Find the pH of a mixture of 25 cm3 0.100 mol dm-3 sulfuric acid and 80 cm3 of 0.200 mol dm-3 sodium hydroxide.

Answer 1

H2SO4 + 2 NaOH ==> Na2SO4 + 2 H2O ... balanced equation

moles H2SO4 present = 25 cm3 x 1L/1000 cm3 x 0.100 mol/L = 0.0025 moles H2SO4

moles NaOH present = 80 cm3 x 1L/1000 cm3 x 0.200 mol/L = 0.016 moles NaOH

Which is present in excess? NaOH is because it takes 2 NaOH per 1 H2SO4

NaOH used up: 0.0025 mol H2SO4 x 2 mol NaOH/mol H2SO4 = 0.005 moles NaOH used up.

NaOH left over after reaction: 0.016 moles NaOH - 0.005 moles NaOH = 0.011 moles left over.

Total volume after reaction = 25 cm3 + 80 cm3 = 105 cm3 = 0.105 L

Final [NaOH] = 0.011 moles/0.105 L = 0.1048 M

pOH = -log [OH-] = 0.9798

pH = 14 - pOH

pH = 13.0