Thanks for the reminder. I don't think you have approached the problem right (at least not completely right). I'll redo some of what you've described. You have the mmols right.
.......NH3 + HCl ==> NH4Cl
I......1.5....0.......0
add...........4............
C.....-1.5...-1.5.....+1.5
E......0......2.5......1.5
To that point you're ok. The remainder is where you erred.
At the equilibrium point you have no free NH3 left and you have a solution of NH4Cl in HCl. And yes, M HCl = 2.5/50 = 0.05
So the pH is determined by the HCl and not by the Henderson-Hasselbalch equation. You have no NH3 present so you can't have a buffer. The NH4Cl will hydrolyze, of course, but the H3O^+ produced by that hydrolysis will be miniscule compared to the strong acid, HCl, present, so I would ignore any hydrolysis of the NH4Cl and calculate the pH of the solution based on pH = -log(HCl). I would think 1.3 for pH (or whatever that 15 digit program gives you for that log value.
Hey Bob,
What is the pH at the point during a titration
when 20 mL of 0.200 M HCl has been added
to 30 mL of 0.0500 M NH3 solution? HCl is
strong acid, Kb = 1.8 × 10−5
for NH3.
Answer in units of pH.
I've gone about getting the mmol of HCl and NH3, getting 4mmol and 1.5 mmol respectively, then subtracted the 1.5 from 4, getting 2.5 mmol for the HCl then dividing that by the total number of milliters (50) to get the molarity of HCL, then I used that 1.5 mmol in the other side of the dissociation equation, using the Hendrickson Hassebach equation to find the pH, which I got about 9 pH. But my answer is wrong. Please remember my system is UT Quest and significant digits are not a factor in the answers.
2 answers
1.301029996, thanks Bob
2 answers