A 25.00-mL sample of an H2SO4 (2 and 4 are subscripted) solution of unknown concentration is titrated with a .1328 M KOH solution. A volume of 38.33 mL of KOH was required to reach the endpoint. What is the concentration of the unknown H2SO4 (again, the 2 and 4 are subscripted) solution?

my work:
38.33 mL KOH* (1L/1000mL)*(.1328 M KOH/L KOH)*(1 mol H2SO4/1mol KOH)=.005090 mol H2SO4

M= mol solution/L= .005090mol/.025L
=.2036M

However, the answer in the book says it should equal .1018M. I'm confused.

Answered above. If you balance the equation you will find it takes 2 KOH for 1 mole H2SO4.

To determine the concentration of the unknown H2SO4 solution, you are correct in using the stoichiometry between H2SO4 and KOH in the balanced chemical equation. However, it seems that you made a calculation error during the conversion.

Let's go through the calculation step by step to find the correct answer:

1. Calculate the number of moles of KOH used:
Volume of KOH solution used = 38.33 mL
Concentration of KOH solution = 0.1328 M

Moles of KOH = (Volume of KOH solution used) x (Concentration of KOH solution)
= 38.33 mL x (1 L / 1000 mL) x 0.1328 mol/L
= 0.005087 mol KOH

2. Use the stoichiometric ratio from the balanced equation to find the moles of H2SO4 present in the solution:
From the balanced equation: H2SO4 + 2 KOH -> K2SO4 + 2 H2O
The ratio of KOH to H2SO4 is 2:1

Moles of H2SO4 = (0.005087 mol KOH) / (2 mol KOH / 1 mol H2SO4)
= 0.002544 mol H2SO4

3. Calculate the concentration of the H2SO4 solution:
The volume of the H2SO4 solution is 25.00 mL = 0.025 L

Concentration of H2SO4 = (moles of H2SO4) / (volume of H2SO4 solution)
= 0.002544 mol / 0.025 L
= 0.1018 M

Therefore, the concentration of the unknown H2SO4 solution should be 0.1018 M, which matches the answer provided in the book.