Standard solutions and acid–base titrations

What you’ll learn

• Follow amount ratios through a standard solution and an aliquot.
• Select valid titres and explain rinsing, transfer and endpoint errors.

Understand the idea

Core practicals 3 and 4 connect concentration measurements to accurate transfer. A standard solution made from a suitable pure solid has its concentration determined by the weighed amount and the final solution volume. Quantitative transfer includes material left on the weighing or dissolving apparatus. Dissolving a solid in a stated volume of water does not necessarily produce that final volume of solution; the volumetric flask defines the final volume.

For a titration calculation, find the amount in the measured aliquot, use the balanced reaction ratio, and then divide the reacting amount by its own solution volume in $\mathrm{dm^3}$. Use the titres that satisfy the stated concordance rule; an initial rough titre helps locate the endpoint but is not automatically included in the mean. The endpoint is an observed indicator change, whereas the equivalence point is the stoichiometric point. A suitable indicator changes within the steep pH change around equivalence.

Residual water in a burette or pipette dilutes the solution that the apparatus is meant to deliver, so the usual final rinse uses that solution. Water left in a conical flask changes the total volume but not the amount transferred into it, and therefore does not itself change the required titre. Reading the meniscus consistently, mixing near the endpoint and avoiding overshoot all matter. For the corrosive solutions in a school titration, splash protection and pipette fillers address specific risks.

Key formulae

$$c=\frac{m}{MV}$$

$$n=cV\quad(V\text{ in }\mathrm{dm^3})$$

$$\frac{n_A}{a}=\frac{n_B}{b}\quad\text{for }aA+bB\rightarrow\text{products}$$

Worked example

A pure monoprotic acid standard has molar mass $204.2\ \mathrm{g\,mol^{-1}}$. A $5.105\ \mathrm{g}$ sample makes $250.0\ \mathrm{cm^3}$ of solution. A $25.00\ \mathrm{cm^3}$ aliquot requires $20.00\ \mathrm{cm^3}$ of sodium hydroxide. The acid and hydroxide react $1:1$. Find the hydroxide concentration.

1. The flask contains $5.105/204.2=0.02500\ \mathrm{mol}$ of acid, so its concentration is $0.02500/0.2500=0.1000\ \mathrm{mol\,dm^{-3}}$.
2. The aliquot contains $0.1000\times0.02500=0.002500\ \mathrm{mol}$ of acid and reacts with the same amount of hydroxide.
3. The hydroxide volume is $0.02000\ \mathrm{dm^3}$, so its concentration is $0.002500/0.02000=0.1250\ \mathrm{mol\,dm^{-3}}$.

Answer: $0.1250\ \mathrm{mol\,dm^{-3}}$.