# Quantitative Analysis of Acids and Bases

> Standard solutions and primary standards, titration technique and calculations, choosing an indicator from a titration curve, conductometric titrations, back titrations and buffers.

From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/quantitative-acid-base-analysis/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: How much acid is in the vinegar?
A vinegar bottle in Ethan's kitchen says "4% acidity". His chemistry teacher asks: how could you check that, using nothing but a burette, a flask and a bottle of sodium hydroxide of known strength?

The answer is a **titration**. You add the base drop by drop until it has exactly cancelled out the acid, and you measure how much it took. From that one volume, you can work out the concentration of the acid. Food companies, water boards and hospitals use this same method every day.

**In short:** A titration finds the concentration of an acid or base by reacting it exactly with a solution of known concentration, using an indicator to show the end point. Use $n = cV$ and the mole ratio from the balanced equation to calculate the unknown.

## Standard solutions

A **standard solution** has an accurately known concentration. It is made from a **primary standard**, a substance that is:

- very **pure**,
- **stable** (does not react with air or absorb water from it),
- has a **high molar mass** (so weighing errors are small),
- **soluble** in water.

Good primary standards: anhydrous **sodium carbonate** ($\mathrm{Na_2CO_3}$) and **oxalic acid dihydrate** ($\mathrm{H_2C_2O_4 \cdot 2H_2O}$).

**NaOH is not** a primary standard: it absorbs water and carbon dioxide from the air. A NaOH solution must itself be **standardised** by titrating it against a primary standard.

**Making a standard solution:**

1. Weigh the solid accurately on an electronic balance.
2. Dissolve it in a beaker of distilled water.
3. Pour into a **volumetric flask**, rinsing the beaker and funnel into the flask.
4. Add distilled water until the bottom of the meniscus is on the mark.
5. Stopper and invert to mix.

## Titration equipment and rinsing

| Equipment | Use | Rinse with |
|---|---|---|
| Volumetric flask | Make up an exact volume | Distilled water |
| Pipette | Deliver an exact volume (the **aliquot**) | The solution it will hold |
| Burette | Deliver a measured volume (the **titre**) | The solution it will hold |
| Conical flask | Holds the aliquot and indicator | Distilled water only |

Rinsing the burette or pipette with **water** dilutes the solution in it. Rinsing the conical flask with the **solution** adds extra moles. Both give wrong answers.

## Titration words

- **Equivalence point**: the reactants have been added in exactly the mole ratio of the equation.
- **End point**: the point where the indicator changes colour. A good indicator makes this match the equivalence point.
- **Concordant titres**: titres that agree within 0.10 mL. Average only these. Leave out the rough first trial and any outliers.

## Titration calculations

Follow the same steps every time:

1. Moles of the known solution: $n = cV$ (with $V$ in **litres**).
2. Use the **mole ratio** from the balanced equation.
3. Concentration of the unknown: $c = \dfrac{n}{V}$.
4. If the sample was **diluted** first, scale back up.

### Worked example: Standardising hydrochloric acid
1.325 g of $\mathrm{Na_2CO_3}$ (105.99 g/mol) is made up to 250.0 mL. 25.00 mL aliquots need an average of 23.80 mL of HCl.

$\mathrm{Na_2CO_3} + \mathrm{2HCl} \rightarrow \mathrm{2NaCl} + \mathrm{H_2O} + \mathrm{CO_2}$

$c(\mathrm{Na_2CO_3}) = \dfrac{1.325 \div 105.99}{0.2500} = 0.05000$ mol/L

$n(\mathrm{Na_2CO_3})$ in the aliquot $= 0.05000 \times 0.02500 = 1.250 \times 10^{-3}$ mol

$n(\mathrm{HCl}) = 2 \times 1.250 \times 10^{-3} = 2.500 \times 10^{-3}$ mol

$c(\mathrm{HCl}) = \dfrac{2.500 \times 10^{-3}}{0.02380} = 0.1050$ mol/L

## Titration curves and indicators

A **titration curve** plots pH against the volume added. The steep jump marks the equivalence point. Choose an indicator whose colour change range lies **inside the steep part**.

| Titration | pH at equivalence | Indicator |
|---|---|---|
| Strong acid + strong base | 7 | Bromothymol blue (or almost any) |
| Weak acid + strong base | Above 7 (about 8 to 9) | Phenolphthalein |
| Strong acid + weak base | Below 7 (about 4 to 5) | Methyl orange |
| Weak acid + weak base | No sharp jump | No good indicator; use a pH meter or conductivity |

For a **weak acid** titration, the curve has a flat **buffer region** before equivalence. Halfway to equivalence, $\mathrm{pH} = \mathrm{p}K_a$.

