Calculate the pH after 0.010 mol gaseous HCl is added to 258.9 mL of each of the following buffered solutions.

Question

0.050 M NH3 / 0.15 M NH4Cl=?
0.50 M NH3 / 1.50 M NH4Cl=?

Answer 1

NH3 + H2O <==> NH4+ + OH-, pKb = 4.75

Case 1:
0.050 M NH3 / 0.15 M NH4Cl
Before adding HCl:
0.2589L*0.050 M=0.0129 mole (NH3)
0.2589L*0.150 M=0.0388 mole (NH4Cl)
After adding 0.010 mol gaseous HCl:
(0.0129mole - 0.01mole)/0.2589L = 0.0114M (NH3)
(0.0388mole + 0.01mole)/0.2589L = 0.189M (NH4Cl)
pKb = 4.75 = -log([NH4+]*[OH-]/[NH3])
pH = 14+log([OH-])= 14 - 4.75 - log([NH4+]/[NH3])
= 9.25 - log(0.189/0.0114) = 8.03

Case 2:
0.50 M NH3 / 1.50 M NH4Cl
With the same method, we have:
NH3: (0.2589*0.50-0.01)/0.2589 = 0.461(M)
NH4Cl: (0.2589*1.50+0.01)/0.2589 = 1.54(M)
pH = 9.25 - log(1.54/0.461) = 8.73