This exam focus covers Exam Focus within Titrations (HT) for GCSE Chemistry. Revise Titrations (HT) in Quantitative Chemistry for GCSE Chemistry with 22 exam-style questions and 20 flashcards. This topic appears regularly enough that it should still be part of a steady revision cycle. It is section 11 of 13 in this topic. Treat this as a marking guide for what examiners are looking for, not just a fact list.
Topic position
Section 11 of 13
Practice
22 questions
Recall
20 flashcards
🎯 Exam Focus
Frequently Examined
Titrations appear regularly in AQA, Edexcel and OCR papers at Higher Tier. Expect questions on:
- Required Practical: Describe the method, including how to achieve concordant results (4-6 marks)
- Calculation: Given titre, concentration of one solution, volume in flask — calculate the other concentration (3-4 marks)
- Error analysis: "Suggest a reason why the titration results were not concordant" (2 marks)
- Indicator choice: Why phenolphthalein is used for strong acid/strong base titrations (1-2 marks)
- Mole ratio questions: Non-1:1 reactions (e.g., H₂SO₄ + 2NaOH) — you must use the ratio correctly (3-4 marks)
Required practical tip: You must describe rinsing the burette with the solution it will contain, reading the meniscus at eye level, and repeating to get concordant results. Missing any of these loses marks.
Quick Check: Why do you rinse the burette with the solution it will contain, rather than with distilled water?
If the burette is wet with distilled water, the solution inside becomes diluted, making the concentration inaccurate. Rinsing with the actual solution ensures the concentration remains correct.
Quick Check: Three titration results are: 24.80, 24.75, 25.30 cm³. Which results should be averaged and what is the mean titre?
The concordant results are 24.80 and 24.75 cm³ (within 0.10 cm³ of each other). The 25.30 result is anomalous and is discarded. Mean titre = (24.80 + 24.75) ÷ 2 = 24.775 cm³ ≈ 24.78 cm³.
Quick Check: How many moles of HCl are in 18.0 cm³ of 0.200 mol/dm³ hydrochloric acid?
n = c × V = 0.200 × (18.0 ÷ 1000) = 0.200 × 0.0180 = 0.00360 mol. Always convert cm³ to dm³ first (÷ 1000).