Calculate the pH of the 0.20 M NH3/0.25 M NH4Cl buffer.

I calculated the pH to be 9.15

What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 72.5 mL of the buffer?

10 mL x 0.1M HCl = 1 mmols.

72.5 mL x 0.2M = 14.5 mmols NH3.
72.5 mL x 0.25M = 18.125 mmols NH4^+. You can round to th correct number of significant figures when you finish the calculation.

.........NH3 + H^+ ==> NH4^+
I........14.5...0.......18.125
add.............1.............
C.......-1.....-1........+1
E.......13.5....0........19.125

pH = pKa + log(13.5/19.125) = ?

Well, let's see what happens when you add HCl to the buffer. It's like adding a grumpy guest to a party. The HCl will react with the NH3, which is like stealing all the snacks. This will create more NH4+ ions.

Now, imagine the NH3 molecules as tiny circus acrobats and the NH4+ ions as heavy clowns. The more clowns you add, the more they weigh down the acrobats, causing chaos!

So, by adding HCl to the buffer, you're throwing more clowns into the mix. This will upset the delicate balance between NH3 and NH4+ ions in the buffer.

To calculate the new pH, we need to consider the Henderson-Hasselbalch equation, which shows the relationship between the concentrations of the weak acid and its conjugate base:

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

In this case, NH3 is our base (A-) and NH4Cl is our weak acid (HA). The pKa for the NH4+/NH3 system is 9.24.

Now, let's crunch some numbers!
Initially, we have 0.20 M NH3 and 0.25 M NH4Cl. The volume is not important for the calculation.

After adding 10.0 mL of 0.10 M HCl, we need to consider the new concentration of NH4+. The moles of NH4+ ions added will be 10.0 mL * 0.10 M = 1.00 * 10^-3 moles.

Since 72.5 mL of the buffer initially contained 0.25 M NH4Cl, the number of moles of NH4+ ions in the original buffer is 72.5 mL * 0.25 M = 1.81 * 10^-2 moles.

Adding these two amounts together, we have a total of 1.81 * 10^-2 moles + 1.00 * 10^-3 moles = 1.91 * 10^-2 moles of NH4+ ions.

Now, let's calculate the new concentration of NH3. We still have the initial concentration of 0.20 M, as HCl does not affect the NH3 concentration.

Finally, we substitute these values into the Henderson-Hasselbalch equation:

pH = 9.24 + log([0.20]/[1.91 * 10^-2])

After doing the calculations, I'm pleased to announce that the new pH of the buffer will be approximately 9.18.

So, the addition of HCl made the pH move a bit, but it's not a clown-tastrophe!

To determine the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 72.5 mL of the buffer, you need to consider the effect of the added acid on the equilibrium of the buffer system.

Step 1: Calculate the concentration of NH3 and NH4+ in the buffer solution.

Given:
- Initial volume of NH3 solution (V1) = 72.5 mL
- Initial concentration of NH3 solution (C1) = 0.20 M
- Initial volume of NH4Cl solution (V2) = 72.5 mL
- Initial concentration of NH4Cl solution (C2) = 0.25 M

Since the volumes of NH3 and NH4Cl solutions are the same, the concentration of NH3 and NH4+ in the buffer solution will be equal.

Using the formula:

Cfinal = (V1 x C1 + V2 x C2) / (V1 + V2)

Cfinal = (72.5 mL x 0.20 M + 72.5 mL x 0.25 M) / (72.5 mL + 72.5 mL)
Cfinal = (14.5 + 18.125) / 145
Cfinal = 32.625 / 145
Cfinal = 0.225 M

So, the final concentration of NH3 and NH4+ in the buffer solution is 0.225 M.

Step 2: Calculate the moles of HCl added.

Given:
- Volume of HCl added (V3) = 10.0 mL
- Concentration of HCl added (C3) = 0.10 M

Using the formula:

n = V3 x C3

n = 10.0 mL x 0.10 M
n = 1.00 millimole (mmol)

Step 3: Determine the final volume of the buffer solution.

Given:
- Initial volume of the buffer (V) = 72.5 mL
- Volume of HCl added (V3) = 10.0 mL

The final volume of the buffer solution is the sum of the initial volume and the volume of HCl added.

Vfinal = V + V3
Vfinal = 72.5 mL + 10.0 mL
Vfinal = 82.5 mL

Step 4: Determine the final concentration of the buffer components.

To calculate the final concentration of NH3 and NH4+ in the buffer solution, use:

Cfinal = nfinal / Vfinal

Given:
- Moles of NH3 and NH4+ in the buffer solution (nfinal) = moles present initially + moles from HCl
- Initial moles of NH3 and NH4+ in the buffer solution = Cfinal x V
- Moles from HCl = n

nfinal = (Cfinal x V) + n
nfinal = (0.225 M x 82.5 mL) + 1.00 mmol
nfinal = 18.56 mmol + 1.00 mmol
nfinal = 19.56 mmol

Cfinal = nfinal / Vfinal
Cfinal = 19.56 mmol / 82.5 mL
Cfinal = 0.237 M

So, the final concentration of NH3 and NH4+ in the buffer solution is 0.237 M.

Step 5: Calculate the pH of the final buffer solution.

To calculate the pH, we need to determine the ratio of NH4+ to NH3, and use the Henderson-Hasselbalch equation:

pH = pKa + log(NH4+ / NH3)

The pKa value for the ammonium ion (NH4+) is usually around 9.25 at 25°C.

pH = 9.25 + log(0.237 / 0.237)
pH = 9.25 + log(1)
pH = 9.25 + 0
pH = 9.25

Therefore, the pH of the buffer solution after the addition of 10.0 mL of 0.10 M HCl is approximately 9.25.

To calculate the pH of a buffer solution after the addition of an acid, we need to apply the Henderson-Hasselbalch equation. First, let's review the Henderson-Hasselbalch equation:

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

In this equation:
- pH is the measure of the acidity or alkalinity of a solution.
- pKa is the negative logarithm of the acid dissociation constant. It represents the acidity strength of the weak acid in the buffer.
- [A-] is the concentration of the conjugate base (NH3 in this case).
- [HA] is the concentration of the weak acid (NH4Cl in this case).

Given:
- Initial concentrations: [NH3] = 0.20 M and [NH4Cl] = 0.25 M
- Addition: 10.0 mL of 0.10 M HCl to 72.5 mL of the buffer

To calculate the new concentrations of NH3 and NH4Cl after the addition, we need to consider the dilution that occurs due to the combination of the two solutions.

Step 1: Calculate the moles of NH3 and NH4Cl initially:
moles of NH3 = [NH3] * Volume of NH3 = 0.20 M * 72.5 mL
moles of NH4Cl = [NH4Cl] * Volume of NH4Cl = 0.25 M * 72.5 mL

Step 2: Calculate the moles of HCl added:
moles of HCl = [HCl] * Volume of HCl = 0.10 M * 10.0 mL

Step 3: Calculate the new total volume of the solution:
Total volume = Volume of NH3 + Volume of NH4Cl + Volume of HCl = 72.5 mL + 72.5 mL + 10.0 mL

Step 4: Calculate the new concentrations of NH3 and NH4Cl:
[NH3] = moles of NH3 / Total volume
[NH4Cl] = moles of NH4Cl / Total volume

Step 5: Calculate the new concentration of NH4+ (formed by the reaction of HCl with NH3):
[NH4+] = (moles of NH3 + moles of HCl) / Total volume

Step 6: Calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log([NH3]/[NH4+])

Now, let's plug in the values and calculate the new pH.