fill in the ice tables

Question

a student was titrating a solution of hydrazine, H2NNH2, with a strong acid solution. determine the pH at a particular point in the titration. do this by constructing a BCA table, constructing an ice table, writing an equilibrium constant expression, and use this information to determine pH. 40 mL of a .200 M H2NNH2 solution was titrated with a 80 mL of a .100 mHNO3 ( a strong acid) fill in the ice table with the appropraite value for each involved species to determine the moles of reactant and product after the reacting of the acid and base

Answer 1

H₂NNH₂ + HNO₃ ==> H₂NNH₃⁺ + NO₃^-

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

Initial mols of HNO₃ = 80 ml x 1 L / 1000 ml x 0.100 mol / L = 0.008 mols

Setting up an ICE table, we have (all values are in moles)....

H₂NNH₂ + HNO₃ ==> H₂NNH₃⁺ + NO₃^-

0.008........0.008..............0..............0...........Initial

-0.008......-0.008...........+0.008.....+0.008....Change

0................0..................0.008.......0.008......Equilibrium

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

Final concentrations @ equilibrium:

[H₂NNH₃⁺] = [NO3-] = 0.008 mol / 0.120 L = 0.0667 M

To find the pH, we need to look at hydrolysis of H₂NNH₃⁺. We can predict an acidic pH because the salt is formed from a weak base and a strong acid, so pH should be <7.

H₂NNH₃⁺ + H2O ==> H₃O⁺ + H₂NNH₂

Next, we need to know the Ka for H₂NNH₃⁺. A tabular value was found to be 7.94x10^-9

Ka = [H₃O⁺] [H₂NNH₂] / [H₂NNH₃⁺]

7.94x10^-9 = (x)(x) / 0.0667-x (assume x is small relative to 0.0667 and ignore it in the denominator)

x² = 5.30x10^-10

x = [H₃O⁺] = 2.30x10^-5 M (note: this is insignificant compared to 0.0667 so above assumption was valid)

pH = -log [H₃O⁺] = 4.64