Calculate the PH of 32 mmol of HCl in 75cm^3 solution. Assume HCl fully dissociates.

32 mmol = 0.032 mol

75cm³ = 0.075dm³

n = c x v    so   c = n / v

c = 0.032mol/0.075 dm³ = 0.426moldm^-3

pH = -log[H+]

pH = -log(0.426) = 0.37 (no units)