4g NaOH is dissolved in 100mL of 2.5M H2SO4 will be required for its complete neutralization?
I don't see a question here.
20ml
To determine the amount of 2.5M H2SO4 required to completely neutralize 4g of NaOH, we need to calculate the number of moles of both substances involved.
First, let's calculate the number of moles of NaOH:
Molar mass of NaOH (Na = 22.99 g/mol, O = 16.00 g/mol, H = 1.01 g/mol):
Na = 22.99 g/mol, O = 16.00 g/mol, H = 1.01 g/mol
NaOH = (1 x Na) + (1 x O) + (1 x H) = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Number of moles of NaOH = mass / molar mass = 4g / 40.00 g/mol = 0.1 mol
Now, since H2SO4 is a diprotic acid, we need to consider its reaction stoichiometry. The balanced chemical equation for the neutralization of NaOH and H2SO4 is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
From the equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4. Therefore, the number of moles of H2SO4 required is twice the number of moles of NaOH.
Number of moles of H2SO4 = 2 x 0.1 mol = 0.2 mol
Now, let's calculate the volume of the 2.5M H2SO4 required using the given concentration and the number of moles:
Molarity (M) = moles / volume (L)
Rearranging the equation, we have:
Volume (L) = moles / Molarity
Volume (L) = 0.2 mol / 2.5 mol/L = 0.08 L = 80 mL
Therefore, 80 mL of 2.5M H2SO4 will be required to completely neutralize 4g of NaOH.