times 10^{-5}$ mol L⁻¹
[SO₄²⁻] = $1.07 \times 10^{-5}$ mol L⁻¹

Step 4: Write the equilibrium expression

For this dissolution equilibrium: $K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$

Step 5: Substitute and calculate

$K_{sp} = [1.07 \times 10^{-5}] \times [1.07 \times 10^{-5}] = [1.07 \times 10^{-5}]^2 = 1.14 \times 10^{-10}$

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Worked Example: Calculating $K_{sp}$ for Zinc Hydroxide

Let's apply this process to zinc hydroxide, which has a solubility of 0.001 g/100 g water at 298 K.

Setting up the problem:

The dissolution equation is: $\text{Zn(OH)}_2(s) \rightleftharpoons \text{Zn}^{2+}(aq) + 2\text{OH}^-(aq)$

Converting units:

Solubility = 0.001 g/100 g H₂O = 0.01 g L⁻¹

Molar mass of Zn(OH)₂ = 99.4 g

$n = \frac{0.01}{99.4} = 1.01 \times 10^{-4} \text{ mol}$

Solubility = $1.01 \times 10^{-4}$ mol L⁻¹

Determining ion concentrations:

Notice the stoichiometry: 1 mole of Zn(OH)₂ produces 1 mole of Zn²⁺ ions but 2 moles of OH⁻ ions.

Therefore:

[Zn²⁺] = $1.01 \times 10^{-4}$ mol L⁻¹
[OH⁻] = $2 \times 1.01 \times 10^{-4} = 2.02 \times 10^{-4}$ mol L⁻¹

Calculating $K_{sp}$ :

$K_{sp} = [\text{Zn}^{2+}][\text{OH}^-]^2$

$K_{sp} = [1.01 \times 10^{-4}] \times [2.02 \times 10^{-4}]^2 = 4.12 \times 10^{-12}$

chatImportant

Always pay careful attention to the stoichiometric coefficients in your balanced equation. When a coefficient is 2 (or higher), you must multiply the solubility by that coefficient to find the ion concentration, and the concentration appears as a squared term (or higher power) in the $K_{sp}$ expression.

Calculating solubility from $K_{sp}$

Working in the opposite direction, if you know the $K_{sp}$ value, you can calculate the solubility of the salt and the concentrations of its constituent ions. This process requires some algebraic manipulation.

Using algebra with variable 's'

Since we don't know the solubility initially, we assign it the variable 's' (representing solubility in mol L⁻¹). We then express all ion concentrations in terms of 's' using the stoichiometric ratios.

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Worked Example: Manganese(II) Carbonate Ion Concentrations

For manganese(II) carbonate with $K_{sp} = 2.24 \times 10^{-11}$ at 25°C:

Step 1: Write the balanced equation

$\text{MnCO}_3(s) \rightleftharpoons \text{Mn}^{2+}(aq) + \text{CO}_3^{2-}(aq)$

Step 2: Express concentrations in terms of 's'

The stoichiometry shows a 1:1:1 ratio. If the solubility is $s$ mol L⁻¹:

[Mn²⁺] = $s$ mol L⁻¹
[CO₃²⁻] = $s$ mol L⁻¹

Step 3: Write the equilibrium expression $K_{sp} = [\text{Mn}^{2+}][\text{CO}_3^{2-}]$

Step 4: Substitute and solve

$2.24 \times 10^{-11} = [s] \times [s]$

$s^2 = 2.24 \times 10^{-11}$

$s = \sqrt{2.24 \times 10^{-11}}$

$s = 4.73 \times 10^{-6}$

Therefore:

[Mn²⁺] = $4.73 \times 10^{-6}$ mol L⁻¹
[CO₃²⁻] = $4.73 \times 10^{-6}$ mol L⁻¹

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Worked Example: Calcium Iodate Ion Concentrations

For calcium iodate, Ca(IO₃)₂, with $K_{sp} = 6.47 \times 10^{-6}$ at 25°C, the process is slightly more complex due to the stoichiometry.

