A buffer solution is 1.00 M in NH3 and 1.10 M in NH4Cl. If 0.230 moles of NaOH are added to 1.00 L of the buffer, what is its pH? Assume the volume remains constant. Kb of NH3 = 1.8 multiplied by 10-5.

Answer 1

Use the Henderson-Hasselbalch equation.
NH3 + H^+ ==> NH4^+
initial:
NH3 = 1.00 M from the problem.
H^+ = whatever, but can be calculated if you want to know.
NH4^+ = 1.10 M from the problem.
Just for the fun of it I plugged in 1.00 for B and 1.10 for acid and found pH = 9.21.
For the problem, if we add 0.230 moles NaOH, we must REMOVE 0.230 H^+, which shifts the above equilibrium to the left. NH4^+ is decreased by 0.230 and NH3 is increased by 0.230.
Final moles NH3 = 1.00 + 0.230 = ??
Final moles NH4^+ = 1.10 - 0.230 = ??
Substitute into HH equation and solve. I get something close to 9.4 but check me out.