[confusedcollegestudent]

Calculate the pH of the 0.39 M NH₃/0.73 M NH₄Cl buffer system. The K_b of NH₃ is 1.8 x 10^-5.
I found that the pH was 8.98 and that was right.
However with part b, I'm very confused.
What is the pH after the addition of 20.0 mL of 0.075 M NaOH to 80.0 mL of the buffer solution?
Can someone show their work and explain how this is solved?
I thought maybe the Henderson-Hasselbalch equation, but I don't know how to set it up?

[Babcock_Hall]

It is a forum rule that you should show your work or give your thoughts before we can help you.  Hint:  write an equation starting with NaOH reacting with something.

[confusedcollegestudent]

It is a forum rule that you should show your work or give your thoughts before we can help you.  Hint:  write an equation starting with NaOH reacting with something.

Sorry! I'm new!
I know that for the first part of the problem, the reaction would proceed like so:
NH₃ NH₄⁺ + OH^- when it reacts with water and will produce hydroxide ions, hence the K_b.
And from that I made an ICE chart with the provided concentrations of the buffers to find pOH, then found pH.
But what I don't understand with the second part is with the addition of NaOH and how that changes everything.
I figure that it would react with the NH₄⁺ like so:
NH₄⁺ + OH^-  NH₃ + H₂O
But beyond this point, I don't really know how to manipulate the ICE chart, if one is even used? And how to change the Henderson-Hasselbalch equation. Since there are more H⁺ on the right side, I realize that K_a would be used.
Since there is 0.39 M of NH₃ present and 0.73 M of NH₄⁺, am I supposed to use both in the chart during the initial portion? Or am I only supposed to use one? I also don't understand what to do with the concentration or the volume of the NaOH that's used.

[Babcock_Hall]

You might start by calculating the number of moles of NaOH from its volume and concentration.