How do you calculate the pH when the following substances are added to a buffer.(all of the solutions are at .10M conc.)

The buffer: 50mL NH3 + 50mL NH4NO3
1) 10mL buffer + 6mL water
2)10mL buffer + 5mL water +1mL HCl
3)10mL buffer + 6mL HCl
4)10mL buffer + 5mL water + 1mL NaOH

I'm really confused on how the equations are set up for these types of problems (buffer solns).I'm posting this for the third time and really need some help..
Thanks

1).

50 mL of 0.1 M NH3 + 50 mL of 0.1 M NH4NO3.
(NH3) = 0.1 x (50/100) = 0.05 M.
(NH4NO3) = 0.1 x (50/100) = 0.05 M in the buffer.
1)So we start with 10 mL 0.05 M buffer. We dilute NH3 and NH4NO3 with 6 mL water.
The new concn (NH3) = 0.05 x (10/16) = ??
The new concn (NH4NO3) = 0.05 x (10/16) = ??
pH = pKa + log [(base)/(acid)]
Solve for pH. (The bottom line to dilution with water only is that the pH doesn't change but you go through a lot of concn changes to show that.)

2).
10 mL buffer gives me how many moles NH3? That is M x L = 0.05 x 0.01 L = 0.0005 moles NH3.
How many moles NH4NO3. 0.05 x 0.01 L = 0.0005 moles NH4NO3.
How much HCl is being added? That is M x L = 0.1 M x 0.001 L = 0.0001 moles HCl.
How much of the NH3 is used? 0.0001 moles HCl will react with 0.0005 moles NH3 to form 0.0001 moles NH4NO3 (to make a total of 0.0001+0.0005 = 0.0006 moles NH4NO3) and it will leave 0.0005 - 0.0001 = 0.0004 moles NH3).

pH = pKa + log(base/acid)
pH = 9.26 + log(0.0004/0.0006) = 9.08
Check my work. It's so easy to omit a zero. It is much easier to work with millimoles and not mols. M x mL = mmoles and M = mmoles/mL. For example 10 mL x 0.05 M = 0.5 mmol NH3. We add 1 mL x 0.1 M HCl = 0.1 mmole. So we form 0.1 + 0.5 = 0.6 mmole NH4NO3 and we leave 0.5-0.1 = 0.4 mmole NH3. Now, I ignored the water which was added. If you want to put in concn, then the new NH3 concn = 0.4 mmole/16 mL = ?? M and the new NH4NO3 concn = 0.6 mmole/16 mL = ?? M. You can see the 16 mL cancels.The others are done essentially the same way.

To calculate the pH of a buffer solution, you need to consider the equilibrium between the weak acid/base and its conjugate. In this case, the weak acid is NH4+ and the weak base is NH3.

To start, let's determine the concentration of NH4+ and NH3 in the buffer solution. Since the initial concentration of both substances is 0.10M and the volumes are equal (50 mL each), the final concentration in the buffer is also 0.10M for both NH4+ and NH3.

Now, let's consider each case separately:

1) 10mL buffer + 6 mL water:
Since water is neutral, it will not react with NH4+ or NH3. Therefore, the concentrations of NH4+ and NH3 remain the same as in the buffer solution. Now, you can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, [A-] is the concentration of NH3 and [HA] is the concentration of NH4+. The pKa for NH4+/NH3 is usually around 9.25.

2) 10mL buffer + 5 mL water + 1 mL HCl:
HCl is a strong acid that completely dissociates in water. It reacts with NH3 to form NH4+ and Cl-. To calculate the new concentrations after the reaction, you can use stoichiometry. Since the volume of the final solution is 16 mL (10 mL buffer + 5 mL water + 1 mL HCl), you can calculate the new concentration of NH4+ and NH3 using the initial concentration and volume.
After finding the new concentrations, you can calculate the pH using the Henderson-Hasselbalch equation mentioned earlier.

3) 10 mL buffer + 6 mL HCl:
Similar to the previous case, HCl reacts with NH3 to form NH4+ and Cl-. Again, you will need to use stoichiometry to determine the new concentrations of NH4+ and NH3 after the reaction. Then, calculate the pH using the Henderson-Hasselbalch equation.

4) 10 mL buffer + 5 mL water + 1 mL NaOH:
NaOH is a strong base that completely dissociates in water. It reacts with NH4+ to form NH3 and H2O. Similar to the previous cases, you will need to determine the new concentrations of NH4+ and NH3 using stoichiometry. After that, calculate the pH using the Henderson-Hasselbalch equation.

Remember to plug in the concentrations of NH4+ and NH3 (obtained after the reactions) into the Henderson-Hasselbalch equation to calculate the pH for each case.

I hope this explanation helps you understand how to set up the equations and solve these types of problems involving buffer solutions.