What volume of 0.744 M NaOH is required to titrate 10.00 mL of 0.526 M H2SO4?

Question

I can only help with Molar Mass, that's it...

Molar Mass of NaOH is 39.998
Na: 22.99 + O: 16.00 + H: 1.008

Molar Mass of H2SO4 is 98.086
H: (2) 1.008 + S: 32.07 + O: (4) 16.00

Please show me how you do it step by step, thanx.

Answer 1

NO, NO, NO. We use the VN equation:
Vol 1 x normality1 = Vol 2 x normality 2
where "1" refers to acid and "2" to the base.

In this case, Vol 1 = 10 mL and normality1= 2x0.526 . normality 2= 0.744. So you plug in and solve for Vol 2.

Sulfuric acid is "dibasic", in that one molecule of it supplies 2 H+ ions. That why we multiply by 2 when dealing with normality.

Answer 2

volume 1 x molarity 1 = volume 2 x molarity 2
so 10 x 0.526 = vol NaOH x 0.744
so vol NaOH = 7.07 ml to 1st end-point
NaOH + H2SO4 = NaHSO4 + H2O

and 2 x 7.07 = 14.14 ml to 2nd end point
NaOH + NaHSO4 = Na2SO4 + H2O

Answer 3

mols H2SO4 = mols NaOH

M1V1 = M2V2

(.01L H2SO4)*(.526M H2SO4) / (.744M NaOH) = volume required to titrat H2SO4 to the equivalence point.