Question

In one of your titrations, 50.0mL of 0.063M KOH was titrated with 0.325M H2SO4. Calculate the number of moles of KOH remaining after 2.50mL of H2SO4 was added. For the same titration, calculate the number of moles of H2SO4 remaining after 2.50mL of H2SO4 was added. For the same titration, calculate the pOH. For the same titration, calculate the pH.

Answer 1

We have,

0.063 M KOH = 0.063 N KOH

0.325 M H₂SO₄ = (0.325 x 2) N H₂SO₄ = 0.65 N H₂SO₄

(Note: For acids Normality = Molarity x Basicity. where basicity is the number of H⁺ ions an acid can give. For bases. Normality = Molarity x Acidity where acidity is the number of OH^- ions a base can give)

Here, volume of KOH = V₁ = 50 mL

Normality of KOH = N₁ =0.063 N

Therefore, N₁ x V₁ = (50 x 0.063) = 3.15 milliequivalent

Volume of H₂SO₄ =V₂= 2.5 mL

Normality of H₂SO₄ = N₂ = 0.65 N

Therefore, N₂V₂ = 1.625 milliequivalent

Moles of KOH left = (3.15 -1.625) x 10^-3 = 1.525 x 10^-3 moles

Since, H₂SO₄ is limiting reagent, therewill be no H₂SO₄ left after the reaction

Therefore, moles of H₂SO₄ left = 0

We have,

[(N₁ x V₁) - (N₂ x V₂)] = N_R x V_R

N_R= [(N₁ x V₁) - (N₂ x V₂)] / V_R

=1.525 / 52.5 = 0.029

[OH-] = 0.029

Therefore, p^OH = -log[OH^-] = -log 0.029 = 1.537

pOH = 1.537

pH= 14 - pOH = 14 -1.537 = 12.47

pH = 12.47