Are my calculations good?

Question

Questions: A 5-mL aliquot of sulfuric acid was titrated with 0.124 M sodium hydroxide, and the initial and final volume readings from the burette were 1.29 mL and 12.64 mL, respectively. What was the concentration of the sulfuric acid?

What I did:

Neutralization equation :

H2SO4 + 2NaOH---> 2H20 + Na2SO4

Vacid * Cacid = Vbase * Cbase

to find the Cacid you need to isolate Cacid so

Cacid = (Vbase * Cbase) / Vacid

Vacid = 5 * 10^-3 L

Cacid= ?

Vbase= 0.01135 L

Cbase= 0.124 mol/L

Cacid =( 0.01135 L * 0.124 mol/L) / (5 * 10^-3 L) = 0.28 mol/L

Answer 1

I'm afraid you neglected to look at the balanced equation; 1 mol H2SO4 = 2 mol NaOH

(12.64-1.29) mL x 0.124 M = 1.4074 mmol NaOH

1.4074 mmol NaOH will neutralize 0.7037 mmol H2SO4 (1.4074/2=0.7037)

5.00 mL H2SO4 x M = 0.7037

M of H2SO4 = 0.14 (0.7037/5.00 = 0.1407)

Answer 2

You left out a factor of 2.

Each molecule of H2SO4 gives up TWO H+.

NaOH is single shot though.

You show this in your balanced chemical equation - you have to adjust your concentration calculation. What you calculated is the concentration of H+, which is half the concentration of H2SO4

So Vacid x Cacid x 2 = Vbase x Cbase