A 28.00 mL sample of an H2SO4 solution of unknown concentration is titrated with

c(H2SO4)*V(H2SO4) = 1/2c(KOH)*V(KOH)
c(H2SO4)= 0,0636M

These titration/concentration problems are all solved about the same way.

Step 1: A balanced equation H2SO4 + 2KOH K2SO4 + 2 HOH

STep 2: Write what you know and need to know under the equation.

H2SO4 + 2KOH K2SO4 + 2 HOH
C = ? C = 0.1105
V = 28 V = 32.22

Step 3: You are always given enough information to find the moles of one of the reactants that are used.
In this case it is the KOH

moles = C x V 0.1105 mol/L x 32.22 mL = 3.56031 moles

Step 4: Look at the mol ratio between KOH and H2SO4. Can you see that for every 2 mol of KOH you use you only need 1/2 that many moles of H2SO4?

So mol of H2SO3 needed will be 3.56031 x 1/2 = 1.780155 mol

Step 5: Now that you have both the moles and volume of H2SO4 used you can calculate C

C = n / V 1.780155 mol / 28.00 mL = 0.0636 mol/L

As I said most stoichiometric problems can be solved this way. Good Luck.

Balance your equation:
H2SO4 + 2KOH ? K2SO4 + 2H2O

Find the moles of KOH used:
.03222 L of KOH * 0.1105 M (moles/liters) KOH = 0.003560 moles of KOH

Convert from moles of KOH to moles of H2SO4 (need molar ratios from balanced equation here):
0.003560 moles of KOH x (1 mole H2SO4/2 moles KOH) = 0.00178 moles of H2SO4

Find the concentration of H2S04:
28ml x (1 liter/1000ml)= 0.028L
0.00178 moles of H2SO4/0.028L = 0.0636 M H2SO4