Find/Determine pH

Question

Determine the pH at the point in the titration of 40 mL of 0.200 M H2NNH2 with 0.100 M HNO3 after 80.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0x10^-6

Answer 1

H₂NNH₂ + HNO₃ ==> H₂NNH₃⁺ + NO₃^-

Initial moles H₂NNH₂ = 40 ml x 1 L / 1000 ml x 0.200 mol /L = 0.008 moles H₂NNH₂

moles HNO3 added = 80 ml x 1 L / 1000 ml x 0.100 mol/L = 0.008 moles HNO₃

moles H₂NNH₃⁺ formed = 0.008 moles H₂NNH₃⁺

Final volume = 40 ml + 80 ml = 120 ml = 0.120 L

Final [H₂NNH₃⁺] = 0.008 mols / 0.120 L = 0.0667 M

H₂NNH₃⁺ + H2O ==> H₃O⁺ + H₂NNH₂

Ka = [H3O+][H₂NNH₂] / [H₂NNH₃⁺]

Ka = 1x10^-14 / 3.0x10^-6 = 3.33x10^-9

3.33x10^-9 = (x)(x) / 0.0667

x² = 2.22x10^-9

x = 4.72x10^-5 M = [H₃O⁺]

pH = -log 4.72x10^-5

pH = 4.33