What is the ph of a solution resulting from 2.50 ml of 0.075 m hcl being added to 35.00 ml of pure water?

HCl is a strong base therefore completely dissociates to give out H⁺ ions. 
Therefore [HCl] = [H⁺]
Number of HCl moles added - 0.075 M/1000 mL/L x 2.50 mL = 0.0001875 mol
total volume of solution - 35.00 mL + 2.50 mL = 37.50 mL 
Concentration of HCl = number of HCl moles / volume 
[HCl] = 0.0001875 mol / 0.03750 L = 0.005 M
pH of solution is;
pH = -log[H⁺]
since [HCl] = [H⁺]
pH = -log(0.005)
pH = 2.30
Therefore pH of solution is 2.30