Your method is a titration of a strong base with a strong acid. The idea is that 1 mol (equivalent) NaOH has exactly reacted with 1 mol (equivalent) HCl at the moment when pH drops below pH=7.

In your case:

The reagent (2M HCl) contains 2 equiv/L = 2 mequiv/mL).

If you titrate with volume Ve (in mL) of HCl, this volume contains

2 mequiv/mL x Ve (mL) of HCl

and has neutralised the same quantity (in mequiv) of NaOH. The sample concentration (i.e, quantity/volume) was

Csample = 2 mequiv/mL x Ve (mL) /25 mL NaOH

Example: Ve was 10.0 mL.

Csample = 2 mequiv/mL x 10.0 (mL) /25 mL NaOH =

= 0.80 mequiv NaOH/mL = 0.80 equiv NaOH/L

Cornel Radu is correct and I go with his answer

Dear Wynand Zwart ,

I think you have had an adequqte answer an explanation by Cornel Radu and no need to have more. Good luck to you.

Cornel Radu made a very good explanation of the mix of strong acid and strong base. I will just add that the general formula is: C1*V1=C2*V2, where V1 is the volume in the beaker, C2 is the concentration of the acid, V2 is the volume in the burette, and C1 is the concentration of unknow base (brine).

Thank you for the answer Cornel Radu