im sure its really simple but i just cant seem to get sensible answer/ am not sure if my answers are sensible so any coments will be greatfully recieved !!
TITRATIONthis is what ive got in my report so far but ive got myself in a bit of a pickle!!!
25cm3 of sodium carbonate put into a conical flask. A 0.2 M solution of hydrochloric acid solution was added . Methyl orange was the indicator turned a grey colour when 26.37ml of the hydrochloric acid solution had been added
So I know the ratio (no of moles) of sodium carbonate and its volume
I can calculate the number of moles of sodium carbonate in my standard solution
then I know the number of moles of sodium carbonate and the volume of the solution I made up (250mL) so I can work out the concentration of the 25ml of sodium carbonate in the conical flask.
Sodium carbonate (aq) + Hydrochloric acid (aq) ? Sodium chloride (aq) + Water (l) + Carbon dioxide (g)
Na2CO3 (s) + 2 HCl (aq) ? 2 NaCl (aq) + CO2 (g) + H2O (l)
1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid
(250cm3 is 0.25 dm3).
C = N/V I will calculate the moles of sodium carbonate =1/0.250 = 4 mol dm-3.
The number of moles in 25ml of my standard solution is
N = C x V
0.4 mol dm-3 x 25ml= 10
Where:
N = number of moles,
C = concentration in mol dm-3
And V = volume of solution in dm3
im in a right mess people !!!!deperatly need a pointer in the right direction as so as u throw numbers in to the situation im completely lost !!