Question

0.02 Moles of HCl were added to a buffer solution, prepared by fluxing 20 L of gaseous NH3 and 10 L of gaseous HCl, both measured in normal conditions, in 2L of water. Calculate the pH of the solution before and after the HCl addition. (Kb NH3 = 1.8 * 10-5).
You have made a buffer by reacting NH3 and HCl. I assume "normal" conditions mean STP.
mols HCl = 10/22.4 = approx 0.45
mols NH3 = 20/22.4 = approx 0.89
You need to redo all of these calculations.

.......NH3 + HCl ==> NH4Cl
I....0.89....0.45.....0
C....-0.45..-0.45.....0.45
E....0.44.....0......0.45

Now you add another 0.02 mols HCl.
I....0.44.....0......0.45
add..........0.02........
C...-0.02...-0.02....0.47
E.....0.42....0.......0.47

Substitute the E line into the HH equation and solve for pH.
Note: You can calculate pKa (needed for the HH equation) from pKa*pKb = pKw = 14.

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