Calculating the volume of H2SO4 required to neutralize NaOH

To calculate the volume of 5 Normal H2SO4 required to neutralize a solution containing 2.50 g of NaOH, we need to follow these steps:

1. Write the balanced chemical equation for the reaction between H2SO4 and NaOH:

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

2. Calculate the number of moles of NaOH in the solution:

molar mass of NaOH = 23 g/mol + 16 g/mol + 1 g/mol = 40 g/mol
number of moles of NaOH = mass of NaOH / molar mass of NaOH
number of moles of NaOH = 2.50 g / 40 g/mol
number of moles of NaOH = 0.0625 mol

3. Determine the stoichiometric ratio between H2SO4 and NaOH from the balanced chemical equation:

1 mole of H2SO4 reacts with 2 moles of NaOH

4. Calculate the number of moles of H2SO4 required to neutralize the NaOH:

number of moles of H2SO4 = 0.0625 mol NaOH / 2 mol H2SO4
number of moles of H2SO4 = 0.03125 mol H2SO4

5. Convert the number of moles of H2SO4 to volume of 5 Normal H2SO4:

5 Normal H2SO4 means 5 moles of H2SO4 per liter of solution
number of liters of 5 Normal H2SO4 = number of moles of H2SO4 / concentration of H2SO4
number of liters of 5 Normal H2SO4 = 0.03125 mol H2SO4 / 5 mol/L
number of liters of 5 Normal H2SO4 = 0.00625 L or 6.25 mL

Therefore, 6.25 mL of 5 Normal H2SO4 is required to neutralize a solution containing 2.50 g of NaOH.

Normality of H2SO4 = 5N
m NaOH =2.5gm
M NaOH = 40gm/mol
n = m/M =(2.5gm)/(40gm/mol)=0.0625mol
N1V1 = N2V2
5N(V1) = zM2V2
5N(V1) = (1)(0.0625mol)
V1=0.0625mol/5N=0.0125L=12.5ml