Solving for equilibrium concentrations often leads to a quadratic or worse. When $K$ is very small, a shortcut avoids it: so little reacts that $x$ hardly changes the starting amount.
Key idea
If $K$ is small compared with the starting concentration, replace $C - x$ by $C$, solve the simpler equation, and then check: if $x$ is less than 5% of $C$, the answer is good. If not, go back and solve exactly.
Why it works
Subtracting a small number from a large one barely changes it: $0.10 - 0.0003 = 0.0997$, within 0.3% of 0.10. Dropping $x$ there changes the answer by about the same small fraction, which is less than the uncertainty in most values of $K$. The 5% rule is a common agreed limit for "small enough." The approximation is only for $x$ beside a larger number; $x$ standing alone, as in $x^2$, must stay.
Worked examples
Worked example: The approximation works
For $\mathrm{A \rightleftharpoons B + C}$, $K = 1.0 \times 10^{-6}$, and $[\mathrm{A}]_0 = 0.10\ \text{M}$. Find $x = [\mathrm{B}]$ at equilibrium.
- $\dfrac{x^2}{0.10 - x} = 1.0 \times 10^{-6}$. Assume $0.10 - x \approx 0.10$: $x^2 = 1.0 \times 10^{-7}$, so $x = 3.2 \times 10^{-4}\ \text{M}$.
- Check: $\dfrac{3.2 \times 10^{-4}}{0.10} = 0.32\%$, well under 5%. The approximation holds.
Answer: $x = 3.2 \times 10^{-4}\ \text{M}$.
Worked example: The approximation fails
Same reaction, but $K = 1.0 \times 10^{-3}$ and $[\mathrm{A}]_0 = 0.010\ \text{M}$.
- Approximate: $x^2 = (1.0 \times 10^{-3})(0.010) = 1.0 \times 10^{-5}$, $x = 3.2 \times 10^{-3}$. Check: $3.2 \times 10^{-3}/0.010 = 32\%$. Far too big.
- Solve exactly: $x^2 = (1.0 \times 10^{-3})(0.010 - x)$, so $x^2 + (1.0 \times 10^{-3})x - 1.0 \times 10^{-5} = 0$.
- $$b^2 - 4ac = (1.0 \times 10^{-3})^2 + 4.0 \times 10^{-5} = 4.1 \times 10^{-5}$$ whose square root is $6.40 \times 10^{-3}$. So $x = \dfrac{-1.0 \times 10^{-3} + 6.40 \times 10^{-3}}{2} = 2.7 \times 10^{-3}\ \text{M}$.
Answer: $x = 2.7 \times 10^{-3}\ \text{M}$ (the shortcut was 17% too high).
A quick test before you start: if $C/K$ is larger than about 400, the 5% rule will usually pass for a reaction like these. The check afterward is still the rule.
You will use this approximation constantly in weak-acid and weak-base problems, where $K_a$ or $K_b$ is usually small and the 5% check usually passes. When it fails, the quadratic formula is always there, and a calculator's equation solver gives the same root in seconds; the approximation's real value is that it shows how the answer depends on $K$ and $C$.
Common mistake
Dropping $x$ when $K$ is large, or never checking whether the approximation held. A large $K$ means most of the reactant reacts, so $x$ is not small at all. Check: always divide $x$ by the starting amount and compare with 5%.