Step 1: Write the balanced equation $\text{Ca(IO}_3)_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{IO}_3^-(aq)$

Step 2: Express concentrations in terms of 's'

The stoichiometry shows 1 mole of solid produces 1 mole of Ca²⁺ and 2 moles of IO₃⁻:

[Ca²⁺] = $s$ mol L⁻¹
[IO₃⁻] = $2s$ mol L⁻¹

Step 3: Write the equilibrium expression

$K_{sp} = [\text{Ca}^{2+}][\text{IO}_3^-]^2$

Step 4: Substitute and solve

$6.47 \times 10^{-6} = [s] \times [2s]^2$

$6.47 \times 10^{-6} = s \times 4s^2$

$4s^3 = 6.47 \times 10^{-6}$

$s^3 = \frac{6.47 \times 10^{-6}}{4}$

$s = \sqrt[3]{\frac{6.47 \times 10^{-6}}{4}}$

$s = 0.0117$

Therefore:

[Ca²⁺] = $s$ = 0.0117 mol L⁻¹
[IO₃⁻] = $2s$ = $2 \times 0.0117$ = 0.0234 mol L⁻¹

infoNote

When the equilibrium expression involves a squared or cubed term, remember to take the appropriate root (square root or cube root) to solve for 's'. Be comfortable using your calculator's root functions.

Converting solubility between units

Often you need to express solubility in different units, such as grams per litre (g L⁻¹) or grams per 100 grams of water (g/100 g H₂O). You can also classify salts based on their solubility values.

Classification of solubility

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Classification of Salts by Solubility:

Insoluble salts: Solubility less than 1 g L⁻¹
Sparingly soluble salts: Solubility between 1 and approximately 10 g L⁻¹
Soluble salts: Solubility greater than 10 g L⁻¹

This classification system helps predict whether a precipitate will form in chemical reactions and is useful in qualitative analysis and industrial applications.

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Worked Example: Converting Calcium Iodate Solubility to Different Units

From the previous calculation, we found that calcium iodate has a molar solubility of 0.0117 mol L⁻¹. Let's convert this to g L⁻¹ and classify the salt.

Converting mol L⁻¹ to g L⁻¹:

Use the formula: $m = n \times MM$

Where MM is the molar mass. For Ca(IO₃)₂, the molar mass is 389.9 g/mol.

$\text{Solubility} = 0.0117 \times 389.9 = 4.56 \text{ g}$

Therefore, the solubility is 4.56 g L⁻¹.

Classifying the salt:

Since 4.56 g L⁻¹ is greater than 1 g L⁻¹, calcium iodate is classified as sparingly soluble rather than insoluble.

Ionic product and predicting precipitation

When a solution is not at equilibrium, we cannot use the term $K_{sp}$ . Instead, we use the ionic product, denoted as $Q_{sp}$ , which represents the reaction quotient for the dissolution equilibrium.

Understanding $Q_{sp}$ vs $K_{sp}$

The key difference is:

$K_{sp}$ : Only applies when the system is at equilibrium (saturated solution with both solid and dissolved ions present)
$Q_{sp}$ : Applies to any solution, whether at equilibrium or not

By comparing $Q_{sp}$ to $K_{sp}$ , we can predict what will happen in the solution:

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Predicting Precipitation by Comparing $Q_{sp}$ and $K_{sp}$ :

When $Q_{sp} < K_{sp}$ :

The solution is unsaturated
The forward reaction is favoured (more solid can dissolve)
No precipitate will form
The system will shift toward equilibrium by dissolving more solid

When $Q_{sp} = K_{sp}$ :

The solution is saturated
The system is at dynamic equilibrium
Both solid and dissolved ions are present at constant concentrations

When $Q_{sp} > K_{sp}$ :

The solution is supersaturated
The reverse reaction is favoured (ions will come out of solution)
A precipitate will form
The system will shift toward equilibrium by forming solid

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Worked Example: Predicting Barium Hydroxide Precipitation

Problem: If 0.500 g of Ba(OH)₂ is added to 200 mL of water at 25°C and stirred, will a precipitate form? The $K_{sp}$ for Ba(OH)₂ at 25°C is $2.55 \times 10^{-4}$ .

