Question 22N.2.hl.TZ0.c.iii
| Date | November 2022 | Marks available | [Maximum mark: 4] | Reference code | 22N.2.hl.TZ0.c.iii |
| Level | hl | Paper | 2 | Time zone | TZ0 |
| Command term | Determine | Question number | c.iii | Adapted from | N/A |
Ammonium nitrate, NH4NO3, is used as a high nitrogen fertilizer.
A 0.20 mol dm−3 solution of ammonium nitrate is prepared.
A 20.00 cm3 sample of the 0.20 mol dm−3 solution of ammonium nitrate is titrated with a 0.20 mol dm−3 solution of sodium hydroxide. Determine the pH at the equivalence point, to two decimal places using section 1 and 21 of the data booklet.
[4]
«n(NH4NO3) = 0.20 mol dm−3 × 0.02000 dm3 =» 0.0040 «mol NH4NO3» ✔
«[NH3] at equivalence point 0.10 «mol dm−3» ✔
«Kb = 10−pKb = 10−4.75 = 1.8 × 10−5»
« «mol dm−3» ✔
«pOH = –log (0.0013) = 2.89»
«pH = 14.00 – pOH =» 11.11✔
Award [4] for correct final answer.
Accept a range of 11.11 – 11.14.
Probably the most difficult question on the paper with an average mark of about 1½/4. Few candidates, were capable of working their way through this long calculation, though quite a number gained partial marks, showing the importance of setting out workings clearly. Many tried to, inappropriately, employ the Henderson-Hasselbach equation.