## Conductometric titrations

Instead of an indicator, measure the **electrical conductivity** as the titrant is added. The equivalence point is where the graph changes direction.

- **HCl + NaOH**: conductivity **falls** (fast-moving $\mathrm{H^+}$ is replaced by slower $\mathrm{Na^+}$), reaches a minimum at equivalence, then **rises** as extra $\mathrm{OH^-}$ and $\mathrm{Na^+}$ build up.
- **Weak acid + NaOH**: conductivity starts **low** (few ions), rises slowly as salt forms, then rises faster after equivalence.

Conductometric titrations work for coloured solutions and for weak acid + weak base, where indicators fail.

## Back titrations

Some samples will not dissolve, or react slowly, like chalk or antacid tablets. In a **back titration**:

1. Add a **known excess** of acid to the sample.
2. Let it react completely.
3. Titrate the **leftover** acid with a standard base.
4. Acid that reacted = acid added − acid left over.

### Worked example: Antacid tablet
A crushed tablet of $\mathrm{CaCO_3}$ reacts with 25.0 mL of 0.600 mol/L HCl. The leftover HCl needs 12.1 mL of 0.100 mol/L NaOH.

HCl added $= 0.600 \times 0.0250 = 0.0150$ mol. HCl left over $= 0.100 \times 0.0121 = 0.00121$ mol.

HCl reacted $= 0.01379$ mol. $\mathrm{CaCO_3} : \mathrm{HCl} = 1 : 2$, so $n(\mathrm{CaCO_3}) = 0.006895$ mol.

Mass $= 0.006895 \times 100.09 = 0.690$ g.

## Buffers

A **buffer** is a solution that **resists changes in pH** when small amounts of acid or base are added. It contains a **weak acid and its conjugate base** in similar amounts, such as $\mathrm{CH_3COOH}$ / $\mathrm{CH_3COO^-}$.

$$\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H_3O^+}$$

- Add a little acid: the extra $\mathrm{H_3O^+}$ reacts with $\mathrm{CH_3COO^-}$, so the equilibrium shifts left and the pH barely changes.
- Add a little base: the $\mathrm{OH^-}$ reacts with $\mathrm{CH_3COOH}$ (or with $\mathrm{H_3O^+}$, which the acid then replaces), so the pH barely changes.

A buffer has a limited **capacity**. Add too much acid or base and it is used up. A strong acid and its salt (HCl / NaCl) is **not** a buffer.

The **carbonic acid / hydrogen carbonate** buffer ($\mathrm{H_2CO_3}$ / $\mathrm{HCO_3^-}$) keeps blood between pH 7.35 and 7.45.

**Mind map**

```text
Quantitative analysis
  Standards
    Pure, stable, high molar mass
    Na2CO3, oxalic acid
    NaOH must be standardised
  Titration
    Rinse burette and pipette with solution
    Concordant titres
    n = cV, mole ratio
  Curves
    Strong-strong: pH 7
    Weak acid: phenolphthalein
    Weak base: methyl orange
  Conductometric
    Turning point = equivalence
  Back titration
    Excess acid, titrate leftover
  Buffers
    Weak acid + conjugate base
    Resist pH change
    Blood: H2CO3 and HCO3−
```

### Year 11: Dilution
When a solution is diluted, the moles of solute stay the same:

$$c_1 v_1 = c_2 v_2$$

**Example.** 25.00 mL of vinegar is made up to 250.0 mL. The diluted vinegar is 0.0925 mol/L. The original was $0.0925 \times \dfrac{250.0}{25.00} = 0.925$ mol/L.

As a mass per volume: $0.925 \times 60.05 = 55.5$ g/L, or **5.55% w/v** (g per 100 mL).

**Diprotic acids.** Watch the mole ratio. $\mathrm{H_2SO_4} + \mathrm{2NaOH} \rightarrow \mathrm{Na_2SO_4} + \mathrm{2H_2O}$, so moles of acid are **half** the moles of NaOH. 24.60 mL of 0.1000 mol/L NaOH is $2.460 \times 10^{-3}$ mol, so 20.00 mL of acid holds $1.230 \times 10^{-3}$ mol: 0.06150 mol/L.

