Buffer solution

Question

Calculate the pH of a buffer that is 0.02M NH3 and 0.03M NH4CI. What is the pH after addition 1ml of 0.1M NaOH to 100 mL of this buffer solution?

Answer 1

Henderson Hasselbalch equation for basic buffer:

pOH = pKb + log [conjugate acid] / [base]

pOH = 4.72 + log (0.03 M / 0.02 M) ... looked up pKb for NH3 and found it to be 4.72

pOH = 4.72 + 0.18

pOH = 4.90

pH = 14 - 4.90

pH = 9.10

Adding 1 ml to 100 ml of buffer, we can ignore the change in volume (101 ml vs 100 ml is 1% change)

moles OH- added = 1 ml x 1 L / 1000 ml x 0.1 mol / L = 0.001 moles OH-

This OH- will react with the NH4+ to produce NH3 and H2O, thus reducing moles NH4+ and increasing mols NH3

Final concentration NH4+ = 0.03 - 0.001 = 0.029 M

Final concentration NH3 = 0.02 + 0.001 = 0.021 M

pOH = 4.72 + log (0.029 / 0.021)

pOH = 4.72 + 0.14

pOH = 4.86

pH = 14 - 4.86

pH = 9.14