# A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3 with 50 mL of 0.300 NH4Cl. The pKb of NH3 is 4.74.

NH3 + H2O-> NH4+ +OH-

7.50 mL of 0.125 M HCl is added to the 100 mL of the buffer solution. Calculate the concentration of NH3 and NH4Cl for the buffer solution. Calculate the pH of the solution.

## To calculate the concentration of NH3 and NH4Cl in the buffer solution after adding 7.50 mL of 0.125 M HCl, we need to follow these steps:

1. Determine the initial moles of NH3 and NH4Cl before adding HCl:
- Moles of NH3 = volume (L) x concentration (M) = 0.050 L x 0.300 M
- Moles of NH4Cl = volume (L) x concentration (M) = 0.050 L x 0.300 M

2. Calculate the moles of HCl added:
- Moles of HCl = volume (L) x concentration (M) = 0.0075 L x 0.125 M

3. Determine the net change in moles of NH3 and NH4Cl after the reaction with HCl:
- NH3 reacts with HCl to form NH4+:
- The stoichiometric ratio is 1:1, so the moles of NH3 decrease by the same amount as the moles of HCl added.
- NH4Cl does not react with HCl, so its moles remain the same.

4. Calculate the final moles of NH3 and NH4Cl after the reaction:
- Final moles of NH3 = initial moles of NH3 - moles of HCl added
- Final moles of NH4Cl = initial moles of NH4Cl

5. Calculate the final concentration of NH3 and NH4Cl:
- Final concentration of NH3 = final moles of NH3 / total volume (L)
- Final concentration of NH4Cl = final moles of NH4Cl / total volume (L)

Now, let's plug in the values:

1. Initial moles of NH3:
- Moles of NH3 = 0.050 L x 0.300 M = 0.015 mol

2. Initial moles of NH4Cl:
- Moles of NH4Cl = 0.050 L x 0.300 M = 0.015 mol

- Moles of HCl = 0.0075 L x 0.125 M = 0.0009375 mol

4. Final moles of NH3:
- Final moles of NH3 = 0.015 mol - 0.0009375 mol = 0.0140625 mol

5. Final moles of NH4Cl:
- Final moles of NH4Cl = 0.015 mol

6. Total volume of the solution:
- Total volume = initial volume of NH3 + initial volume of NH4Cl + volume of HCl added
- Total volume = 0.050 L + 0.050 L + 0.0075 L = 0.1075 L

7. Final concentration of NH3:
- Final concentration of NH3 = 0.0140625 mol / 0.1075 L = 0.1307 M

8. Final concentration of NH4Cl:
- Final concentration of NH4Cl = 0.015 mol / 0.1075 L = 0.1395 M

To calculate the pH of the solution, we can use the Henderson-Hasselbalch equation:

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

The pKa can be calculated from the pKb of NH3:

pKa = 14 - pKb

Now, let's calculate the pH:

pKa = 14 - 4.74 = 9.26

[A-] = concentration of NH4Cl = 0.1395 M
[HA] = concentration of NH3 = 0.1307 M

pH = 9.26 + log (0.1395 / 0.1307)
pH = 9.26 + log (1.064)
pH = 9.26 + 0.027
pH = 9.29

Therefore, the concentration of NH3 is 0.1307 M and the concentration of NH4Cl is 0.1395 M in the buffer solution after adding 7.50 mL of 0.125 M HCl. The pH of the solution is 9.29.

## millimols NH3 = 50 mL x 0.3M = 15.0

(NH3) = mmols/mL = 15.0/100 = 0.15M initially.

millimols NH4Cl = 50 x 0.3 = 15.0
(NH4Cl) = 15.0/100 = 0.15M initially

millimols HCl = 7.50 x 0.125 = 0.9375
pKb NH3 = 4.74; pKa = 14-4.74 = 9.26

I prefer to work with this equilibrium in mmols.
...........NH3 + H^+ ==> NH4^+
I..........15....0........15