[mariam amro]

4.90g of pure sulphuric acid was dissolved in water; the resulting total volume was 200 cm3. 20.7 cm3 of this solution was found on titration, to completely neutralise 10.0 cm3 of a sodium hydroxide solution. [Atomic masses: S = 32, O = 16, H = 1)
Calculate the concentration of the sodium hydroxide in mol/liter (molarity).

I tried to solve this an I got 4.16 M for NAOH
Is it right?? Can you please help me solve it, I think my mistakes is in the units because I need to solve it by using L and other SI units...

Thanks for your help 

[Borek]

That's not what I got, so one of us is wrong.

Please show what you did. Your result - when compared to mine - is exactly four times off. That's a strong suggestion problem lies in the stoichiometry.

[mariam amro]

H2SO4 +2NaOH ---> 2H2O +Na2SO4

H2SO4:
mass=4.9 grams... M.M= 98 g/mol... moles= mass/M.M=4.9/98=0.05 moles
volume= 20.7 ml = 0.0207 L
Molarity= moles/volume = 0.05/0.0207= 2.415

2NaOH:
volume= 24 ml = 0.024 L

Stiochiometry: ratio... H2SO4: NaOH... 1:2... (MaVa)/1=(MbVb)/2...((2.415)(0.0207))/1=(Mb(0.024))/2... Mb=4.16 M

What is wrong??

[Borek]

volume= 20.7 ml = 0.0207 L
Molarity= moles/volume = 0.05/0.0207= 2.415

20.7 mL is not the volume containing 0.05 moles of the acid, please reread the question.

2NaOH:
volume= 24 ml = 0.024 L

No idea where you got 24 mL from.

Apparently the factor of 4 is just a random combination of several mistakes.

[mariam amro]

Ya sorry, I don't know where I got this from. The volume is 10 ml so 0.01 L
But what do you mean that it is not 20.7 ml??

[Borek]

But what do you mean that it is not 20.7 ml??

Please reread the question. You took 20.7 mL of what?

[mariam amro]

....20.7 cm3 of this solution was found on titration, to completely neutralise 10.0 cm3 of a sodium hydroxide....
the 20.7 ml is for the acid because it neutralises the base which is 10 ml??

[Borek]

....20.7 cm3 of this solution

Which solution?