For the titration of 50.00 mL of 0.106 M NH3 with 0.225 M HCl, calculate

(A) the initial pH and when the pH neutralization is
(B) 25.0% complete
(C) 50.0% complete
(D) 100.0% complete

You need to be able to recognize where you are on the titration curve with these problems.

50.00 x 0.106 = millimoles NH3
x mL x 0.225M HCl = ? mmols HCl

a. At the beginning you have a solution of NH3 and nothing else. So you know how to solve for the pH of a 0.106 M NH3 soln. Set up the ICE chart and go from there.
d. At the equivalence point you have no acid and no base; the pH is determined by the hydrolysis of the salt. The salt is NH4Cl and it is the NH4^+ that is hydrolyzed. You will need to determine the (NH4Cl). You do that by mmoles NH3 at the start and divide by mL base(at start)+mL acid to get to the eq point). I'll call the (NH4^+) = 0.2 but that will not be right.
.......NH4^+ + H2O ==> H3O^+ + NH3
initial.0.2............0........0
change..-x.............x.......x
equil...0.2-x...........x......x

Then Ka for NH4^+ = (Kw/Kb for NH3) = (H3O^+)(NH3)/(NH4^+)
Substitute from the ICE chart and solve for x and convert that to pH.

b and c.
For all points between the beginning and the equivalence point, use the Henderson-Hasselbalch equation.

(A) Well, if you combine ammonia (NH3) with hydrochloric acid (HCl), you get a bunch of chemists holding their noses! But let's get to the calculation. The initial pH can be found by considering the concentration of OH- ions produced by the reaction. Since NH3 is a weak base, it will react with the HCl to form NH4+ and Cl- ions. However, NH4+ is a weak acid too, so it will partially dissociate to form NH3 and H+. This means that the H+ ions from the HCl and the NH4+ will determine the acidity of the solution. I'm not gonna lie, things can get a bit complicated here, but you can use the Henderson-Hasselbalch equation to calculate the pH. I'll leave the actual calculation to you, but just remember to put on your chemist hat and handle those equations carefully!

(B) When the pH neutralization is 25% complete, you may start to hear some groans from the chemists because they're halfway between being happy and neutralized. To figure out the pH at this point, you need to consider the amount of acid and base that have reacted. You can use the mole ratio of HCl to NH3, along with the initial concentrations and volumes, to determine how much of each is consumed. Then, you can use the Henderson-Hasselbalch equation once again to calculate the pH. Remember, laughter is the best pH-neutralizer, but in this case, equations are your best friends!

(C) Now we're getting halfway to the finish line, just like a marathon runner who stops halfway to eat a taco. When the neutralization reaction is 50% complete, you'll find that the amount of HCl consumed is equal to the amount of NH3 titrated. Once again, you can use the mole ratio and the initial concentrations and volumes to determine the consumed amounts. Put those values into the Henderson-Hasselbalch equation, and voila! You'll have your pH. Just remember, if you're feeling 50% neutralized, a good joke can always perk you up!

(D) Congratulations, you did it! One hundred percent complete, just like an embarrassing dance move at a high school prom. At this stage, all the HCl and NH3 have reacted, so you're left with only the resulting products. Without getting into too much detail, you'll find that the solution is now acidic because you have excess H+ ions leftover. To calculate the pH, you can use the concentration of the remaining H+ ions and do one final round with the Henderson-Hasselbalch equation. It's like the grand finale of a fireworks show, where the final burst of chemistry happens. And now, your pH awaits!

To calculate the initial pH, we need to consider that NH3 is a weak base and HCl is a strong acid. So, initially, we have a solution of NH4Cl formed by the reaction of NH3 and HCl.

(A) Initial pH:
First, we need to calculate the moles of NH3 and HCl:
Moles of NH3 = Volume (L) × Concentration (M) = 0.05000 L × 0.106 M = 0.00530 moles
Moles of HCl = Volume (L) × Concentration (M) = 0.05000 L × 0.225 M = 0.01125 moles

NH3 acts as a base and reacts with HCl to form NH4+ and Cl- ions. Since NH3 is a weak base, we can assume that it doesn't fully ionize.

NH3 + HCl → NH4+ + Cl-

Since the moles of HCl are greater than the moles of NH3, the reaction is limited by NH3. This means that all the NH3 will react, and we will be left with excess HCl.

The moles of NH4+ formed will be the moles of NH3 reacted, so we have:
Moles of NH4+ = 0.00530 moles

To calculate the concentration of NH4+, we divide the moles by the volume of the solution at the start of the reaction (50.00 mL = 0.05000 L):
Concentration of NH4+ = Moles / Volume = 0.00530 moles / 0.05000 L = 0.106 M

To calculate the initial pH, we consider that NH4+ is the conjugate acid of NH3 with the Kb value of NH3.
Kb = [NH4+][OH-] / [NH3]

Since we have a solution of NH4Cl, which will dissociate as NH4+ and Cl-, and the concentration of Cl- ions is the same as NH4+ ions, we can ignore the Cl- contribution to the Kb expression.

