.70 L of a buffer solution that is 0.120 M in NH3 and 0.450 M in NH4Cl are diluted to 5.80 L with distilled water. What is the pH of the buffer? Kb of NH3 = 1.8 x 10^-5

do i work this out the same way i do a pH problem with one volume?

I don't know how YOU work it out. That may work depending upon how you do it.

I would use the Henderson-Hasselbalch equation.

I have tried using that equation but I keep getting 4.something. My answer should be around 9.

I expect you are using the wrong Ka.

Ka for NH3 is NOT 4.76. It is 9.25.
Using the HH equation, I obtained 8.68 with a quickie calculation but check me out on that.

Thank you, I was using the wrong one.

I'm glad things worked out for you. By the way, I meant, of course, PKa = 9.25 (not Ka) and pKb = 4.76

To find the pH of a buffer solution, we first need to determine the concentration of hydroxide ions (OH-) in the solution.

In this case, we're given the concentrations of NH3 (ammonia) and NH4Cl (ammonium chloride) in the buffer, as well as the Kb value for NH3. We can use this information to find the concentration of OH- ions.

1. First, convert the volumes of the original solution and the added water to the same units. Since we want to find the concentration of OH- ions in the final solution, we'll use the final volume of the diluted buffer solution: 5.80 L.

2. Calculate the initial moles of NH3 and NH4Cl in the original buffer solution. To do this, multiply the volume of the original buffer solution (0.70 L) by their respective concentrations:
Moles of NH3 = 0.70 L × 0.120 M
Moles of NH4Cl = 0.70 L × 0.450 M

3. Calculate the total moles of NH3 and NH4Cl in the original buffer solution by summing their individual moles:
Total moles = moles of NH3 + moles of NH4Cl

4. Calculate the final concentration of NH3 and NH4Cl in the diluted buffer solution. To find this, divide the total moles obtained in step 3 by the final volume of the solution (5.80 L):
Concentration of NH3 = Total moles / 5.80 L
Concentration of NH4Cl = Total moles / 5.80 L

5. Since NH4Cl fully dissociates in water, the concentration of NH4+ ions is the same as the concentration of NH4Cl:
Concentration of NH4+ = Concentration of NH4Cl

6. Now, let's set up an equilibrium expression using the Kb value and the concentration of NH3 and NH4+:
Kb = [OH-][NH4+]/[NH3]

Since OH- is produced when NH3 reacts with water, we can assume that [OH-] is equal to [NH4+]. Let's denote the concentration of [OH-] as x.

Therefore, the equilibrium expression can be simplified to:
Kb = x^2 / [NH3]

7. Rearrange the equation in step 6 to solve for x, which represents the concentration of [OH-]. In this case, we're given the Kb value of NH3, so we can substitute that value and solve for x:
x^2 = Kb * [NH3]
x = √(Kb * [NH3])

8. Finally, we can calculate the pOH of the diluted buffer solution using the concentration of OH-:
pOH = -log10([OH-])

And since pH + pOH = 14, we can find the pH of the buffer solution:
pH = 14 - pOH