# Homework Help: Mr of a hydrated sodium carbonate

1. Dec 1, 2009

### nobahar

1. The problem statement, all variables and given/known data
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 was made up to 250cm3. In a titration, 25cm3 portion of this solution required 24.3cm3 of 0.2 mol dm-3 hydrochloric acid for complete reaction.

2. Relevant equations

Na2CO3 + 2HCl is the important bit

3. The attempt at a solution
I got the number of moles of HCl to be 4.86 x 10-3 mol; therefore the number of moles of Na2CO3 to be 2.43 x 10-3 mol in 25cm3 and so 0.0243 mol in the orginal 250cm3.
This must be the number of moles of Na2CO3 in the original 3.01g. 0.0243 moles of Na2CO3 weighs 2.5758g, and so the water must make up 0.4342g (derived from 3.01g - 2.5758g). The number of moles of H2O is therefore 0.024. I figured that the number of moles of Na2CO3 to H2O is approximately 1, and so the orginal formula must be Na2CO3.H2O (i.e. x is 1).
The Mr is therefore106+(1 x 18) = 124g/mol.
Is this correct?

2. Dec 2, 2009

### nobahar

I fear may question may appear too long. This is the important part:

This must be the number of moles of Na2CO3 in the original 3.01g. 0.0243 moles of Na2CO3 weighs 2.5758g, and so the water must make up 0.4342g (derived from 3.01g - 2.5758g). The number of moles of H2O is therefore 0.024. I figured that the number of moles of Na2CO3 to H2O is approximately 1, and so the orginal formula must be Na2CO3.H2O (i.e. x is 1).
The Mr is therefore106+(1 x 18) = 124g/mol.

3. Dec 2, 2009

### Staff: Mentor

Looks OK to me.

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4. Dec 3, 2009

### nobahar

Thanks Borek.