a 23ml H2SO4 solution is titrated by 50 ml 0.10 M NaOH solution, what is the molarity of H2SO4 solution? (Note that H2SO4 is a diprotic acid)

H2SO4 + 2NaOH ==> Na2SO4 + 2H2O

moles NaOH used = M x L = ??
Convert moles NaOH to moles H2SO4 using the coefficients in the balanced equation. 1 mol NaOH = 1/2 mol H2SO4.
M H2SO4 = moles H2SO4/L H2SO4.

To find the molarity of the H2SO4 solution, we can use the equation for neutralization reaction between H2SO4 and NaOH:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

From the equation, we can see that the mole ratio between H2SO4 and NaOH is 1:2. Therefore, for complete neutralization, 1 mole of H2SO4 reacts with 2 moles of NaOH.

Given that the volume of NaOH solution used is 50 mL and its concentration is 0.10 M, we can calculate the number of moles of NaOH:

Moles of NaOH = Volume (in L) × Concentration
= 50 mL × (1 L/1000 mL) × 0.10 M
= 0.005 moles

Since the mole ratio between H2SO4 and NaOH is 1:2, we know that the number of moles of H2SO4 is half of the number of moles of NaOH:

Moles of H2SO4 = 0.005 moles ÷ 2
= 0.0025 moles

Finally, we can calculate the molarity of the H2SO4 solution by dividing the moles of H2SO4 by the volume of the solution in liters:

Molarity of H2SO4 = Moles of H2SO4 / Volume of Solution (in L)
= 0.0025 moles / (23 mL × (1 L/1000 mL))
≈ 0.109 M

Therefore, the molarity of the H2SO4 solution is approximately 0.109 M.