E2.4 Solve equilibrium-concentration, solubility, and pH calculations

Ontario Grade 12 Chemistry

Chemical Systems and Equilibrium

SCH4U E2.4 study guide

A clear solution can contain dissolved particles that cannot be seen. When a small amount of an ionic solid dissolves, ions spread through the water. If some solid remains, ions can also rejoin the solid. At equilibrium, these changes continue at equal rates, so the concentrations stay constant. This lesson uses equilibrium relationships to calculate concentrations, solubility, and pH.

What you will learn

Use equilibrium expressions to calculate an unknown equilibrium concentration.
Use a solubility-product expression to calculate molar solubility.
Calculate pH from hydrogen-ion concentration and report results with appropriate significant figures.

1. Prerequisite bridge: concentration and equilibrium

Concentration describes the amount of dissolved substance in a volume of solution. Molar concentration is the amount in moles divided by the solution volume in litres. Its unit is moles per litre.

A reversible reaction can proceed in both directions. At dynamic equilibrium, the forward and reverse reactions continue at equal rates. Concentrations remain constant, but they are not necessarily equal.

An equilibrium expression relates the concentrations of products and reactants at equilibrium. Balance the reaction first. Each coefficient becomes an exponent on that substance’s concentration. Pure solids and pure liquids are left out of the expression.

a\mathrm{A}(aq)+b\mathrm{B}(aq)\rightleftharpoons c\mathrm{C}(aq)+d\mathrm{D}(aq),\quad K_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

Convert volume to litres when calculating molar concentration.
Equilibrium means constant concentrations, not equal concentrations.
Use the balanced equation to set the powers in an equilibrium expression.

2. Equilibrium-concentration calculations

The equilibrium constant,

K_c

, relates equilibrium concentrations for a reaction at a stated temperature. Use concentrations at equilibrium, rather than starting concentrations, when substituting into the expression.

Balance the reaction and write its equilibrium expression. Substitute the known equilibrium concentrations and the value of

K_c

. Rearrange to find the unknown concentration. A concentration must be positive.

Coefficients affect the calculation as powers. For example, if a product has coefficient 2, its concentration is squared in the expression. Keep concentration units consistent.

K_c=\frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}

Use equilibrium concentrations in the expression.
Use coefficients from the balanced equation as exponents.
Check that the calculated concentration is positive and has appropriate units.

3. Solubility and the solubility product

Solubility describes how much of a substance dissolves. Molar solubility is the amount of solid that dissolves per litre of solution, measured in

\mathrm{mol\,L^{-1}}

. If undissolved solid remains, the solid and its dissolved ions may be at equilibrium.

The solubility-product constant,

K_{sp}

, describes this equilibrium. Write the balanced dissolving equation, omit the solid, and raise each aqueous-ion concentration to its coefficient.

To find molar solubility, let

s

represent the amount of solid dissolved per litre. Use the equation coefficients to express each ion concentration in terms of

s

. Substitute those concentrations into the

K_{sp}

expression and solve for the positive value of

s

. The example below works through this process for a solid that produces one cation and two anions.

K_{sp}=\prod [aqueous ion]^{coefficient}

Write a balanced dissolving equation before writing the $K_{sp}$ expression.
Omit the solid; include the aqueous ions.
Relate each ion concentration to molar solubility using the equation coefficients.

4. pH calculations and careful reporting

The pH of an aqueous solution is calculated from its hydrogen-ion concentration, written

[\mathrm{H^+}]

. Use concentration in

\mathrm{mol\,L^{-1}}

. A greater hydrogen-ion concentration gives a lower pH.

To calculate pH from concentration, take the negative base-10 logarithm. To calculate concentration from pH, raise 10 to the negative pH. Hydrogen-ion concentration must be positive.

For pH reporting, the number of decimal places matches the number of significant figures in the hydrogen-ion concentration. Keep extra digits during the calculation and round only the final result.

\mathrm{pH}=-\log_{10}[\mathrm{H^+}],\qquad [\mathrm{H^+}]=10^{-\mathrm{pH}}

Use hydrogen-ion concentration in $\mathrm{mol\,L^{-1}}$ .
A larger hydrogen-ion concentration corresponds to a lower pH.
Match pH decimal places to the concentration’s significant figures.

Worked example

Calculate equilibrium concentration and molar solubility

At a stated temperature, the equilibrium constant for

\mathrm{H_2}(g)+\mathrm{I_2}(g)\rightleftharpoons 2\mathrm{HI}(g)

is

K_c=4.0

. At equilibrium,

[\mathrm{H_2}]=0.20\ \mathrm{mol\,L^{-1}}

and

[\mathrm{I_2}]=0.10\ \mathrm{mol\,L^{-1}}

. Calculate

[\mathrm{HI}]

. Then, at the same stated temperature, a sparingly soluble solid has

K_{sp}=3.2\times10^{-11}

and dissolves according to

\mathrm{M X_2}(s)\rightleftharpoons\mathrm{M^{2+}}(aq)+2\mathrm{X^-}(aq)

. Calculate its molar solubility in pure water.

