Quadratics & Equilibria — Maths for General Chemistry

When the unknown is squared

Equilibrium problems are quadratics in disguise

A quadratic has the form $a x^{2} + b x + c = 0$ . Its graph is a parabola, and the solutions are where it crosses the x-axis. The quadratic formula always works:

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The discriminant $b^{2} - 4 a c$ tells you how many real roots there are: positive → two distinct roots, zero → one repeated root, negative → no real roots. In chemistry you will almost always have two real roots but only one is physically meaningful (e.g. concentrations cannot be negative).

Weak-acid and solubility equilibria are the most common source of quadratics. Setting up an ICE (Initial / Change / Equilibrium) table gives $K_{a} = x^{2} / (C_{0} - x)$ , which rearranges to the standard form $x^{2} + K_{a} x - K_{a} C_{0} = 0$ . When x is small (less than 5% of C₀) the approximation $x \approx K_{a} C_{0}$ saves effort; otherwise use the full formula.

Quadratic formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Weak-acid ICE

K_{a} = \frac{x ^{2}}{C _{0} - x} \Rightarrow x^{2} + K_{a} x - K_{a} C_{0} = 0

5% rule

x / C_{0} < 5%

, use

x \approx K_{a} C_{0}

; else solve fully.

Root selection

Always reject the negative root — a concentration cannot be negative.

The 5% rule is a short-cut, not a law. For a weak-acid problem with $K_{a} = x^{2} / (C_{0} - x)$ , if you first solve approximately ( $x \approx K_{a} C_{0}$ ) and that value is less than 5% of $C_{0}$ , the approximation is safe. If it exceeds 5%, solve the full quadratic. A useful alternative is the successive-approximation method: substitute $x_{0} = K_{a} C_{0}$ into $C_{0} - x$ in the denominator, getting a better $x_{1}$ , and repeat until convergence — usually two or three cycles.

Common-ion problems add a twist: if the solution already contains $[A^{-}]_{0} = S$ from a salt, the ICE table for $K_{a} = x (S + x) / (C_{0} - x)$ rearranges to $x^{2} + (K_{a} + S) x - K_{a} C_{0} = 0$ . The common ion suppresses dissociation, so x is much smaller — but you still need the quadratic formula or the 5% check to be sure.

Common pitfalls

Choosing the negative root. The quadratic formula always gives two roots; for concentration problems only the positive root is physical. Always check both and discard the negative (or unphysical) one.
Dropping x when the 5% rule fails. If $x / C_{0} > 5%$ , setting $C_{0} - x \approx C_{0}$ introduces a significant error. Use the full quadratic.
Forgetting to rearrange into standard form first. $K_{a} = x^{2} / (C_{0} - x)$ must be multiplied out to $x^{2} + K_{a} x - K_{a} C_{0} = 0$ before applying $a, b, c$ to the quadratic formula.
Common-ion: wrong ICE setup. If a common ion is already present at concentration S, the equilibrium expression for A⁻ is $S + x$ , not just x. Using x alone overestimates dissociation.

The classic case

Weak-acid and solubility equilibria give Ka = x²/(C₀ − x). When x isn't tiny you can't drop it, and the problem becomes a quadratic. Always reject the negative root — a concentration can't be negative. Try the solver and the Ka lab.

Quadratic solver

Adjust a, b, c to see the parabola and its roots. Watch the discriminant change sign when the parabola lifts off the axis.

Shape the parabola — its roots are where it crosses the x-axis

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

b-1

c-2

discriminant b²−4ac9.00

root 12.000

root 2-1.000

Worked example 1Solving the standard quadratic

Solve $2 x^{2} - 3 x - 2 = 0$ and identify which root would be physically rejected in a chemistry context where x represents a concentration.

Equilibrium (Ka)

Drag the pKa and initial concentration to see when the 5% approximation breaks down.

Weak-acid equilibrium: exact quadratic vs the small-x shortcut

K_{a} = \frac{x ^{2}}{C _{0} - x} \Rightarrow x^{2} + K_{a} x - K_{a} C_{0} = 0

pKₐ (Kₐ = 10^−pKa)4.76

C₀ initial concentration0.100 M

exact x = [H⁺]1.31e-3 M

approx √(KₐC₀)1.32e-3 M

error of approximation0.7%

pH2.88

x is 1.3% of C₀. Under 5%, so the shortcut x ≈ √(KₐC₀) is safe.

Worked example 2Full ICE-table treatment for acetic acid

Acetic acid (CH₃COOH) has $K_{a} = 1.8 \times 1 0^{- 5}$ . Find $[H^{+}]$ and pH in a $0.0100 M$ solution.

Worked examples

Solubility equilibrium and a discriminant that determines whether precipitation occurs.

Worked example 3Solubility product and quadratic

Silver chloride dissolves: $AgCl (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$ , $K_{sp} = 1.77 \times 1 0^{- 10}$ . Find the molar solubility s in pure water.

Worked example 4HF — when the 5% rule fails

Hydrofluoric acid has $K_{a} = 6.8 \times 1 0^{- 4}$ . Solve for $[H^{+}]$ in a $0.0100 M$ solution, and show that the approximation fails the 5% test.

Worked example 5Buffer pH using the Henderson–Hasselbalch form (verifying with quadratic)

A buffer contains 0.100 M acetic acid and 0.100 M sodium acetate ( $K_{a} = 1.8 \times 1 0^{- 5}$ ). Show that the Henderson–Hasselbalch approximation gives pH = 4.74, and confirm that the exact quadratic treatment agrees to within 0.01 pH units.

Worked example 6Common-ion effect — AgCl solubility in NaCl solution

Find the molar solubility of AgCl ( $K_{sp} = 1.77 \times 1 0^{- 10}$ ) in a solution already containing $0.0100 M$ NaCl (which provides a common Cl⁻ ion). Compare with the solubility in pure water.

Worked example 7Weak acid Ka solved exactly — no approximation

Chloroacetic acid (ClCH₂COOH) has $K_{a} = 1.36 \times 1 0^{- 3}$ . Find the exact $[H^{+}]$ and pH in a $0.0500 M$ solution, showing that the 5% approximation is invalid here.

ChallengeChallenge — common-ion buffer where x is not negligible

A buffer is made by adding sodium formate (HCOONa) to 0.0200 M formic acid (HCOOH, $K_{a} = 1.77 \times 1 0^{- 4}$ ) so that $[HCO O^{-}]_{0} = 0.00200 M$ . This gives a low buffer capacity, so x (the protons released) is NOT negligible compared with the formate concentration. Set up and solve the exact quadratic to find $[H^{+}]$ .

Check yourself

Four questions linking the quadratic formula to equilibrium chemistry.

Question 1 of 4 · Score 0

The quadratic formula solves ax² + bx + c = 0. What is it?

Choose an answer.