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