A student titrates 25.0 cm325.0\,\mathrm{cm^{3}}25.0cm3 portions of sodium hydroxide solution with 0.0500 mol dm−30.0500\,\mathrm{mol\,dm^{-3}}0.0500moldm−3 sulfuric acid: H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\mathrm{H_{2}SO_{4}(aq) + 2NaOH(aq) \rightarrow Na_{2}SO_{4}(aq) + 2H_{2}O(l)}H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)
| Titration | Rough | 1 | 2 | 3 |
|---|---|---|---|---|
| Volume of acid added / cm3\mathrm{cm^{3}}cm3 | 23.60 | 22.40 | 22.95 | 22.45 |
Titres within 0.10 cm30.10\,\mathrm{cm^{3}}0.10cm3 of each other are concordant. What is the concentration of the sodium hydroxide solution?
0.0224 mol dm−30.0224\,\mathrm{mol\,dm^{-3}}0.0224moldm−3
0.0449 mol dm−30.0449\,\mathrm{mol\,dm^{-3}}0.0449moldm−3
0.0897 mol dm−30.0897\,\mathrm{mol\,dm^{-3}}0.0897moldm−3
0.0904 mol dm−30.0904\,\mathrm{mol\,dm^{-3}}0.0904moldm−3
51 exam-style questions on Edexcel GCSE Chemistry 6.2 Quantitative analysis, covering 6.2.1 Concentration in mol dm⁻³ and titration calculations, 6.2.2 Percentage yield and why actual yield falls short, 6.2.3 Atom economy of a reaction, 6.2.4 Choosing a reaction pathway, and 6.2.5 Molar volume of gases and Avogadro's law. Each one has a worked solution and a mark scheme showing where the marks go.