CaCO3 % Purity pH 1.61 TRICK MADE EASY | May/June 2025 Q7.4 | Grade 12 P2

This question WILL come in your exams. DBE May/June 2025 Question 7.4 - IMPURE CaCO3 + HCl + pH. In this question they combine 3 topics in 1: 1. pH to concentration 2. Limiting reagent / excess 3. % Purity / Impurity mass Many learners get 0/9 because they forget excess HCl from pH. FULL SOLUTION: Question: 1,5g impure CaCO3 + 200 cm3 of 0,15 mol/dm3 HCl → pH = 1,61. Find mass of impurities. Step 1: Initial moles HCl n = C x V = 0,15 x 0,2 = 0,03 mol Step 2: Excess HCl from pH pH = 1,61 → [H+] = 10^-1,61 = 0,0245 mol/dm3 n(excess) = 0,0245 x 0,2 = 0,0049 mol Step 3: Reacted HCl n(reacted) = 0,03 - 0,0049 = 0,0251 mol Step 4: Pure CaCO3 CaCO3 + 2HCl → ... so n(CaCO3) = 0,0251/2 = 0,01255 mol m(pure) = 0,01255 x 100,09 = 1,256 g Step 5: Impurities m(impurities) = 1,5 - 1,256 = 0,244 g % purity = 83,7% WATCH NEXT: 👉 Back Titration % Purity May 2021 Q5.5 - 30% CTR Winner (6:56) 👉 Equilibrium Graph - Rank 1 Video (338 views) DOWNLOAD: Clean Question Paper in video - screenshot at 00:05 #PhysicalSciences #Grade12 #Prelims2025 #DBE #Chemistry #AcidsAndBases #PercentPurity #Matric2025 #P2 #SouthAfrica TIMESTAMPS: 00:00 - Question analysis 00:45 - Initial moles 02:10 - pH trick (most important) 04:30 - Back calculation 06:00 - Final answer & common mistake