Question

A solution of sulfuric acid is produced by placing 0.500 L of water in contact with 0.100 atm of H2SO4 gas. The solution is then used to titrate NaOH. What volume (L) of the H2SO4 solution will be required to neutralize 100.0 mL of 0.500 M NaOH? Use a Henrey's Law constant for H2SO4 of 35.3 mol/L-atm

Answer 1

According to Henrey's Law H = c/p

Where

c = concentration = ?

p = atmosphere = 0.100 atm

H = Henrey's Law constant for H₂SO₄ = 35.3 mol/L-atm

Plug the values we get c = H x p

                                    = 35.3(mol/L-atm) x 0.100 atm

                                    = 3.53 mol/L

                                    = (3.53/0.5) mol/(0.500L)

                                    = 7.06 (mol/0.500L)

The number of moles of NaOH is , n = Molarity x volume in L

                                                     = 0.500 M x 0.100 L

                                                     = 0.05 mol

The reaction between NaOH & H2SO4 is

2NaOH + H₂SO₄ Na₂SO₄ + 2H₂O

According to the balanced equation ,

2 moles of NaOH requires 1 mole of H₂SO₄

0.05 mol of NaOH requires M mole of H₂SO₄

M = (0.05x1)/2

    = 0.025 moles

So the volume of H₂SO₄ is , V = number of moles / molarity

                                              = 0.025 mol / 7.06 M

                                             = 3.54x10^-3 L

Therefore the required answer is 3.54x10^-3 L