What is the ph of a .05 L solution containing .075 mol NH3 and 6.80 g of NH4+ when .025 L of 2 M is added to it? Kb= 1.8 x10^-5

Answer 1

When 0.025 L of 2M WHAT is added to it.
Use the Henderson-Hasselbalch equation and set up an ICE chart with it. If you will show your work I can help you through it.

Answer 2

Do i use Ka or Kb?

Answer 3

The HH equation uses pKa. Don't have pKa? Remember to convert Kb to pKb, then
pKa + pKb = pKw = 14