Sulphuric acid (H2SO4) is titrated against 25.00cm3 of 0.200 mol/dm3 Sodium hydroxide solution (NaOH). It is found that 23.20 cm3 of sulphuric acid is required for neutralization. Calculate the concentration of Sulphuric acid

My answer,
Approximately 0.108 mol dm^-3

Explanations,
1. Make the balance chemical equation
2. Convert the 25 cm^3 to 0.025 dm^3 by multiplying 25 with 10^-3 (or 25 / 1000)
3. Concentration is equal to = moles of substance over volume of solution
4. Calculate how many number of moles present in 0.025dm^3 of sulphuric acid.

Mole = concentration x volume
^ rearranged from conc = mole/volume

Mole = 0.200 x 0.025 = 0.005 mol

5. Use the balance chemical equation ad mole ratio. In this case,

H2SO4 + 2NaOH —> 2H2O + Na2(SO4)

Thus the ratio between Sulphuric acid and Sodium Hydroxide is 1:2

Hence, the mole of H2SO4 is 0.005/2 = 0.0025 mol

6. Calculate the concentration of the sulphuric acid by using conc. = mole/volume

Hence the answer is 0.0025/(23.20 x 10^-3) = 0.108 mol dm^-3 rounded off at the 3 Sigfigs

I hope u understand sorry for any mistakes done.

Tropical easterlies is mostly likely your answer correct me if i’m wrong