What volume of 0.4m sulphuric acid is needed to neutralize 25cm³ of 0.5m sodium hydroxide?

To determine the volume of 0.4M sulfuric acid needed to neutralize 25cm³ of 0.5M sodium hydroxide, we can use the equation:

M1V1 = M2V2

Where:
M1 = concentration of sulfuric acid = 0.4M
V1 = volume of sulfuric acid
M2 = concentration of sodium hydroxide = 0.5M
V2 = volume of sodium hydroxide = 25cm³

First, we need to calculate the amount of moles of sodium hydroxide present in 25cm³ of 0.5M solution:

Moles of NaOH = Molarity x Volume
Moles of NaOH = 0.5M x 25cm³ = 12.5 mmol

Since sulfuric acid is a diprotic acid, it will require twice the amount of moles of sodium hydroxide to neutralize it. Therefore, the moles of sulfuric acid needed will be:

Moles of H2SO4 = 2 x Moles of NaOH
Moles of H2SO4 = 2 x 12.5 mmol = 25 mmol

Now, we can calculate the volume of 0.4M sulfuric acid needed to provide 25 mmol of sulfuric acid:

V1 = Moles / Molarity
V1 = 25 mmol / 0.4M = 62.5 cm³

Therefore, 62.5 cm³ of 0.4M sulfuric acid is needed to neutralize 25cm³ of 0.5M sodium hydroxide.