Question

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.

a) 73.4 mL

Answer 1

a) 73.4 mL

total volume = 30.0 ml + 73.4 ml= 103.4 ml or 0.1034 L

a balanced reaction is as follows:

NH3 + HCl = NH4Cl

First calculate the number of mole of NH3 anad HCl as follows:

Number of mole = molarity * volume in L

30.0 mL /1000 * 0.0050 M NH3 = 0.0015 mol NH3

73.4 mL /1000 * 0.025 M HCl = 0.001835 mol HCl

In this reaction; we have 0.0015 mol HCl reacting with 0.0015 mol NH3, which leaves 0.001835 mol HCl - 0.0015 mol HCl = 0.000335 mol HCl.

Now calculate the molarity of HCl as follows:

Molarity = Number of moles / total Volume in L

=0.000335 mol HCl.   /0.1034 L

= 0.00324 M

OR 0.00324 H+

and the pH is:

pH = -log [H+]

pH = -log(0.00324)

= 2.50