Sodium carbonate exists in hydrated form, Na2CO3.xH2O, in the solid state. 3.5 g of a sodium carbonate sample was dissolved in water and the volume made up to 250 cm3. 25.0 cm3 of this solution was titrated against 0.1 moldm-3 HCl and 24.5 cm3 of the acid were required. Calculate the value of x given the equation:
Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
NO. of Moles = Conc. x volume/1000
The Attempt at a Solution
I have attempted this question in this way,
Moles of HCI = 0.1 x 24.5 / 1000 = 0.00245
Then looking at the molar ratio, I see 1:2 ratio of Sodium carbonate and HCI.
So, Moles of Na2CO3, is 0.00245/2 = 0.001225 moles in 25 cm3
Then I did the NO. of moles for Na2CO3 in 250cm3 so 250cm3/25cm3 = 10 which is the multiplier, then i did 10 x 0.001225 = 0.01225 mol
I did the Molar Mass of Na2CO3 which is 3.5/0.01225 = 285.7142.. which rounds to 286.
I assumed the Mr of Na2CO3.xH2O is 286, as the mark scheme assumed that
I don't know what to do after this, can someone please help me, as this is Homework!