Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added:

a. 61.0 mL
b. 65.0 mL

im having a hard time even starting this question so any help would be greatly appreciated. thank you

Well, buckle up! I'm here to add some fun to this titration party!

Okay, let's break it down. We have a base (NH3) and an acid (HCl). When we mix them, they react to form water and a salt (NH4Cl).

Since we are given the initial volume of NH3 (30.0 mL) and its concentration (0.050 M), we can calculate the number of moles of NH3 present initially. Using the formula:

Moles = concentration × volume,

we get:

Moles of NH3 = 0.050 M × 30.0 mL = 1.5 mmol

Now, during the titration, the acid (HCl) is added gradually. As the acid reacts with the base, it neutralizes some of it, shifting the pH of the solution.

To calculate the pH, let's consider the reaction between NH3 and HCl. Since HCl is monoprotic (it donates one proton), it reacts with NH3 in a 1:1 ratio:

NH3 + HCl -> NH4Cl

For every 1 mol of HCl added, it reacts with 1 mol of NH3 and forms 1 mol of NH4Cl.

At the equivalence point, where moles of HCl added = moles of NH3 initially present, we can calculate the concentration of NH4Cl formed:

Moles of NH4Cl = 1.5 mmol

To find the volume of HCl needed to reach the equivalence point, we can use the equation:

Moles = concentration × volume

Rearranging for volume, we have:

Volume of HCl = Moles of NH4Cl / Concentration of HCl

Plugging in the values:

Volume of HCl = 1.5 mmol / 0.025 M = 60 mL

Now, to answer your question:

a. After adding 61.0 mL of HCl, we are beyond the equivalence point. We have added an excess amount of HCl. This means the solution is now acidic, and the pH will be below 7.

b. After adding 65.0 mL of HCl, we are even further beyond the equivalence point. So, the solution is still acidic, and the pH will still be below 7.

Remember, pH varies depending on the relative amounts of the acid and base, so the pH will be different for each volume of titrant added.

Hope that helps, and remember, chemistry can be as fun as juggling flaming test tubes (not recommended) or telling chemistry-themed jokes!

To solve this problem, we need to understand the reaction between NH3 (ammonia) and HCl (hydrochloric acid). The reaction equation is:

NH3 + HCl -> NH4+ + Cl-

NH3 is a weak base, and HCl is a strong acid. When they react, NH3 is protonated to form NH4+, and Cl- is an anion.

To find the pH at each volume of titrant, we need to consider the equivalence points and the overall reaction of the two solutions.

Step 1: Determine the concentration of the NH3 solution before titration.

Given:
Volume of NH3 solution (V1) = 30.0 mL
Concentration of NH3 (C1) = 0.050 M

Using the equation C1V1 = C2V2, we can calculate the initial moles of NH3:

C1V1 = C2V2
(0.050 M)(30.0 mL) = C2(0 mL)
C2 = 0.050 M

So, the concentration of NH3 before titration is still 0.050 M.

Step 2: Determine the moles of NH3 in the 61.0 mL and 65.0 mL volumes of titrant.

For a. volume = 61.0 mL:
Moles of HCl (nHCl) = Molarity x Volume = 0.025 M x 61.0 mL
Moles of HCl (nHCl) = 1.525 mmol

The reaction is 1:1, so the moles of NH3 consumed are also 1.525 mmol.

For b. volume = 65.0 mL:
Moles of HCl (nHCl) = Molarity x Volume = 0.025 M x 65.0 mL
Moles of HCl (nHCl) = 1.625 mmol

Again, the moles of NH3 consumed are equal to 1.625 mmol.

Step 3: Determine the remaining moles of NH3.

For a. volume = 61.0 mL:
Initial moles of NH3 = 0.050 M x 30.0 mL = 1.50 mmol
Remaining moles of NH3 = Initial moles - Moles consumed = 1.50 mmol - 1.525 mmol = -0.025 mmol
Since moles cannot be negative, we set the remaining moles to 0.

For b. volume = 65.0 mL:
Initial moles of NH3 = 0.050 M x 30.0 mL = 1.50 mmol
Remaining moles of NH3 = Initial moles - Moles consumed = 1.50 mmol - 1.625 mmol = -0.125 mmol
Again, we set the remaining moles to 0.

Step 4: Calculate the final volume of the solution.

For a. volume = 61.0 mL:
Final volume of the solution = Initial volume of NH3 solution + Volume of HCl added = 30.0 mL + 61.0 mL = 91.0 mL

For b. volume = 65.0 mL:
Final volume of the solution = Initial volume of NH3 solution + Volume of HCl added = 30.0 mL + 65.0 mL = 95.0 mL

Step 5: Calculate the concentration of NH4+ in the final solution.

For a. volume = 61.0 mL:
Concentration of NH4+ (CH+) = Moles of NH4+ / Final volume of solution = 1.525 mmol / 91.0 mL = 0.0167 M

For b. volume = 65.0 mL:
Concentration of NH4+ (CH+) = Moles of NH4+ / Final volume of solution = 1.625 mmol / 95.0 mL = 0.0171 M

Step 6: Calculate the pOH and pH.

