First, it's always a good idea to write the balanced chemical reactions:
1) HNO3 (aq) + NaOH (aq) ––�?gt; NaNO3 (aq) + H2O (l)
Strong Acid Strong Base Neutral Salt
2) HNO3 (aq) + NH3 (aq) ––�?gt; NH4NO3 (aq)
Strong Acid Weak Base NH4+ = Weak Acid (NO3�?/SUP> = neutral)
For both questions, first calculate the moles of each reactant. Which is the limiting reactant? How many moles of the excess reactant will remain?
For Question 1, that's all you need. Either HNO3 or NaOH will remain in excess. The moles divided by the total volume gives the molarity of H+ if HNO3 was in excess, or OH�?/SUP> if NaOH was in excess. From this, you can calculate the pH.
For Question 2, you also need to calculate how many moles of NH4NO3 are formed. It will turn out in this problem that HNO3 is the limiting reactant, so after reaction there will be excess NH3 remaining, plus the product NH4NO3. NH4+ is the important part of the product, because it is the conjugate acid of NH3 and a weak acid also. NO3�?/SUP> is the conjugate base of a strong acid, and as such, it is such a weak base that it is neutral.
Thus you will have in this solution a weak base (NH3) plus its conjugate acid (NH4+). That is a buffer solution. A "shortcut" for calculating the pH of buffer solutions is the Henderson-Hasselbalch equation:
pH = pKa + log{[weak base] / [conjugate acid]}
or, pH = –log(Ka) + log{[NH3] / [NH4+]}
^
Note that you have to convert Kb to Ka.
You can also take the "ICE" approach, a little longer, but it gives exactly the same result. But first, go ahead and work Question 1, to be sure to get the "bugs" worked out, and then tackle Question 2. 
Steve
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