[Sis290025]

25.00 mL of 0.100 M HCl is going to be neutralized by titrating against 0.123 M NaOH. Find the pH at:

i) the equivalence point
ii) 2.00 mL before the equivalence point
ii) 2.00 mL after the equivalence point

HCl + NaOH <-> NaCl + H2O

25.00 mL HCl*0.100 M = 2.5 mmol

i) The pH is 7.00 but how do you show this mathematically if you do not know the volume of NaOH added to reach the equivalence point?
For parts ii and iii, how can you calculate the pH if you do not know the exact volume of NaOH used to reach the equivalence point? How do you find the volume (or mmol) of NaOH for the equivalence point if a method is possible?

Any help on setting up each part is greatly appreciated and needed.

[Albert]

i) The pH is 7.00 but how do you show this mathematically if you do not know the volume of NaOH added to reach the equivalence point?

At the equivalence point, the moles of H⁺ are the same as those of OH^-.

Hence:   Molarity acid : Volume acid = Molarity base : Volume base

You can easily calculate the moles of acid, can't you?