Chemical Forums
Specialty Chemistry Forums => Citizen Chemist => Topic started by: 19kevin87 on June 14, 2010, 11:45:03 PM

I've calculated and my answer is 1.66 mL.
But I'm not sure bout the ans plus my working is too complicated, can you show me how u get the ans?
Correct me if I'm wrong.
Thanks! :)
my version:
1) to change % to concentration
For H2SO4: 4.74*10/MW_{H2SO4} = 0.48 mol L^{1}
For NaOH: 48.5*10/MW_{NaOH} = 12.13 mol L^{1}
2) 0.48mol of 21mL H2SO4 is eq to 0.01008 mol
3) since acid to alkali ratio is 1:2, therefore moles of NaOH needed is 0.02016 mol
4) therefore volume needed for 48.5% NaOH is
0.02016*1000/12.13 = 1.66 mL
Thanks again!

Densities of both solutions are needed

Densities of both solutions are needed
Ya, from the formula, the densities are needed.. since both sulphuric acid and caustic soda have density around 1gcm^{3}, i treat it as 1

since both sulphuric acid and caustic soda have density around 1gcm^{3}, i treat it as 1
That's true only for diluted solutions, one of the solutions in question have density around 1.5 g/mL.