What would the [H3O+] be after addition of 5.15 mL of 0.1000 M HCl to 25.00 mL of 0.1000 B (a weak base, Kb = 6.31e-5)?
millimols BOH = 25 mL x 0.1M = 2.5
millimols HCl = 5.25 mL x 0.1M = 0.515
..........BOH + HCl ==> BCl + H2O
I.........2.5....0.......0......0
added.........0.515...............
C......-0.515..-0.515...+0.515
E...... 1.985....0.......0.515
I would substitute the E line into the Henderson-Hasselbalch equation and solve for pH, then H^+.
Note: you have Kb and you need pKa. Convert Kb to pKb, the pKa + pKb = pKw = 14 and solve for pKa.
You don't need to use the HH equation; it's just easier to do it that way.