After a tumultuous year between high school and sixth form, I've started my A-levels four weeks late. If someone could go through this question with extensive methodology, it'd be beyond appreciated. I'm familiar with the general concepts but I haven't had time to refresh myself properly.
Sodium carbonate forms a number of hydrates of general formula Na2CO3.xH2O. A 3.01g sample of one of these hydrates was dissolved in water and the solution made up to 250cm³.In a titration, a 25.0cm³ portion of this solution required 24.3cm³ of 0.200 mol dm-³ hydrochloric acid for complete reaction.
The equation for this reaction is shown below.
Na2CO3 + 2HCl 2NaCl + 2H2O + CO2
- Calculate the number of moles of HCl in 24.3cm³ of 0.200 mol dm-³ hydrochloric acid.
- Deduce the number of moles of Na2CO3 in 25.0cm³ of the Na2CO3 solution.
- Hence deduce the number of moles of Na2CO3 in the original 250cm³ of solution.
- Calculate the Mr of the hydrated sodium carbonate.
In an experiment, the Mr of a different hydrated sodium carbonate was found to be 250. Use this value to calculate the number of molecules of water of crystallisation, x, in this hydrated sodium carbonate, Na2CO3.xH2O.
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- Thread Starter
- 09-10-2006 12:17
- 09-10-2006 12:48
Let's do the first part:
1. Using concentration = amount/volume (in units mol.dm-³ = mol/dm³),
The no. of moles of HCl used to neutralise the 25 cm³ of Na2CO3 = (24.3/1000)dm³ * 0.200 mol.dm-³ = 0.00486 mol
2. Looking at the reaction given, you can see that 1 mole of Na2CO3 reacts with 2 moles of HCl.
So, the number of moles of Na2CO3 in 25cm³ = 0.00486/2 = 0.00243 mol
3. No. of moles in the 250cm³ = 0.00243 * 10 = 0.0243 mol
4. This part may be a bit tricky:
First, you should know that mass = amount/Mr
Secondly, 1 mole of Na2CO3.xH2O generates 1 mole of Na2CO3.
so it follows that amount of Na2CO3.xH2O in 250 cm³ = 3.01/(23*2 + 12 + 16*3 + x(2+16)) = 3.01/(106 + 18x)
but this also equals 0.0243
Solving for x, you get x = 1.00 (2 d.p.)
Therefore, the hydrated form you used was Na2CO3.H2O, and the Mr of this is 124 mol/g.
- 09-10-2006 12:51
The second part is easy:
Mr = 250 = 106 + 18x
Solve for x, x = 8