Question

Consider the titration of 30.0 mL of 0.050 M NH3with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.

a) 72.3 ml

Answer 1

we have:

Molarity of HCl = 0.025 M

Volume of HCl = 72.3 mL

Molarity of NH3 = 0.05 M

Volume of NH3 = 30 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.025 M * 72.3 mL = 1.8075 mmol

mol of NH3 = Molarity of NH3 * Volume of NH3

mol of NH3 = 0.05 M * 30 mL = 1.5 mmol

We have:

mol of HCl = 1.8075 mmol

mol of NH3 = 1.5 mmol

1.5 mmol of both will react

excess HCl remaining = 0.3075 mmol

Volume of Solution = 72.3 + 30 = 102.3 mL

[H+] = 0.3075 mmol/102.3 mL = 0.003 M

we have below equation to be used:

pH = -log [H+]

= -log (3.006*10^-3)

= 2.52

Answer: 2.52