Question
In: Chemistry
What is the pH at the equivalence point of the titration of 50.0 mL of 0.100...
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)
Solutions
Expert Solution
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
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