The standard calculation
3 marks · EasyA student has the following titration results:
| Titration | Volume of NaOH (cm³) | Concentration of NaOH (mol/dm³) |
|---|---|---|
| 1 | 22.0 | 0.150 |
| 2 | 21.5 | 0.150 |
| 3 (repeat) | 21.7 | 0.150 |
The student used 25.0 cm³ of hydrochloric acid of unknown concentration each time. Use the results to calculate the mean number of moles of sodium hydroxide used to neutralise the acid.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
0.00323 mol (±0.0001)
Also accept: Accepts working with mean volume 21.7 cm³
Common mistakes: Not using mean volume.; Not converting to dm³.
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Finds mean volume of NaOH: (21.5 + 21.7)/2 = 21.6 cm³ or uses correct method. (Ignoring first titration as rough)
M2 · 1 mark · Converts mean volume to dm³: 21.6 cm³ = 0.0216 dm³ (or correct mean). (follow-through)
A1 · 1 mark · Calculates (0.0216) × 0.150 = 0.00324 mol (accept 0.00323 depending on mean). (follow-through)
Worked solution
Mean volume (excluding first – rough): (21.5 + 21.7)/2 = 21.6 cm³
Convert to dm³: 21.6/1000 = 0.0216 dm³