[19kevin87]

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!

[AWK]

Densities of both solutions are needed

[19kevin87]

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

[Borek]

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.