Question

A 40.00 mL aliquot of 0.100 M NH3 is titrated with 20.00 mL of 0.0700 M HCl. Calculate the pH. Kb = 1.76 x 10−5

Answer 1

First,

find each species in solution

mmol of NH3 = Mbase*Vbase = 40*0.1 = 4 mmol of NH3

mmol of HCl = Macid*Vacid = 20*0.07 = 1.4 mmol of HCl

there will be a partial reaction:

HCl + NH3 = NH4+ + Cl-

so

mmol of NH3 left after reaction = 4-1.4 = 2.6 mmol of NH3 left

mmol of NH4+ formed = 0 + 1.4 = 1.4 mmol of NH4+

Note that this is a buffer!

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

the buffer equation is given by

pOH = pKb + log(NH4+/NH3)

pKb = -log(Kb) = -log(1.76*10^-5) = 4.7544

now, substitute data

pOH = pKb + log(NH4+/NH3)

pOH = 4.75+ log(1.4/2.6)

pOH = 4.48115

pH = 14-pOH

pH = 14-4.48115

pH = 9.51885

pH = 9.52 approx