Equilibrium Numerical Review

Introduction

Equilibrium problems share a structure: a balanced equation, a table, a K expression and some algebra. The same structure means the same mistakes recur. Rather than learning a separate trick for each problem type, it is more efficient to finish every calculation with a fixed set of checks. This review collects those checks and applies them to examples that draw on reaction quotients, gas equilibria, weak acids and solubility.

Core explanation

Check 1: direction before magnitude. If both reactants and products are present at the start, compare Q with K first. Q < K means the reaction moves forward and x is positive in the product direction; Q > K means it moves backward. Setting up the table in the wrong direction produces a negative x or an impossible root.

Check 2: roots within bounds. Every species must end with a non-negative amount, and no reactant can be used beyond what was supplied. A quadratic usually has one root inside this window and one outside. Write the window down (for example 0 < x < 0.050) before solving, then reject the other root with a sentence.

Check 3: approximation tests. After using c − x ≈ c, calculate x / c. Under about 5% the shortcut is acceptable; above it, solve exactly. The same logic applies to common-ion problems, where s must be small beside the added ion.

Check 4: back-substitution. Put the equilibrium concentrations back into the K expression. You should recover K to within rounding. This single step catches most algebraic slips.

Check 5: the model itself. Some answers are numerically correct yet chemically absurd because a hidden assumption has failed. The standard example is very dilute strong acid: 1.0 × 10⁻⁸ mol dm⁻³ HCl does not have pH 8.00, since an acid cannot make water alkaline. Including water's ionisation through the charge balance [H⁺] = [Cl⁻] + [OH⁻] gives [H⁺]² − 1.0 × 10⁻⁸[H⁺] − 1.0 × 10⁻¹⁴ = 0, so [H⁺] = 1.05 × 10⁻⁷ mol dm⁻³ and pH = 6.98.

Check 6: size sense. A K much greater than 1 should give mostly products; a K much less than 1 mostly reactants. If your numbers show the opposite, look for an inverted expression.

Step-by-step reasoning

1. Balance the equation and compare Q with K to find the direction. 2. Build the table and state the allowed range of x. 3. Try an approximation if K is small; otherwise solve exactly. 4. Select the root inside the allowed range. 5. Test any approximation with the 5% criterion. 6. Back-substitute to confirm K, and ask whether the result is chemically sensible.

Visual explanation

Draw a number line for x from zero to the amount of the limiting species. Mark both quadratic roots on it. Only one lies on the shaded segment; the other sits off the end, where it would demand negative amounts. The diagram makes the reason for rejecting a root visible at a glance.

Real-world analogy

A pilot runs through a checklist before every landing, however experienced they are. The checks take seconds but catch rare, costly mistakes. The six checks here play the same role for equilibrium answers: quick, routine and occasionally decisive.

Real-world example

Environmental chemists model dissolved metals in rivers using many linked equilibria. Their software solves the equations numerically, but analysts still check that total metal is conserved, that the charge balance closes and that no concentration is negative, because solvers can converge on non-physical roots.

Why?

Why does a quadratic give an unphysical root at all? The algebra only encodes the K expression; it does not know that concentrations must be positive or that amounts are limited by the feed. Those physical constraints must be applied by you after solving.

Common misconception

"If the calculator gives a positive number, the root is valid." A positive x can still exceed the starting amount of a reactant and leave a negative concentration. Every species must be checked, not just x.

Worked example

Question: For N₂O₄(g) ⇌ 2NO₂(g), Kc = 4.6 × 10⁻³ mol dm⁻³ at 298 K. Starting from 0.100 mol dm⁻³ N₂O₄ only, find the equilibrium concentrations and check the result.

Reasoning: Table: N₂O₄ = 0.100 − x, NO₂ = 2x, with 0 < x < 0.100. Kc = 4x² / (0.100 − x) = 4.6 × 10⁻³ gives 4x² + 4.6 × 10⁻³x − 4.6 × 10⁻⁴ = 0. The discriminant square root is √(7.38 × 10⁻³) = 0.0859, so x = (0.0859 − 0.0046) / 8 = 0.0102 or x = −0.0113. The negative root would give negative [NO₂] and is rejected. Back-substitution: 0.0203² / 0.0898 = 4.6 × 10⁻³, confirming the answer. The small-x approximation would give x = 0.0107, which is 10.7% of c and fails the test.

Answer: [N₂O₄] = 0.090 mol dm⁻³ and [NO₂] = 0.020 mol dm⁻³.

Quick check

1. A weak-acid calculation gives pH 7.3 for a very dilute acid solution. What is wrong? Answer: An acid cannot give a pH above 7; water's ionisation has been ignored and must be included through the charge balance.

Exam focus

Examiners award marks for stating assumptions, for the reason a root is rejected and for a final check. Write one line for each: "assume x ≪ c; x/c = 1.3%, valid", "reject x = 1.13 as it exceeds the I₂ supplied", "check: K recalculated = 64".

Advanced insight

For systems with several coupled equilibria, no simple quadratic exists. The standard strategy is to write mass balances for each element, a charge balance and every K expression, then solve numerically. The same checks still apply: non-negative concentrations, closure of the mass and charge balances, and recovery of each K.

Summary

Finish every equilibrium calculation with the same routine: determine direction from Q, bound x, choose the physical root, test approximations, back-substitute into K and ask whether the answer makes chemical sense. These checks catch sign errors, invalid shortcuts and hidden model failures such as ignoring water's ionisation.

Practice questions

1. For H₂ + I₂ ⇌ 2HI with Kc = 64, a mixture contains 0.10 mol dm⁻³ of each of H₂, I₂ and HI. In which direction does it move? Answer: Q = 0.10² / (0.10 × 0.10) = 1, which is less than 64, so the reaction moves forward to form more HI. 2. A quadratic for a reaction starting with 0.050 mol dm⁻³ of the limiting reactant gives roots 0.032 and 0.071. Which is valid? Answer: 0.032, because 0.071 exceeds the 0.050 supplied and would leave a negative concentration. 3. Calculate the pH of 1.0 × 10⁻⁸ mol dm⁻³ HCl at 25 °C. Answer: Solving [H⁺]² − 1.0 × 10⁻⁸[H⁺] − 1.0 × 10⁻¹⁴ = 0 gives [H⁺] = 1.05 × 10⁻⁷ mol dm⁻³, so pH = 6.98. 4. After solving a Ksp problem, back-substitution gives 3.6 × 10⁻¹⁰ instead of the given 1.8 × 10⁻¹⁰. Suggest a likely error. Answer: A factor of 2 has slipped in, for example by doubling an ion concentration that should not have been doubled or by using the wrong stoichiometric coefficient.