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)
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)
