I notice you omitted the concn of NH4NO3. Yes, the HH equation will work this too. NH3 is the base and NH4NO3 is the acid. The equation is
pH = pKa + log(base)/(acid). Technically, this is concn base and concn acid in M or moles/L. So if you wanted to do this, without thinking about it, you would note that the NH3 concn is 0.1M x (50mL/100 mL) = 0.05M and the NH4NO3 is whatever the concn x (50/100), then plug those into the HH equation and solve. However, let me show you a couple of short cuts.
Since pH = pKa + log(base)/(acid) and remember (base) = moles/L and (acid) = moles/L, and remember moles = M x L.
So moles NH3 = 0.1M x 0.05L = ?? and moles NH4^+ = M x L = ??.
Then (NH3) = (0.05L x 0.1M)/0.1L total volume. Notice that (NH4^+) = M x L/total volume also. Now, since the total volume will ALWAYS be the same in these problems because it's always the same solution these acids and bases find themselves in, the volume will always cancel and you can use just moles. It saves a lot of calculations. A further shortcut is to use millimoles and not moles. To do that mmoles = mL x M.
So mmoles NH3 = 50 x 0.1M = ?
mmoles NH4^+ = 50 mL x ?M = ?
and you substitute mmoles for (base) and mmoles for (acid) and solve. You never need to reconvert to moles because the mmoles/mmoles unit cancels. I hope I've not confused you.
So is it a problem if the conch of NH4NO3 was not given to me. Or do I just assume it's the same as the NH3
If the problem is to calculate the pH of the buffer, then you must have the concn of NH4NO3. Or if you were given a pH of the buffer, you can calculate the concn of NH4NO3 needed to produce that pH. You must know one of them to calculate the other.
Don't think I know how to calculate the M of NH4NO3
Second thought I just don't understand what to do.
pH= pKa + log [NH3]/[NH4NO3]
pH= -l log ? + log (0.005/?)
As I explained previously, this problem can't be solved with the information given.