Library / Foundations: Chemistry / FCH-30
Foundations: Chemistry
Concentration of excess acid or base after mixing
Concentration of whichever strong reagent is left over after a strong acid and a strong base have neutralised each other, with the volumes added together.
- Ref
- FCH-30
- Rev
- v1
- Field
- Foundations: Chemistry
- Result
- cexcess [mol/dm³]
A strong acid and a strong base react one mole for one mole, H⁺ + OH⁻ → H₂O. The moles are n = cV (with V in dm³; using cm³ gives millimoles). The excess of whichever is left, divided by the total volume, is the concentration cexcess = |nH⁺ − nOH⁻|/Vtotal. At the equivalence point the two are equal and the excess is zero.
Concentration of whichever strong reagent is left over after a strong acid and a strong base have neutralised each other, with the volumes added together. It is zero at exact equivalence.
Calculator
- Moles of acid, nA2.5 mmol
- Moles of base, nB2 mmol
- Net moles left (acid minus base; positive means acid in excess)0.5 mmol
- Total volume45 cm³
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Mixing 25 cm³ of 0.1 M acid with 20 cm³ of 0.1 M base leaves 0.5 mmol of acid in 45 cm³: 0.01111 mol/dm³.
- Neutralisation needs 25 cm³ of this base; adding more than that turns the mixture from acid to base.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mol/dm³) | Local step | Change in result |
|---|---|---|---|---|---|---|
| cA Concentration of the strong acid | 0.001 to 2 mol/dm³ | 0.003205 | 1.067 | 1.063 | +10% | +50 % |
| cB Concentration of the strong base | 0.001 to 2 mol/dm³ | 7.948e-4 | 0.8333 | 0.8325 | +10% | −40 % |
| VA Volume of the acid | 1 to 200 cm³ | 0.0022 | 0.09048 | 0.08828 | +10% | +42.1 % |
| VB Volume of the base | 1 to 200 cm³ | 0.004822 | 0.09231 | 0.08749 | +10% | −42.6 % |
Sweep of Volume of the base
| Volume of the base (cm³) | Excess concentration (mol/dm³) |
|---|---|
| 1 | 0.09231 |
| 25.88 | 0.00172 |
| 50.75 | 0.03399 |
| 75.63 | 0.05031 |
| 100.5 | 0.06016 |
| 125.4 | 0.06675 |
| 150.3 | 0.07147 |
| 175.1 | 0.07502 |
| 200 | 0.07778 |
Sweep one variable
Variables and how each one contributes
Excess concentration
What the equation gives you. Everything else feeds it.
Concentration of the strong acid
The concentration of the acid, in mol/dm³. Raises the excess while acid is in excess, because there is more acid left over. Raising it by 0.01 (to 0.11) moves the result by +50 %, from 0.01111 to 0.01667 mol/dm³.
Volume of the acid
The volume of acid, in cm³. Raises the excess while acid is in excess, though less than in proportion because the total volume also rises. Raising it by 5 cm³ (to 30 cm³) moves the result by +80 %, from 0.01111 to 0.02 mol/dm³.
Concentration of the strong base
The concentration of the base, in mol/dm³. Lowers the excess while acid is in excess, because the base neutralises more of the acid; once the base is the one in excess, raising it raises the excess. Raising it by 0.01 (to 0.11) moves the result by −40 %, from 0.01111 to 0.006667 mol/dm³.
Volume of the base
The volume of base, in cm³. Lowers the excess while acid is in excess; once the base is the one in excess, raising it raises the excess. Raising it by 2 cm³ (to 22 cm³) moves the result by −42.6 %, from 0.01111 to 0.006383 mol/dm³.
Worked example with the values above
- Moles (in mmol): nA = 0.1 × 25 = 2.5; nB = 0.1 × 20 = 2.
- Excess: |2.5 − 2| = 0.5 mmol, in a total volume of 45 cm³.
- cexcess = 0.5 / 45 = 0.01111 mol/dm³ of acid.
Assumptions and limits
- A strong monoprotic acid and a strong monobasic base, each fully dissociated.
- Volumes add on mixing, and the temperature is 25 °C so that Kw = 10⁻¹⁴.
Common mistakes
- Forgetting that the total volume includes both solutions.
- Using the volumes in cm³ with concentrations as if they were in dm³ without keeping the units consistent.
- Applying it to a weak acid or base.
Statistics and uncertainty
Near the equivalence point the excess is a small difference of two large numbers, so it is very sensitive to measurement error: a 1 % error in either volume can change a small excess by a large fraction.
Where it is used
- Concentration of acid or base left after a neutralisation or titration step.
- Step towards finding the pH of the mixture.
Maths you need
- Moles: concentration times volume
- Neutralisation: H⁺ + OH⁻ → H₂O, one mole for one mole