Question

Calculate the equivalence point in the titration of 20.00 mL of 0.1252M HCl with 0.1008M NaOH.

I know we have to do:

20ml ( 0.1252M ) = 2.5mmol

But I dont understand why we have to divide 2.5 mmol over .1008M . Im trying to understand the concept so I can understand how to do other problems similar to this one. So if someone can explain that would be very helpful :)

Answer 1

There is molarity equation used for this.

Equivalence point is the fórut where all the acid moles are neutralized by Base moles. for the neutralisation, one need equal amount of moles of acid and base. for equivalence Points Number of moles of Acid = Number of moles of Base. Definition of nolarity says molarity moles 5 moles Molaci ty x volume (Molarity of Acid) (Volume of Acid) = (Molanity Base) Nolume of Gaso) In short, it can be written as; : M,V, 2 M 2 7 IcAcid) Base troca y que 7209 Son Here we need to lese the given nalues; Acid ; M = 0.1252 m ✓ = 20.00 ml Mga 0.1008 M (0.125M) (20.00ml) = (01008m) (V) V. 2 (0.1252 M) (20.00 ml) V, > 70.1008 m) [V₂ = 24.84 mL ! Auswer. So, the equivalence Point is at 24.84 mL added of NaOH. 1.