Question

Using only 0.200 mol / L NH4OH, 0.400 mol / L HCl solutions and a pH meter, propose a procedure for the preparation of 250 mL NH3 / NH4 + buffer solution of pH = 9.0 where the sum of NH3 + NH4+ concentrations = 0.10 mol / l.

Step by step and to show the procedure in details

Answer 1

According to the Henderson-Hasselbulch equation:

pH = pKa + Log[NH3/NH4+]

Here, [NH3] + [NH4+] = 0.1 mol/L * (250/1000) L = 0.025 mol, i.e. [NH3] = 0.025 - [NH4+]

The pKa of NH3 = 9.26

i.e. 9 = 9.26 + Log[(0.025-NH4+)/NH4+]

i.e. Log[(0.025-NH4+)/NH4+] = -0.26

i.e. (0.025-NH4+)/NH4+ = 0.5495

i.e. [NH4+] = 0.016 mol

Therefore, the volume of given HCl solution needed for prepartion of buffer = 0.016 mol/(0.4 mol/L) = 0.040 L or 40 mL

And [NH3] = 0.025 - 0.016 = 0.009 mol

Therefore, volume of given NH4OH solution needed for preparation of buffer = 0.025 mol/(0.2 mol/L) = 0.125 L =125 mL