### Year 12: Calculating the pH of a buffer
Use the $K_a$ expression with both concentrations from the mixture. Because little acid ionises, $[\mathrm{HA}]$ and $[\mathrm{A^-}]$ are the amounts you put in:

$$[\mathrm{H_3O^+}] = K_a \times \frac{[\mathrm{HA}]}{[\mathrm{A^-}]} \qquad \text{or} \qquad \mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$

**Example.** 0.0250 mol of sodium acetate is dissolved in 500 mL of 0.0700 mol/L acetic acid ($K_a = 1.76 \times 10^{-5}$).

$[\mathrm{A^-}] = 0.0250 \div 0.500 = 0.0500$ mol/L, $[\mathrm{HA}] = 0.0700$ mol/L.

$\mathrm{pH} = 4.754 + \log\dfrac{0.0500}{0.0700} = 4.754 - 0.146 = 4.61$.

When $[\mathrm{HA}] = [\mathrm{A^-}]$, $\mathrm{pH} = \mathrm{p}K_a$. Buffers work best within about 1 pH unit of the $\mathrm{p}K_a$.

**Ka from the half-way point.** Half-way to equivalence, $[\mathrm{HA}] = [\mathrm{A^-}]$, so $\mathrm{pH} = \mathrm{p}K_a$. If that pH is 4.20, $K_a = 10^{-4.20} = 6.3 \times 10^{-5}$.

**Making a buffer of a chosen pH.** Rearrange: $\log\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \mathrm{pH} - \mathrm{p}K_a$. For pH 9.00 from 100 mL of 0.100 mol/L $\mathrm{NH_3}$ (0.0100 mol) with $\mathrm{p}K_a = 9.25$: the ratio is $10^{-0.25} = 0.562$, so $n(\mathrm{NH_4^+}) = \dfrac{0.0100}{0.562} = 0.0178$ mol, which is 0.951 g of $\mathrm{NH_4Cl}$.

**pH at the equivalence point.** All the weak acid has become its conjugate base. Titrating 50.0 mL of 0.100 mol/L acetic acid with 0.100 mol/L NaOH makes 0.00500 mol of acetate in 100 mL, so $[\mathrm{A^-}] = 0.0500$. With $K_b = 5.6 \times 10^{-10}$, $[\mathrm{OH^-}] = 5.3 \times 10^{-6}$, pOH = 5.28 and pH = 8.72.

**Back titration for ammonium.** Boiling an ammonium salt with a known excess of NaOH drives off $\mathrm{NH_3}$ ($\mathrm{NH_4^+} + \mathrm{OH^-} \rightarrow \mathrm{NH_3} + \mathrm{H_2O}$). Titrate the leftover NaOH with standard acid. NaOH used = NaOH added − NaOH left, and that equals the moles of $\mathrm{NH_4^+}$. Then scale up for any dilution.

## Quick summary

- A primary standard is pure, stable, soluble and has a high molar mass. NaOH is not one.
- Rinse the burette and pipette with the solution they will hold; rinse the conical flask with water.
- Average concordant titres (within 0.10 mL). Use $n = cV$ and the mole ratio.
- Pick an indicator that changes colour in the steep part of the curve.
- Conductometric titrations find equivalence from where conductivity turns.
- Back titration: excess acid, then titrate what is left.
- A buffer (weak acid + conjugate base) resists pH change.

## Quiz

Try these questions from the chapter quiz.

1. A primary standard must be:
   - A. Very pure and stable
   - B. Coloured
   - C. A strong acid
   - D. A liquid
   - Answer: A. Very pure and stable. Its mass must give an exact number of moles.

2. Why is sodium hydroxide not a primary standard?
   - A. It is too expensive
   - B. It absorbs water and carbon dioxide from the air
   - C. It is not soluble
   - D. It is a weak base
   - Answer: B. It absorbs water and carbon dioxide from the air. Its mass changes as it sits in air, so it cannot be weighed exactly.

3. The burette should be rinsed with:
   - A. Distilled water
   - B. The solution it will hold
   - C. Ethanol
   - D. Tap water
   - Answer: B. The solution it will hold. Water would dilute the titrant.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/quantitative-acid-base-analysis/#quiz