For a. volume = 61.0 mL:
Since NH4+ is a weak acid, we can assume it completely dissociates to NH3 and H+ in water.
NH4+ + H2O -> NH3 + H3O+

The concentration of H+ equals the concentration of NH4+, so the pOH can be calculated using the concentration of NH4+:

pOH = -log[OH-] = -log(CH+)
pOH = -log(0.0167 M) = 1.78
pH = 14 - pOH = 14 - 1.78 = 12.22

For b. volume = 65.0 mL:
pOH = -log[OH-] = -log(CH+)
pOH = -log(0.0171 M) = 1.77
pH = 14 - pOH = 14 - 1.77 = 12.23

Therefore, the pH after adding 61.0 mL of titrant is 12.22, and the pH after adding 65.0 mL of titrant is 12.23.

To calculate the pH during a titration, we need to determine the moles of acid and base, and then use the stoichiometry of the reaction and the volume change to determine the concentrations of the acid and base at each point.

Here's how to approach the problem step-by-step:

1. First, let's write the balanced chemical equation for the reaction between NH3 (base) and HCl (acid):
NH3 + HCl → NH4Cl

2. Determine the number of moles of NH3 initially present:
moles of NH3 = volume (L) x molarity
moles of NH3 = 0.030 L x 0.050 mol/L = 0.0015 mol

3. Determine the number of moles of HCl added at each point:
a. For 61.0 mL of HCl:
moles of HCl = volume (L) x molarity
moles of HCl = 0.061 L x 0.025 mol/L = 0.001525 mol

b. For 65.0 mL of HCl:
moles of HCl = volume (L) x molarity
moles of HCl = 0.065 L x 0.025 mol/L = 0.001625 mol

4. Use the stoichiometry of the reaction to find the number of moles of NH3 remaining:
a. For 61.0 mL of HCl:
moles of NH3 remaining = moles of NH3 - moles of HCl
moles of NH3 remaining = 0.0015 mol - 0.001525 mol = -0.000025 mol

b. For 65.0 mL of HCl:
moles of NH3 remaining = moles of NH3 - moles of HCl
moles of NH3 remaining = 0.0015 mol - 0.001625 mol = -0.000125 mol

5. Since the amount of NH3 remaining cannot be negative, it means that all the NH3 has reacted completely when the volume of HCl reaches 61.0 mL and 65.0 mL.

6. To determine the pH, we need to find the concentration of NH4Cl formed. This solution will be neutral, as NH4Cl is the conjugate acid of a strong base (NH3) and a strong acid (HCl).

7. At the endpoints, when the NH3 is completely neutralized, we can assume that the volume has doubled, and the concentration of NH4Cl formed is equal to the concentration of the base originally present:
a. For 61.0 mL of HCl:
volume of NH3 (base) = 0.030 L
concentration of NH4Cl = concentration of NH3 (base) = 0.050 mol/L

b. For 65.0 mL of HCl:
volume of NH3 (base) = 0.030 L
concentration of NH4Cl = concentration of NH3 (base) = 0.050 mol/L

8. The pH of NH4Cl (a neutral salt) is calculated as the logarithm of the concentration of H+ ions in the solution, which is equal to the logarithm of the concentration of NH4Cl. pH = -log [NH4Cl]

Now you can calculate the pH at each endpoint using the concentration of NH4Cl calculated in step 7.

The first step is to determine where you are on the titration curve because that tells you what you have in the solution. To do that we determine where the equivalence point is located; i.e., what volume of HCl is needed to exactly neutralize the initial 15 millimoles NH3.

mL NH3 x M NH3 = mL HCl x M HCl
30 x 0.05 = mL HCl x 0.025
mL HCl = 60 mL. Let's show that below.

Here are the data:
millimoles NH3 = mL x M = 30 x 0.05 = 15 (initial)
millimoles HCl @ 61 mL = 61 x 0.025 = 15.25
millimoles HCl @ 65 mL = 65 x 0.025 = 16.25
Use these data to calculate the equivalence point as follows:
.........................NH3 + HCl ==> NH4Cl
I........................15.........0...............0
add...............................15..................
C......................-15.........-15............+15
E.........................0..........0...................15
So at the equivalence point you have only the salt, NH4Cl, in solution, with 15 mmols NH4Cl in 90 mL H2O.
Therefore, the problem is asking you to determine the pH of the solution AFTER the equivalence point where you have the salt NH4Cl, an excess of HCl now, and water to dilute everything.
How much NH3 did you start with? That's 15 mmols and 30 mL
How much HCl has been added @ 65 mL? That's 16.25 mmols and 65 mL.
How much HCl is in excess? That's 16.25 - 15 = 1.25 mmols HCl in 95 mL.
(HCl) = 1.25 mmols/95 mL = ?. Convert that to pH = -log(HCl)
The 61 mL is done the same way.
Post your work if you get stuck.