Therefore, we can write:
Kb = [NH4+][OH-] / [NH3]
Kb = [NH4+][OH-] / (Concentration of NH4+)

Since the reaction is in the basic range and the [OH-] concentration is not provided, we need to solve for OH- and convert it to pOH, then to pH.

Using this equation, we can set up the Kb expression:
Kb = [OH-] × [NH4+] / (Concentration of NH4+)

We know that [NH4+] = 0.106 M, and the Kb for NH3 is 1.8 × 10^-5.

Solving for [OH-]:
1.8 × 10^-5 = [OH-] × 0.106 / 0.106
[OH-] = 1.8 × 10^-5 M

To find pOH:
pOH = -log[OH-] = -log(1.8 × 10^-5) = 4.74

Finally, we can calculate the initial pH:
pH = 14 - pOH = 14 - 4.74 = 9.26

So, the initial pH is 9.26.

(B) pH at 25.0% complete:
To calculate the pH at 25.0% completion, we need to determine the moles of NH3 and HCl that have reacted.

25.0% of NH3 has reacted, so the change in moles of NH3 is:
Change in moles of NH3 = 0.25 × moles of NH3 = 0.25 × 0.00530 moles = 0.00133 moles

Since the reaction between NH3 and HCl is 1:1, the moles of NH4+ formed will also be 0.00133 moles.

To calculate the concentration of NH4+, we divide the moles by the total volume of the solution at 25.0% completion. Since the volume of the solution doesn't change during the neutralization, the total volume remains 50.00 mL = 0.05000 L:
Concentration of NH4+ = Moles / Volume = 0.00133 moles / 0.05000 L = 0.0266 M

We can use the same Kb expression as before to calculate the pOH and pH.

Using Kb = [OH-] × [NH4+] / (Concentration of NH4+), and solving for [OH-]:
1.8 × 10^-5 = [OH-] × 0.0266 / 0.0266
[OH-] = 1.8 × 10^-5 M

To find pOH:
pOH = -log[OH-] = -log(1.8 × 10^-5) = 4.74

Finally, we can calculate the pH at 25.0% complete:
pH = 14 - pOH = 14 - 4.74 = 9.26

So, the pH at 25.0% completion is 9.26.

(C) pH at 50.0% complete:
We can follow the same procedure as in part (B). Since the reaction is 1:1, 50.0% completion means that 50.0% of the moles of NH3 have reacted.

Change in moles of NH3 = 0.5 × moles of NH3 = 0.5 × 0.00530 moles = 0.00265 moles

Concentration of NH4+ = 0.00265 moles / 0.05000 L = 0.053 M

Using the Kb expression, we find:
1.8 × 10^-5 = [OH-] × 0.053 / 0.053
[OH-] = 1.8 × 10^-5 M

pOH = -log[OH-] = -log(1.8 × 10^-5) = 4.74

pH = 14 - pOH = 14 - 4.74 = 9.26

So, the pH at 50.0% completion is also 9.26.

(D) pH at 100.0% complete:
When the neutralization reaction is 100.0% complete, all of the NH3 will have reacted with HCl. In this scenario, we can consider the resulting solution to be only NH4Cl, a salt made from the reaction of NH3 and HCl.

NH4Cl dissociates completely in water, so we have NH4+ and Cl- ions in solution.

Since NH4Cl is a neutral salt, it will not affect the pH of the solution. The pH will be determined by the auto-ionization of water, which is pH = 7.

So, the pH at 100.0% completion is 7.

To solve this problem, we need to use the principles of acid-base titration and the concept of stoichiometry. Here is how we can calculate the initial pH and the pH at different stages of the titration:

(A) The initial pH:
To find the initial pH of the solution, we need to determine the concentration of hydroxide ions (OH-) from the ammonia (NH3) solution. For the dissociation of ammonia in water, we have the following equilibrium reaction:
NH3 + H2O ⇌ NH4+ + OH-

Since ammonia is a weak base, we can assume that only a small fraction of it will dissociate. Therefore, we can use the initial concentration of NH3 (0.106 M) as the concentration of OH-.

Next, we can use the pOH formula to find the pOH, and then pH by using the relation pH + pOH = 14. The formula for pOH is:
pOH = -log[OH-]

Substituting [OH-] = 0.106 M into the equation, we get:
pOH = -log(0.106)

Finally, substitute the value of pOH into the pH equation:
pH = 14 - pOH

(B) 25.0% completion:
At 25.0% completion, we assume that 25.0% of the NH3 has reacted with HCl. Since HCl is a strong acid, it will fully dissociate in water, providing H+ ions. The reaction between NH3 and HCl can be written as follows:
NH3 + HCl → NH4+ + Cl-

Since NH3 and HCl have a 1:1 stoichiometric ratio, the amount of HCl reacted would be 25.0% of 0.106 M NH3.

(C) 50.0% completion:
At 50.0% completion, we assume that 50.0% of the NH3 has reacted with HCl. Following the same logic as in part (B), we can find the amount of HCl reacted.

(D) 100.0% completion:
At 100.0% completion, all the NH3 has reacted with HCl. This means that the NH3 has been neutralized by HCl, resulting in a complete transfer from a basic to an acidic solution.

By following these steps, you should be able to calculate the initial pH and the pH at different stages of the titration.