Question

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 K_b = 1.8 x 10⁻⁵)

Answer 1

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