Verified answer

NH3 acts as a base by the equilibrium:

NH3(aq) + H2O(l) <--> NH4+(aq) + OH-(aq)

Kb = 1.8X10^-5 = [NH4+][OH-]/[NH3]

Calculate moles of each reactant:

0.075 L X 0.200 mol/L = 0.0150 mol NH3

0.0150 L X 0.500 mol/L = 7.5X10^-3 mol HNO3

Now, a strong acid will quantitatively convert NH3 to NH4+. So, after this addition:

moles NH4+ = 0.00750 mol

Moles NH3 = 0.0150 - 0.0075 = 0.0075 mol

Using these in the expression for Kb gives:

1.8X10^-5 = (0.0075) [OH-] / (0.0075)

[OH-] = 1.8X10^-5

pOH = 4.74

pH = 14.00 - pOH = 9.26