Calculate the pH after 0.020 mol NaOH is added to 1.00 L of each of the solutions. [Solved]

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HONH2 (Kb = 1.1 10-8) is a weak base.
I would expect that the small amount of its conjugate acid released in its Kb equilibrium would have any sig fig affect on the addition of 0.020 mol NaOH

so since the 0.020 mol NaOH is rather unaffected by the HONH2
its 0.020 moles of OH / 1.00 Litres = 0.020 Molar OH-

pOH = log of [0.020 Molar OH- ]
pOH = 1.70

since pH pOH = 14
the pH should be 12.30

so originally
pOH = pKb = 7.96
so
originally the pH was 6.04