Solving Equilibrium Problems Using K HL Only
Quick Notes
- The equilibrium constant (K) allows calculation of equilibrium concentrations of reactants and products in a homogeneous system.
- ICE tables (Initial, Change, Equilibrium) help organise known and unknown values.
- For very small K values, we can approximate that initial concentrations of reactans are the same as equilibrium concentrations: [Reactant]0 ≈ [Reactant]eqm.
- Only homogeneous equilibria (all species in the same phase) are considered.
Full Notes
Using an ICE Table
The expression for K relates the concentrations of substances at equilibrium, not before or after. To calculate K, we need to use equilibrium concentrations. These can be found using an ICE Table.
ICE stands for Initial, Change, Equilibrium — a structured way to track concentration changes in equilibrium problems.
Example Problem
For the reaction:
H2(g) + I2(g) ⇌ 2HI(g)
At 298 K, the equilibrium constant is K = 50.0.
At equilibrium: [HI] = 0.80 mol dm⁻³ and [I2] = 0.10 mol dm⁻³.
Calculate the equilibrium concentration of H2.
Step 1: Use an ICE table
| Species | Initial | Change | Equilibrium |
|---|---|---|---|
| H2 | a | −0.40 | a − 0.40 |
| I2 | – | −0.40 | 0.10 |
| HI | 0 | +0.80 | 0.80 |
Step 2: Apply the equilibrium expression
- K = [HI]2 / ([H2][I2])
- 50.0 = (0.80)2 / [(a − 0.40)(0.10)]
- 50.0 = 0.64 / [0.10(a − 0.40)]
- 0.10(a − 0.40) = 0.0128
- a − 0.40 = 0.128 → a = 0.528 mol dm⁻³
Step 3: Final Answer
[H2] at equilibrium = a − 0.40 = 0.128 mol dm⁻³
Approximation Use
If K is very small (K << 1), then little product forms at equilibrium, and we can assume the initial (starting) concentrations of reactants will be the same as the equilibrium concentrations:
[Reactant]initial ≈ [Reactant]eqm
Example Approximation for small K
Consider the equilibrium:
A ⇌ B
K = 1.0 × 10⁻⁵
Initially: [A] = 0.20 mol dm⁻³, [B] = 0 mol dm⁻³.
Estimate the equilibrium concentration of B.
| Species | Initial | Change | Equilibrium |
|---|---|---|---|
| A | 0.20 | −x | 0.20 − x |
| B | 0 | +x | x |
- K = [B] / [A] = x / (0.20 − x)
- Since K is very small, assume 0.20 − x ≈ 0.20
- 1.0 × 10⁻⁵ = x / 0.20 → x = 2.0 × 10⁻⁶ mol dm⁻³
Conclusion: At equilibrium, only a tiny amount of B is formed and [A] hardly changes — justifying the approximation.
[A] ≈ 0.20 mol dm⁻³
[B] ≈ 2.0 × 10⁻⁶ mol dm⁻³
Summary
- ICE tables can be used to track equilibrium values.
- Always write the correct K expression.
- Approximate only when K is small and justified.
- All species must be in the same phase (homogeneous equilibria).
- Use mol dm⁻³ units and round appropriately.
Linked Course Question
How does the equilibrium law help us determine the pH of a weak acid, weak base, or buffer solution?
For a weak acid (HA ⇌ H+ + A−):
Ka = [H+][A−] / [HA].
Using the initial concentration of the acid and assuming only slight ionisation, we can estimate [H+] and calculate the pH.
For a weak base (B + H2O ⇌ BH+ + OH−):
Kb = [BH+][OH−] / [B].
From [OH−], we can calculate the pOH and then the pH.
In a buffer solution, both the weak acid and its conjugate base are present in significant amounts. The same equilibrium expression applies, or we can use:
pH = pKa + log([A−]/[HA])
This form (the Henderson–Hasselbalch equation) is especially useful for buffers, where the ratio of base to acid controls the pH.