Calculate the change in pH when 6.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

First what do we have?

100 mL x 0.1M NH3 = 10 millimoles.
100 mL x 0.1M NH4Cl = 10 mmoles.
6.00 mL x 0.1M HCl = 0.6 mmol.
6.00 mL x 0.1M NaOH = 0.6 mmol.
I'll do one to show you how to do it.

.........NH3 + H^+ ==> NH4^+
initial..10.....0.......10
add............0.6
change...-0.6..-0.6.....+0.6
equil....9.4.....0......10.6

pH = pKa + log (base/acid)
pH = pKa + log (9.4/10.6)
Solve for pH and take the difference between that and the pH of the original solution to find the change in pH. pH of original solution is done by HH equation, too.

I would workup a new ICE chart for the addition of NaOH as follows:
.........NH4^+ + OH^- ==> NH3 + H2O
.........10......0.........10
add..............0.6..............
change....-0.6...-0.6.......+0.6
equil.....9.4.....0........0.6
etc.

DrBob222, I understand the work, however why do we use mols? aren't we trying to find the concentrations?

To calculate the change in pH when HCl or NaOH is added to a buffer solution, you need to consider the Henderson-Hasselbalch equation. This equation relates the pH of a buffer solution to the concentrations of its conjugate acid and base. The equation is:

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

Where:
pH is the acidity level of the solution
pKa is the negative logarithm of the acid dissociation constant
[A-] is the concentration of the conjugate base
[HA] is the concentration of the acid

Let's solve the two given problems one by one:

1. Change in pH when 6.00 mL of 0.100 M HCl(aq) is added:
a. First, calculate the final concentrations of NH3 and NH4Cl after adding the HCl.
- Start with the initial concentrations:
[NH3] = 0.100 M
[NH4Cl] = 0.100 M

- Determine the moles of HCl added:
Moles = concentration × volume
Moles = 0.100 M × (6.00 mL / 1000 mL)
Moles = 0.0006 mol

- Calculate the new molar concentrations:
[NH3] = [NH3]initial - moles HCl / (initial volume + added volume)
[NH3] = 0.100 M - 0.0006 mol / (100.0 mL + 6.00 mL)
[NH3] = 0.0943 M

[NH4Cl] = [NH4Cl]initial + moles HCl / (initial volume + added volume)
[NH4Cl] = 0.100 M + 0.0006 mol / (100.0 mL + 6.00 mL)
[NH4Cl] = 0.1057 M

- Now, use the Henderson-Hasselbalch equation to calculate the new pH:
pH = pKa + log([NH3] / [NH4Cl])
(pKa for NH3/NH4Cl = 9.25)

pH = 9.25 + log(0.0943 M / 0.1057 M)
pH = 9.20

Therefore, the change in pH when 6.00 mL of 0.100 M HCl(aq) is added is -0.05.

2. Change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution:
a. Follow the same steps as above to determine the new concentrations of NH3 and NH4Cl after adding NaOH.
- Calculate the moles of NaOH added using the concentration and volume.
- Adjust the concentrations of NH3 and NH4Cl accordingly.

b. Use the Henderson-Hasselbalch equation to calculate the new pH.
- pH = pKa + log([NH3] / [NH4Cl])

Following these steps, you can find the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.