Question

Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq)

Change in pH = _____

Calculate the change in pH when 5.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Change in pH= _______

Answer 1

NH3 + NH4Cl is a basic buffer

pKb for NH3 = 4.75

We know that pH for a basic buffer is given by,

pH = pKb + log (Salt / Base)

Salt = NH4Cl

Base = NH3

Millimoles of NH4Cl = 0.1 x 100 = 10

Millimoles of NH3 = 0.1 x 100 = 10

=> pOH = 4.75 + log (10 / 10) = 4.75

=> pH = 14 - 4.75 = 9.25

pOH after adding HCl

Millimoles of HCl added = 0.1 x 5 = 0.5

New pOH = 4.75 + log (10 + 0.5 / 10 - 0.5) = 4.793

=> pH = 14 - 4.793 = 9.2065

=> Change in pH = 9.2065 - 9.25 = - 0.0435

Millimoles of NaOH added = 0.1 x 5 = 0.5

pOH after adding NaOH = 4.75 + log (10 - 0.5 / 10 + 0.5) = 4.7065

=> pH = 9.2935

=> Change in pH = 9.2935 - 9.25 = 0.0435