You have read the exam question carefully and decided that the following are the key bits of information:
- 3.69 g of Ba(OH)2.xH2O is made into a 250 cm3 solution
- A 25 cm3 sample of the solution is titrated with 23.4 cm3 of 0.1 mol/dm3 HCl.
- Calculate the relative formula mass of Ba(OH)2.xH2O, and hence x.
| Step 1 We know that moles = concentration x volume: n HCl = 0.1 x (23.4 / 1000) = 2.34 x 10-3 mol. Step 2 The balanced symbol equation isn’t given so we need to work it out. This is an acid-base reaction neutralisation: 2 HCl + Ba(OH)2 –> BaCl2 + 2 H2O Step 3 The molar ratio is 2:1 so this means the moles of Ba(OH)2 that reacted in the sample was: n Ba(OH)2 = 1/2 x 2.34 x 10-3 mol = 1.17 x 10-3 mol Step 4 The total is 10x bigger than the sample (i.e., 250 / 25): n Ba(OH)2 = 10 x 1.17 x 10-3 mol = 1.17 x 10-2 mol in total Step 5 This means that the Mr of Ba(OH)2.xH2O is: Mr Ba(OH)2.xH2O = 3.69 / 1.17 x 10-2 = 315 Step 6 We can now subtract the Mr of just Ba(OH)2 (Mr = 171) Mr xH2O = 315 – 171 = 144 Step 7 We can now divide by 18 (the Mr of H2O) to find x x = 144 / 18 = 8 |