A 10.0-mL solution of 0.30 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCl solution:

a) 0.00 mL
b) 10.0 mL
c) 30.0 mL
d) 40.0 mL

Try the equation, ICE chart, etc just like the NaOH + CH3COOH problem you did previously. Then figure what you have in solution an solve accordingly. Post your work if you get stuck and I'll help you through it.

I don't even know where to start. There is no volume for HCl and am not sure how the equation is suppose to look like. I'm just confused about this problem.

Yes, there is a volume for HCl; in fact, that's what the problem is all about. That is, what is the pH after the addition of each of those quantities of 0.1M HCl.

You have 10.00 mL of 0.3M NH3 initially which is 3. mmole.
After 0 mL, that is 0 mmole HCl added.
After 10 mL x 0.1M HCl = 1 mmol added.
After 30 mL x 0.1M HCl = 3 mmole added.
After 40 mL x 0.1M HCl = 4 mmole added.

...............NH3 + HCl ==> NH4Cl
initial........3.....0.......0
added...............1............
added...............2...........
added...............3...........
added...............4...........
equil..........you can complete this.
So look to see where you are at each addition.
initially (at zero mLHCl) you have 10.0 mL of 0.3M NH3. And that's all except for the water. You know NH3 is a weak base, so you write the ionization, Kb, and solve for the OH^- and from there the pOH and pH.

At 30 mL, you have the equivalence point so all of the NH3 has been titrated with HCl and you have NH4Cl so the pH is determined by the hydrolysis of the NH4Cl salt.
All points between the beginning and the equivalence point is a buffer solution and the pH is determined by the Henderson-Hasselbalch equation.
At all point after the equivalence point, the pH is determined by the amount of excess HCl present.
Just complete the ICE chart for each of the quantities of HCl added and plug into those expressions as outlined above.

To calculate the pH after each addition of the HCl solution, we need to consider the reaction that takes place between NH3 (ammonia) and HCl (hydrochloric acid). Ammonia (NH3) is a weak base while hydrochloric acid (HCl) is a strong acid.

The balanced chemical equation for the reaction between NH3 and HCl is:

NH3 + HCl -> NH4+ + Cl-

Initially, we have a 10.0 mL solution of 0.30 M NH3. This means there are (0.30 mol/L) x (0.0100 L) = 0.0030 moles of NH3 present in the solution.

a) 0.00 mL HCl added:
Since no HCl is added, the reaction does not occur. Therefore, the concentration of NH3 remains the same. To calculate the pH, we need to solve for the pOH first using the equation:

pOH = -log[NH4+]

Since no NH4+ is formed in this case, the pOH is 0.0. And since pH + pOH = 14, the pH will be 14 - 0 = 14.

b) 10.0 mL HCl added:
Since the same volume of HCl is added as the NH3 solution, we need to calculate the resulting concentration of NH3 and NH4+ after the reaction.

The reaction is 1:1 between NH3 and HCl, meaning that for every 1 mole of NH3, 1 mole of NH4+ is formed. Thus, all 0.0030 moles of NH3 will react with 0.0030 moles of HCl.

The volume of the resulting solution will be 10.0 mL + 10.0 mL = 20.0 mL = 0.0200 L.

The new concentration of NH3 can be calculated as follows:
NH3: (0.0030 mol) / (0.0200 L) = 0.150 M

To calculate the concentration of NH4+, we need to consider that NH4+ is the product of the reaction. Since 0.0030 moles of NH3 will react with 0.0030 moles of HCl, the concentration of NH4+ will also be 0.150 M.

Now, to calculate the pH, we need to find the pOH again. Since NH4+ is a weak acid, we can use its concentration to calculate the pOH:

pOH = -log[NH4+]
pOH = -log(0.150)
pOH ≈ 0.823

Since pH + pOH = 14, the pH is 14 - 0.823 = 13.18.

c) 30.0 mL HCl added:
Following the same process as in part (b), we find that the final concentration of NH3 and NH4+ after the reaction is 0.090 M. Using the pOH equation, we find that pOH ≈ 0.959, and thus, the pH is 14 - 0.959 ≈ 13.04.

d) 40.0 mL HCl added:
Following the same process as in parts (b) and (c), we find that the final concentration of NH3 and NH4+ after the reaction is 0.060 M. Using the pOH equation, we find that pOH ≈ 1.221, and thus, the pH is 14 - 1.221 ≈ 12.78.

Therefore, after the additions of 0.00 mL, 10.0 mL, 30.0 mL, and 40.0 mL of HCl solution, the pH is approximately 14, 13.18, 13.04, and 12.78, respectively.