Strong Acid Weak Base Titration Problem!?

Equation:
HNO3 + NH3 → NH4NO3
1mol HNO3 reacts with 1 mol NH3

Because you have titrated 0.50M HNO3 with 0.50M NH3 solutions, the 16.75mL of HNO3 solution will have reacted with 16.75mL of the NH3 solution at the end point

If you add only 7.00mL of the acid, you produce a NH3/NH4NO3 buffer solution.

Calculate the molarity of the salt and unreacted base in the solution. You will have solution volume = 16.75+7.00 = 23.75mL solution:

Mol HNO3 in 7mL of 0.5M HNO3 = 7/1000*0.50 = 0.0035mol HNO3 reacted:
This will have reacted with 0.0035mol NH3 to produce 0.0035mol NH4NO3
Molarity of NH4NO3 solution = 0.0035/0.02375 = 0.147 M NH4NO3

There will be 16.75 - 7.00 = 9.75mL of 0.5M NH3 unreacted- dissolved in 23.75mL solution
Molarity of NH3 = 9.75/23.75*0.5 = 0.205M NH3 solution.

You need the pKb for NH3 = 4.75

Equation( Henderson - Hasselbalch)
pOH = pKb + log ([salt]/[base])
pOH = 4.75 + log ( 0.147/0.205)
pOH = 4.75 + log 0.716
pOH = 4.75 - 0.14
pOH = 4.61

pH = 14.00- pOH
pH = 14.00 - 4.61
pH = 9.39

The pH of the titration solution after adding 7.00mL HNO3 will be 9.39.

As a point of reference: If the titration is carried through to completion you will have a solution of NH4NO3 - This is an acidic salt and the pH of the end point will be at the methyl orange end point - pH = 4.5

The pH will not be 7 at the end the equivalence point. With a strong acid and a weak base it will be around 9. Therefore, use Phenolphthalein.