Library / Solution chemistry, equilibria and kinetics / SCH-21
Solution chemistry, equilibria and kinetics
Analyte amount from a back titration
Moles of analyte X from a back titration: a known excess of reagent R reacts with the analyte (a mol per mol), and the leftover is titrated with a titrant T (b mol per mol of R).
- Ref
- SCH-21
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- nX [mol]
In a back titration a known excess of reagent R reacts with the analyte X (a mol of R per mole of X), and the unreacted R is titrated with titrant T (b mol of R per mole of T). The analyte amount is nX = (nR − b cT VT)/a. A negative result means the data are inconsistent, for example because the reagent was not in excess. Use litres for VT and mol/L for cT.
Moles of analyte X from a back titration: a known excess of reagent R reacts with the analyte (a mol per mol), and the leftover is titrated with a titrant T (b mol per mol of R). Both reactions must be complete.
Calculator
- Reagent left over, b cT VT0.002 mol
- Reagent used by the analyte, nR − b cT VT0.008 mol
- Share of the reagent that reacted with the analyte80 %
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.
- 80 % of the reagent reacted with the sample, so it contained 0.004 mol (4 mmol) of analyte.
- For good accuracy, the reagent should be in excess by no more than about a factor of two, so that the analyte is the difference of two comparable amounts, not of two nearly equal large ones.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mol) | Local step | Change in result |
|---|---|---|---|---|---|---|
| nR Amount of reagent R added | 0.01 to 0.5 mol | 0.004 | 0.249 | 0.245 | +10% | +12.5 % |
| a Moles of R per mole of analyte | 1 to 3 | 0.002667 | 0.008 | 0.005333 | +10% | −9.09 % |
| b Moles of R per mole of titrant | 1 to 2 | 0.003 | 0.004 | 0.001 | +10% | −2.5 % |
| VT Titre of the titrant | 0.001 to 0.02 L | 0.004 | 0.00495 | 9.5e-4 | +10% | −2.5 % |
| cT Titrant concentration | 0.01 to 0.1 mol/L | 0.004 | 0.0049 | 9e-4 | +10% | −2.5 % |
Sweep of Titre of the titrant
| Titre of the titrant (L) | Amount of analyte (mol) |
|---|---|
| 0.001 | 0.00495 |
| 0.003375 | 0.004831 |
| 0.00575 | 0.004712 |
| 0.008125 | 0.004594 |
| 0.0105 | 0.004475 |
| 0.01288 | 0.004356 |
| 0.01525 | 0.004237 |
| 0.01763 | 0.004119 |
| 0.02 | 0.004 |
Sweep one variable
Variables and how each one contributes
Amount of analyte
What the equation gives you. Everything else feeds it.
Amount of reagent R added
The amount of reagent R added, in mol. Raises the analyte amount in proportion to 1/a. Raising it by 10 % (to 0.011) moves the result by +12.5 %, from 0.004 to 0.0045 mol.
Moles of R per mole of titrant
The mol of R reacting per mole of titrant. A larger b means the titre accounts for more residual reagent, so the analyte amount falls. Raising it by 10 % (to 1.1) moves the result by −2.5 %, from 0.004 to 0.0039 mol.
Titrant concentration
The titrant concentration, in mol/L. A stronger titrant means more residual reagent measured, so the analyte amount falls. Raising it by 10 % (to 0.11) moves the result by −2.5 %, from 0.004 to 0.0039 mol.
Titre of the titrant
The titre of the titrant, in L. A larger titre means more reagent was left over, so the analyte amount falls. Raising it by 10 % (to 0.022) moves the result by −2.5 %, from 0.004 to 0.0039 mol.
Moles of R per mole of analyte
The mol of R reacting per mole of analyte. A larger a means more reagent used per mole of analyte, so the analyte amount falls in inverse proportion. Raising it by 10 % (to 2.2) moves the result by −9.09 %, from 0.004 to 0.003636 mol.
Worked example with the values above
- Residual reagent: b cT VT = 1 × 0.1 × 0.02 = 0.002 mol.
- Reagent used by the analyte: 0.01 − 0.002 = 0.008 mol.
- nX = 0.008 / 2 = 0.004 mol.
Assumptions and limits
- Complete, fast and well-defined reactions with a clear stoichiometry.
- Volumes in litres or in a single consistent unit, and calibrated glassware and electrodes.
Common mistakes
- Mixing mL and L in one calculation.
- Using the wrong mole ratio of reagent to analyte.
- Using points too close to the equivalence point for a Gran plot.
Statistics and uncertainty
Burette readings are good to ±0.02 to 0.05 mL, so a 25 mL titre has a relative error near 0.1 to 0.2 %, and the endpoint-detection error adds to it.
Where it is used
- Analysing antacids, alkalinity, chloride and other titratable species.
- Estimating the endpoint error of a titration.
Maths you need
- Stoichiometry: mole ratios in a balanced equation
- Equivalence point: where the reagent amounts match exactly
- Titre: the volume of titrant delivered