Urgent Help Needed - Moles!!!

17.16g of hydrated sodium carbonate of formula Na2CO3.xH2O were dissolved in water and made up to exactly 250cm3 in a standard flask. 25cm3 of this solution were then pipetted into a conical flask and titrated with 0.78M hydrochloric acid; 15.4cm3 were requires. Calculate the value of x in the formula.

HELP!!!!!!!!!!!!!!!!!

Reply 1

mubinchoudhury

lucaz

15.4cm3 of acid was used.

This is 0.0154dm3. The concentration of the acid was 0.78M. Therefore, according to the formula:

Mole = Concentration * Volume.

there were (0.78*0.0154) = 0.012012 Mole of acid used.

The reaction ratio is 1:1, therefore 0.012012 moles of Na2CO3 were present in 25cm3. Again using M=CV (or just multiplying by 10), this means that there are 0.12012 Moles of Na2CO3 in 250cm3 of the solution and in the original 17.16g hydrated sample.

Mr. of Na2CO3 is 106. Therefore the mass of .12012 Moles is: (106*0.12012) = 12.73g

This means, that from the original sample, only 17.16-12.73 = 4.43g was water.

4.43g = 0.246 moles of H20 (this is achieved by multiplying by 18, the Mr of water)

Therefore, the ratio of H20 atoms to Na2CO3 atoms is:

0.246:0.12012
2:1

Therefore, for every Na2CO3 there is 2 H20 Molecules, hence, X=2