pH During Titration

Consider the titration of 100 mL of .500 M NH3 (kb=1.8x10^-5) with .500 M HCl. After 50.0 ml of HCL have been added, the [H+] of the solution is a) 1.8x10^-5 b) 5.6x10^-10 c) 1.2x10^-5 d) 1.0x10^-7 e) none of these
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NH_3(aq) + H₂O <==> NH₄⁺(aq) + OH-(aq)
K_b = [ NH4+][ OH-]/[ NH3] = 1.8x10^-5
(0.100L)(0.50mol/L) = 0.050 moles NH3
(0.050L)(0.50mol/L) = 0.025 moles HCl
0.025 moles of NH3 will be converted to NH4+ when NH3 reacts with HCl
We will end up 0.025 moles of each, NH3 and H+, in 150 mLs total volume
The concentrations are:
[NH3] = [H+] = 0.025 moles / 0.150L = 0.167 M
[ 0.167][ OH-]/[0.167] = 1.8x10^-5, SOLVE for [OH-]
[OH-] = 1.8x10^-5
[H+][1.8x10^-5]=1x10^-14
[H+]=5.55x10^-10