Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 100.0 mL of the strong acid has been added. The value of Kb for H₂NNH₂ is 3.0 × 10⁻⁶.

Question

In addition to finding the pH, how would you fill a before, change, and after table?

H2NNH2(aq) + H+(aq) --> H2NNH3+(aq)

Before (mol)

Change (mol)

After (mol)

Answer 1

The Before Change and After table is often referred to as an ICE table (Initial, Change and Equilibrium).

Here is the reaction taking place...

H₂NNH₂ + HNO₃ ==> H₂NNH₃⁺ + NO₃^-

moles H2NNH2 initially present = 40.0 ml x 1 L/1000 ml x 0.200 mol/L = 0.008 mol H2NNH2

moles HNO3 added = 100 ml x 1 L/1000 ml x 0.100 mol/L = 0.01 mol HNO3

Using an ICE table (BCA table) we have...

H₂NNH₂ + HNO₃ ==> H₂NNH₃⁺ + NO₃^-

0.200........0.01...................0...................Initial

0.2-0.01.....-0.01.........+0.01................Change

0.19...........0..................0.01..............Equilibrium

Final volume = 40 ml + 100 ml = 140 ml = 0.140 L

Final [H2NNH2] = 0.19 mol/0.140 L = 1.357 M

Final [H2NNH3+] = 0.01 mol/0.140 L = 0.0714 M

Using the Henderson Hasselbalch equation we have pOH = pKb + log [salt]/[base]

pOH = 5.52 + log (0.0714/1.357)

Solve for pOH and subtract from 14 to get pH

pOH = 4.24

pH = 9.76