A 100.0 mL sample of 0.05 M NH3 is titrated with 0.10 M HCl. Determine the pH of the solution after the addition of 50.0 mL HCl.

Here is my work:
NH3 + H3O+ --> NH4+ + H2O

Before Addition:
NH3= 0.005 mol
H3O= 0 mol
NH4+ = 0 mol

Addition:
NH3 = 0.005 mol
H3O+ = 0.005 mol
NH4+ = 0.005 mol

After Addition:
NH3 = 0 mol
H3O+ = 0 mol
NH4+ = 0.005

ICE Table
NH4+ + H2O --> NH3 + H3O+

INITIAL:
NH3= 0.0333 M (0.005 mol/(0.100 L + 0.0500 L)
NH3 = 0 M
H3O+ = 0 M

CHANGE (respectively):
-x
+x
+x

EQUILIBRIUM (respectively):
0.0333 - x
x
x

Ka expression:

(x^2)/(0.0333-x) = (1 x 10^-14)/(1.8 x 10^-5)

assume x is small approximation...
x= 4.30 x 10^-6

pH= -log (4.30 x 10^-6) = 5.37

FINAL ANSWER: pH = 5.37

Please check.

I didn't go through your work step by step but I worked the problem from scratch. Using 1.8E-5 for Kb, I obtained the same answer you did. Good work.

Your work is mostly correct, but there are a couple of mistakes. Let me help you go through it step by step.

First, let's calculate the moles of NH3 initially present in the solution:

moles of NH3 = 0.05 M * 0.1 L = 0.005 mol

So, you got this part correct.

Next, let's calculate the moles of HCl added:

moles of HCl = 0.10 M * 0.05 L = 0.005 mol

Again, correct.

Now let's determine the limiting reactant. Since the stoichiometric ratio between NH3 and HCl is 1:1, and the moles of NH3 and HCl added are equal, neither is in excess. Therefore, all the NH3 will react with the added HCl.

Now let's calculate the concentration of NH4+ after the reaction:

volume of the solution = 0.1 L + 0.05 L = 0.15 L

concentration of NH4+ = moles of NH4+ / volume of solution
= 0.005 mol / 0.15 L
= 0.0333 M

Your calculation for the concentration of NH4+ after the reaction is correct.

Now, let's use the Henderson-Hasselbalch equation to determine the pH of the solution after the addition of HCl:

pH = pKa + log([A-]/[HA])

In this case, the NH4+ acts as the acid (HA) and water acts as the base (A-). The pKa of NH4+ is 9.24 (at 25 degrees Celsius).

pH = 9.24 + log([NH4+]/[NH3])

pH = 9.24 + log(0.0333/0.0333)
= 9.24 + log(1)
= 9.24 + 0
= 9.24

Therefore, the pH of the solution after the addition of 50.0 mL of HCl is 9.24.

There is an error in your final answer. The pH should be 9.24, not 5.37. Please double-check your calculations and equations to find the mistake.