You have read the exam question carefully and decided that the following are the key bits of information:
- 1.50 g of impure Na2O reacts and dissolves to make a 150 cm3 solution
- A 50 cm3 sample is titrated with 13.7 cm3 of 0.5 mol dm-3 HNO3
- Determine the % by mass purity of Na2O.
| Step 1 We know that moles = concentration x volume: n HNO3 = 0.5 x (13.7 / 1000) = 6.85 x 10-3 mol Step 2 The balanced symbol equation isn’t given so we need to work it out. There are actually two reactions to consider here: Na2O + H2O –> 2 NaOH NaOH + HNO3 –> NaNO3 + H2O Step 3 There is a 1:1 ratio between NaOH and HNO3 n NaOH = 6.85 x 10-3 mol (in 50 cm3 sample) Step 4 The total is 3x bigger than the sample (i.e., 150 / 3): n NaOH = 3 x 6.85 x 10-3 mol = 2.06 x 10-2 mol in total Step 5 There is a 1:2 ratio between Na2O and NaOH n Na2O = 0.5 x 2.06 x 10-2 mol = 0.0103 mol in total Step 6 The relative formula mass (Mr) of Na2O is 62. This means that the actual mass of Na2O in the impure solid is: m Na2O = 0.0103 x Mr = 0.0103 x 62 = 0.64 g Step 7 The % by mass purity is then calculated using: % by mass purity = 0.64 / 1.50 x 100 = 43% |