What is the concentration of a solution of sulphuric acid of which 20ml would be needed to neutralise 20.5ml of 2M sodium hydroxide?

Answer 1

The balanced chemical equation for the reaction between sulphuric acid (H2SO4) and sodium hydroxide (NaOH) is:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH.

Given that 20.5mL of 2M NaOH is used, we can calculate the moles of NaOH:

20.5mL * 2M = 41 mmol NaOH

Since 1 mole of H2SO4 reacts with 2 moles of NaOH, the moles of H2SO4 required would be half of the moles of NaOH:

41 mmol NaOH / 2 = 20.5 mmol H2SO4

Now, if 20mL of the sulphuric acid solution is used, we can determine the concentration of the solution:

Concentration (mol/L) = moles / volume (L)

Concentration = 20.5 mmol / 0.020 L = 1025 mmol/L

Therefore, the concentration of the sulphuric acid solution is 1025 mmol/L.