Question

Calculate the pH of 100.0 mL of a buffer that is 0.050 M NH4Cl and 0.180 M NH3 before and after the addition of 1.00 mL of 5.70 M HNO3.

Before =

After=

Answer 1

Answer- We are given, volume = 100.0 mL , [NH₄Cl] = [NH₄⁺] = 0.050 M

[NH₃] = 0.180 M , [HNO₃] = 5.70 M , volume = 1.00 mL

We know Kb for the NH₃ = 1.8*10^-5

pH before addition -

We are given the base and its conjugate acid, so we need to use the Henderson Hasselbalch equation,

We know Henderson Hasselbalch equation,

pOH = Pkb + log [NH₄⁺] / [NH₃]

Now we need to calculate the pKb from the given Kb

We know, pKb = - log Kb

                         = -log 1.8*10^-5

                         = 4.74

Now plugging the values in the Henderson Hasselbalch equation

pOH = 4.74 + log 0.050 M / 0.180 M

         = 4.19

We know form,

pOH + pH = 14

pH = 14 – pOH

      = 14 -4.19

      = 9.81

pH after addition –

We need to calculate moles of each before addition of HNO₃

Moles of NH₄⁺ = 0.050 M * 0.100 L = 0.0050 moles

Moles of NH_­3 = 0.180 M * 0.100 L = 0.0180 moles

Moles of HNO₃ = 5.7 M *0.001 L = 0.0057 moles

We know when we added the strong acid then there is moles of conjugate acid increase and moles of base decrease , so moles after the addition

Moles of NH₄⁺ = 0.0050 moles + 0.0057 moles = 0.0107 moles

Moles of NH_­3 = 0.0180 moles -0.0057 moles =0.0123 moles

Total volume = 100+1 = 101

New molarity

[NH₄⁺] = 0.0107 moles / 0.101 L

            = 0.106 M

[NH₃] = 0.0123 moles / 0.101 L

           = 0.122 M

Now plugging the values in the Henderson Hasselbalch equation

pOH = 4.74 + log 0.106 M / 0.122 M

         = 4.68

We know form,

pOH + pH = 14

pH = 14 – pOH

      = 14 -4.68

      = 9.32