posted by robyn .
How many moles of HCl must have been present in the 25mL of HCl solution in the 2 trials?
Trial 1: V of HCl= 25.00mL, V of NaOH used=29.50mL and M of NaOH=0.18M
Trial 2: V of HCl=25.00mL, V of NaOH used=28.50mL and M of NaOH=0.18M
Do it this way.
mols NaOH = M x L = ?
mols HCl = mols NaOH.