You have read the exam question carefully and decided that the following are the key bits of information:
- 100 cm3 of 0.1 mol/dm3 HCl is added to 0.240 g of a mixture containing Mg(OH)2 and an inert solid.
- The resulting solution is made up to 250 cm3 with distilled water.
- 25 cm3 of the solution is titrated with 22.1 cm3 of 0.02 mol/dm3 NaOH.
- Calculate the mass of inert solid in the sample.
| Step 1 We know that moles = concentration x volume: n HCl = 0.1 x (100 / 1000) = 0.01 mol. n NaOH = 0.02 x (22.1 / 1000) = 4.42 x 10-4 mol. Step 2 The balanced symbol equations aren’t given so we need work them out. Reaction 1: 2 HCl + Mg(OH)2 –> MgCl2 + 2 H2O Reaction 2: HCl + NaOH –> NaCl + H2O Step 3 The balanced symbol equation for Reaction 2 tells us that HCl and NaOH react in a 1:1 ratio so the HCl moles at the start of Reaction 2 was: n HCl = 1 x 4.42 x 10-4 = 4.42 x 10-4 mol Step 4 The total is 10x bigger than the sample (i.e., 250 / 25) so we can work out the HCl moles at the end of Reaction 1. n HCl = 10 x 4.42 x 10-4 = 4.42 x 10-3 mol (NB: The original sample volume of 100 cm3 is not relevant to the calculation here; this is because the dilution from 100 cm3 to 250 cm3 does not change the number of moles) Step 5 We know that there was 0..01 mol HCl at the start and 4.42 x 10-3 mol HCl at the end of Reaction 1. By subtraction, we can work out the HCl moles used up in Reaction 1: 0.01 – 4.42 x 10-3 = 5.58 x 10-3 mol Step 6 From Reaction 1, we know that HCl and Mg(OH)2 react in a 2:1 ratio, so: n Mg(OH)2 = 0.5 x 5.58 x 10-3 = 2.79 x 10-3 mol Step 7 Now that we have the moles of Mg(OH)2 we can work out the mass of Mg(OH)2 using its Mr (58.3): m Mg(OH)2 = 58.3 x 2.79 x 10-3 = 0.163 g Step 8 The mass of the inert solid is then found by subtracting the mass of Mg(OH)2 from the total mass: 0.240 – 0.163 = 0.077 g |