Chemistry Pros..help needed

Question

Kb for NH3 is 1.8 x 10-5 at 25oC. Calculate the pH of a buffer solution made by mixing 65.7 mL of 0.103 M NH3 with 41.4 mL of 0.183 M NH4Cl at 25oC. Assume that the volumes of the solutions are additive.

Answer 1

We need to calculate moles of NH3 and NH4Cl.

moles NH3 = M NH3 x L NH3 = (0.103)(0.0657) = 0.00677 moles NH3

moles NH4+ = M NH4+ x L NH4+= (0.183)(0.0414) = 0.00758

Ka NH4+ = Kw / Kb NH3 = (1 x 10^-14) / (1.8 x 10^-5) = 5.6 x 10^-10

pH = pKa + log ([NH3] / [NH4+])

Since the NH3 and NH4+ are in the same volume, this simplifies to

pH = pKa + log (moles NH3 / moles NH4+)

= -log (5.6 x 10^-10) + log (0.00677 / 0.00758) = 9.25 - 0.05 = 9.20

Answer 2

This is a direct application of the Hendersen/Haselbach relationship

pH = pKb + log base /acid

pH =  4.74  + log[ NH3]/[NH4+] = 4.74 +{ mmoles NH3/mmolesNH4+}

mmoles of NH3   = mL x M = 6.767

mmoles of NH4+ = mL x M = 7.576

pH = 4.74 + log 6.767/7.576 = 4.74 - 0.05 = 4.69