Write the first equilibrium expression
The hydrogen iodide coefficient is 2, so its concentration is squared. Substitute the stated equilibrium concentrations and isolate the square of the unknown.
$K_c=\frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$
Calculate the hydrogen iodide concentration
Multiply $K_c$ by the two known concentrations, then take the positive square root. The result is rounded to two significant figures, as are the given values.
$[\mathrm{HI}]=\sqrt{(4.0)(0.20\ \mathrm{mol\,L^{-1}})(0.10\ \mathrm{mol\,L^{-1}})}=0.28\ \mathrm{mol\,L^{-1}}$
Relate dissolved ions to molar solubility
Let $s$ be the molar solubility. Each dissolved formula unit produces one $\mathrm{M^{2+}}$ ion and two $\mathrm{X^-}$ ions. Therefore, the equilibrium ion concentrations are $s$ and $2s$ . The solid is omitted from the solubility-product expression.
$K_{sp}=[\mathrm{M^{2+}}][\mathrm{X^-}]^2=s(2s)^2=4s^3$
Solve for molar solubility
Substitute the given value of $K_{sp}$ and take the positive cube root. The units of $s$ are concentration units. The reported value has two significant figures.
$s=\sqrt[3]{\frac{3.2\times10^{-11}}{4}}=2.0\times10^{-4}\ \mathrm{mol\,L^{-1}}$

Answer: The equilibrium hydrogen iodide concentration is

0.28\ \mathrm{mol\,L^{-1}}

. The solid’s molar solubility is

2.0\times10^{-4}\ \mathrm{mol\,L^{-1}}

.

Check: For the first calculation,

[\mathrm{HI}]^2/([\mathrm{H_2}][\mathrm{I_2}])=(0.28)^2/[(0.20)(0.10)]=3.92

, which rounds to

4.0

. For the solubility calculation,

4(2.0\times10^{-4})^3=3.2\times10^{-11}

, matching the given

K_{sp}

.

Common mistakes and how to avoid them

Assuming equilibrium means all concentrations are equal.

Correction: At equilibrium, concentrations stay constant because forward and reverse rates are equal. The concentrations can differ.

Using starting concentrations in an equilibrium expression.

Correction: Use concentrations at equilibrium when calculating with the equilibrium expression.

Including a pure solid in a

K_c

or

K_{sp}

expression.

Correction: Omit pure solids and liquids. Include the relevant aqueous species or gases.

Using reaction coefficients as multipliers rather than powers in an equilibrium expression.

Correction: Use each coefficient as the exponent on that species’ concentration.

Forgetting that a coefficient changes the ion concentration in terms of molar solubility.

Correction: Use the balanced dissolving equation. A coefficient of 2 means the ion concentration is

2s

when the solid’s molar solubility is

s

.

Lesson summary

At equilibrium, concentrations are constant but are not necessarily equal.
Balance the reaction before writing an equilibrium expression.
Use equilibrium concentrations and raise each concentration to its coefficient.
For $K_{sp}$ , omit the solid and relate ion concentrations to molar solubility using the balanced equation.
Calculate pH from hydrogen-ion concentration and report it with appropriate significant figures.

Check your understanding

Question 1

For

\mathrm{CaF_2}(s)\rightleftharpoons\mathrm{Ca^{2+}}(aq)+2\mathrm{F^-}(aq)

, which expression represents

K_{sp}

?

$[\mathrm{Ca^{2+}}][\mathrm{F^-}]$
$[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2$
$[\mathrm{CaF_2}][\mathrm{F^-}]^2$
$[\mathrm{Ca^{2+}}]^2[\mathrm{F^-}]$

Show answer and explanation

[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2

Omit the solid. The coefficient 2 becomes the power on the fluoride concentration.

Question 2

A solution has

[\mathrm{H^+}]=1.0\times10^{-3}\ \mathrm{mol\,L^{-1}}

. What is its pH?

$3.00$
$-3.00$
$1.0\times10^{-3}$
$11.00$

Show answer and explanation

3.00

The negative base-10 logarithm gives pH

3.00

. The concentration has two significant figures, so report two decimal places.

Key terms

Equilibrium concentration: The concentration of a substance when a reversible reaction has reached equilibrium.
Dynamic equilibrium: A state in which forward and reverse reactions continue at equal rates, so concentrations remain constant.
Molar solubility: The amount of a substance that dissolves per litre of solution, expressed in moles per litre.
Solubility-product constant, $K_{sp}$: The equilibrium constant expression for the ions formed when a sparingly soluble ionic solid dissolves.
pH: A value calculated from the hydrogen-ion concentration of an aqueous solution.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Chemistry (SCH4U), expectation E2.4. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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E2.4 · Solve equilibrium-concentration, solubility, and pH calculations