In titration calculations you found the equivalence point. A titration curve shows the pH all the way through. For a strong acid and strong base, each point is a strong-acid or strong-base pH after the neutralization.
Key idea
Why it works
Worked examples
Worked example: Four points on the curve
- 0 mL: $[\mathrm{H_3O^+}] = 0.100\ \text{M}$, pH 1.00.
- 10.0 mL: acid $2.50\ \text{mmol}$, base $1.00\ \text{mmol}$; $1.50\ \text{mmol}$ $\mathrm{H_3O^+}$ left in $35.0\ \text{mL}$: $0.0429\ \text{M}$, pH 1.37.
- 25.0 mL: $2.50\ \text{mmol}$ of each, all neutralized. pH 7.00.
- 30.0 mL: $0.50\ \text{mmol}$ $\mathrm{OH^-}$ left in $55.0\ \text{mL}$: $9.09 \times 10^{-3}\ \text{M}$, pOH 2.04, pH 11.96.
Answer: 1.00, 1.37, 7.00, 11.96.
Worked example: Equivalence volume
- Acid: $(20.0)(0.120) = 2.40\ \text{mmol}$.
- Base needed: $2.40\ \text{mmol} \div 0.150\ \text{M} = 16.0\ \text{mL}$.
Answer: $16.0\ \text{mL}$.
Why is the curve so flat at first and so steep near equivalence? The pH is a logarithm. Early on, adding base cuts the leftover acid by a small fraction, which barely moves its log. Near equivalence, a single drop can cut the leftover acid by a factor of 100 or more, and then switch the solution to excess base: the pH leaps several units. The point where an indicator changes color, the end point, is chosen to fall on that leap, so it lands within a drop of the true equivalence point.
Common mistake
Titrate an acid drop by drop yourself in the Mixing Lab.