[biomed77]

2.12g of pure anhydrous sodium carbonate was dissolved in water in a 100cm^3 volumetric flask. if 25cm^3 of this solution required 33.3cm^3 of hydrochloric acid solution for neutralisation, calculate the molarity of the hydrochloric acid solution...

formulae mass of sodium carbonate = 23*2+12+16*3 = 106moles   2.12g in 106 = 0.02mol in 100cm^3  therefore 0.2 in 1000cm^3 so molarity is 0.2mol dm^-3

moles sodium carbonate used in titration = 25*0.2/1000 = 0.005
moles HCl titrated = 2* moles of sodium carbonate used = 0.01 mol HCl in 33.3cm^3 of the acid solution
molarity of HCl = 0.01*1000/33.3 = 0.300mol dm^-3....

i appologise to everybody and especially to borek because i send questions involving titrations all the time, but i have about 15 to answer for my assignment and i have to hand it in on monday, thanks everybody for their response..  p.s.  is it right???

[AWK]

Yes, it is