In: Chemistry
What is the pH at the equivalence point of the titration of 50.0 mL of 0.100 M HN4+ with 50.0 ml of 0.100 M NaOH? (For NH3 Kb = 1.8 x 10−5)
millimoles of NH4+ = 50 x 0.1 = 5
millimoles of NaOH = 50 x 0.1 = 5
Ka of NH4+ = 10^-14 / 1.8 x 10^-5
= 5.56 x 10^-10
pKa = 9.26
NH4+ + OH- ------------------> NH3 + H2O
5 5 0 0
0 0 5
here base only remains.
base concentration = 5 / 50 + 50 = 0.05 M
pOH = 1/2 (pKb - log C)
= 1/2 (4.74 - log 0.05)
pOH = 3.02
pH = 10.98