Acid-Base Titration Calculations: Step-by-Step Method with Mole Ratios
Titration questions appear in almost every Class 11 and Class 12 chemistry paper, and also in the practical exam. The idea is simple: you add a solution of known strength from a burette until the reaction is just complete, and then you use the volume to find the unknown. Most marks are lost not in the chemistry but in one small step — forgetting the mole ratio of the balanced equation. This article gives one method that works every time.
The core formula
For a reaction a A + b B → products at the end point:
(MA VA) / a = (MB VB) / b
Here M is molarity in mol/L, V is the volume (any unit, but the same unit on both sides), and a and b are the coefficients from the balanced equation. If the acid and the base react 1 : 1 (for example HCl and NaOH), the formula becomes the familiar M1V1 = M2V2. For H2SO4 and NaOH the ratio is 1 : 2, so you cannot use M1V1 = M2V2 directly.
Words you must know
- Titrant: the solution in the burette, of known concentration.
- Analyte: the solution in the conical flask, of unknown concentration (or unknown purity).
- Equivalence point: the point where the moles added are exactly right according to the balanced equation.
- End point: the point where the indicator changes colour. A good indicator is chosen so that the end point is very close to the equivalence point.
- Concordant readings: burette readings that agree within about 0.1 mL. Use the mean of concordant readings only.
Method in four steps
- Write the balanced equation and note the mole ratio.
- Find the moles of the titrant: moles = M × V(L). A shortcut is to keep V in mL; the answer is then in millimoles (mmol).
- Use the mole ratio to get the moles of the analyte.
- Divide by the analyte volume to get molarity, or multiply by molar mass to get mass.
Worked examples
Example 1: strength of NaOH (1 : 1 reaction). 25.0 mL of NaOH solution needs 22.4 mL of 0.100 M HCl for complete neutralisation. Find the molarity of NaOH.
Equation: HCl + NaOH → NaCl + H2O (ratio 1 : 1).
Moles of HCl = 0.100 × 22.4 = 2.24 mmol.
Moles of NaOH = 2.24 mmol (ratio 1 : 1).
Molarity of NaOH = 2.24 mmol / 25.0 mL = 0.0896 M.
Check by M1V1 = M2V2: MNaOH = 0.100 × 22.4 / 25.0 = 0.0896 M. Both routes agree.
If you also want the mass of NaOH per litre: 0.0896 × 40.00 = 3.58 g/L.
Example 2: sulphuric acid (1 : 2 reaction). 25.0 mL of H2SO4 is neutralised by 30.0 mL of 0.100 M NaOH. Find the molarity of the acid.
Equation: H2SO4 + 2 NaOH → Na2SO4 + 2 H2O.
Moles of NaOH = 0.100 × 30.0 = 3.00 mmol.
Moles of H2SO4 = 3.00 / 2 = 1.50 mmol.
Molarity of H2SO4 = 1.50 / 25.0 = 0.0600 M.
If you wrongly used M1V1 = M2V2, you would get 0.120 M, which is double the correct value.
Example 3: percentage purity of sodium carbonate. A 1.00 g sample of impure Na2CO3 is dissolved in water and titrated with 0.500 M HCl. It needs 36.0 mL to reach the methyl orange end point (both carbonate steps complete). Find the percentage purity. (Molar mass of Na2CO3 = 105.99 g/mol.)
Equation: Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2 (ratio 1 : 2).
Moles of HCl = 0.500 × 36.0 = 18.0 mmol.
Moles of Na2CO3 = 18.0 / 2 = 9.00 mmol.
Mass = 9.00 mmol × 105.99 mg/mmol = 953.9 mg = 0.9539 g.
Purity = 0.9539 / 1.00 × 100 = 95.4 %.
Example 4: standard solution and then titration. 0.630 g of oxalic acid dihydrate, H2C2O4·2H2O (molar mass 126.07 g/mol), is dissolved to make 100 mL of solution. 10.0 mL of it needs 10.0 mL of NaOH. Find the molarity of NaOH.
Moles of oxalic acid in the flask = 0.630 / 126.07 = 4.997 × 10−3 mol in 100 mL, so molarity = 0.04997 M.
In 10.0 mL: 0.04997 × 10.0 = 0.4997 mmol.
Oxalic acid is diprotic: H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O, so moles of NaOH = 2 × 0.4997 = 0.9994 mmol.
Molarity of NaOH = 0.9994 / 10.0 = 0.0999 M, about 0.100 M.
Choosing the indicator
The indicator must change colour near the pH of the equivalence point. For a strong acid with a strong base the equivalence point is at pH 7, and both phenolphthalein and methyl orange work. For a weak acid with a strong base (for example ethanoic acid with NaOH) the equivalence point is above 7, so phenolphthalein is used. For a strong acid with a weak base, the equivalence point is below 7, so methyl orange is used. A wrong indicator gives a wrong volume, even if your arithmetic is perfect.
Common mistakes
- Ignoring the mole ratio and using M1V1 = M2V2 for a 1 : 2 or 2 : 1 reaction.
- Mixing units: one volume in mL and the other in litres.
- Reading the burette from the top of the meniscus, or using a reading taken before the air bubble in the tip was removed.
- Using the first rough reading in the average. Only concordant readings are averaged.
- Forgetting that water of crystallisation changes the molar mass (Example 4 uses 126.07, not 90.03).
- Rinsing the conical flask with the analyte; it must be rinsed with distilled water only, while the burette and the pipette are rinsed with their own solutions.
Quick summary
| Situation | Mole ratio | Relation to use |
|---|---|---|
| HCl with NaOH | 1 : 1 | M1V1 = M2V2 |
| H2SO4 with NaOH | 1 : 2 | MacidVacid = MbaseVbase / 2 |
| Na2CO3 with HCl (methyl orange) | 1 : 2 | moles of Na2CO3 = moles of HCl / 2 |
| Oxalic acid with NaOH | 1 : 2 | moles of NaOH = 2 × moles of acid |
| Purity of a sample | from equation | mass found / mass taken × 100 |
In the board exam, write the balanced equation first, show the mole ratio in one line, and then calculate. Check your own board's practical-exam pattern in the latest syllabus, because the marking scheme can differ between boards.
Check your titration answers quickly: the molarity and concentration tool finds moles, molarity and mass from the values you enter.
Open the Molarity and Concentration Calculator →Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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