Step 1: Write the dissolution equation

$\text{Ba(OH)}_2(s) \rightleftharpoons \text{Ba}^{2+}(aq) + 2\text{OH}^-(aq)$

Step 2: Calculate the concentration of Ba(OH)₂

First, convert to g L⁻¹:

$0.500 \text{ g in } 200 \text{ mL} = 0.500 \times 5 = 2.50 \text{ g L}^{-1}$

Then convert to mol L⁻¹ using molar mass (171.3 g/mol): $n = \frac{m}{MM} = \frac{2.50}{171.3} = 0.0146 \text{ mol L}^{-1}$

Step 3: Calculate ion concentrations

If all the Ba(OH)₂ dissolved, the concentrations would be:

[Ba²⁺] = 0.0146 mol L⁻¹ (1:1 ratio)
[OH⁻] = $2 \times 0.0146 = 0.0292$ mol L⁻¹ (1:2 ratio)

Step 4: Calculate $Q_{sp}$

$Q_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2$

$Q_{sp} = [0.0146] \times [0.0292]^2 = 1.24 \times 10^{-5}$

Step 5: Compare $Q_{sp}$ to $K_{sp}$

$Q_{sp} = 1.24 \times 10^{-5}$

$K_{sp} = 2.55 \times 10^{-4}$

Since $Q_{sp} < K_{sp}$ , the solution is unsaturated and no precipitate will form. All of the barium hydroxide will dissolve.

chatImportant

Always complete your calculations before making a conclusion. Don't guess whether a precipitate will form—let the mathematics guide you by comparing $Q_{sp}$ and $K_{sp}$ values.

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Summary of Key Formulas

For converting solubility to molar concentration: $n = \frac{\text{mass}}{\text{molar mass}}$

For converting molar solubility to mass: $m = n \times MM$

General form of $K_{sp}$ expression:

For a salt $A_xB_y$ that dissolves as: $A_xB_y(s) \rightleftharpoons xA^{m+}(aq) + yB^{n-}(aq)$

The solubility product is: $K_{sp} = [A^{m+}]^x[B^{n-}]^y$

Ionic product: $Q_{sp} = [A^{m+}]^x[B^{n-}]^y \text{ (at any point, not necessarily at equilibrium)}$

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Key Points to Remember:

Molar solubility is the concentration of a dissolved salt in mol L⁻¹. It has a direct mathematical relationship with the solubility product constant ( $K_{sp}$ ) when only the salt and water are present.
To calculate $K_{sp}$ from solubility, convert the solubility to mol L⁻¹, determine ion concentrations using stoichiometric ratios, write the equilibrium expression, and substitute your values.
To calculate solubility from $K_{sp}$ , assign the unknown solubility the variable 's', express all ion concentrations in terms of 's', substitute into the $K_{sp}$ expression, and solve using algebra (taking square roots or cube roots as needed).
Salts are classified based on solubility: insoluble salts have solubility less than 1 g L⁻¹, while sparingly soluble salts have solubility greater than 1 g L⁻¹.
The ionic product ( $Q_{sp}$ ) is used when a solution is not at equilibrium. Compare $Q_{sp}$ to $K_{sp}$ : if $Q_{sp} < K_{sp}$ , no precipitate forms; if $Q_{sp} = K_{sp}$ , the solution is saturated; if $Q_{sp} > K_{sp}$ , a precipitate forms.

Solubility Calculations (HSC SSCE Chemistry): Revision Notes