The conc of NaOH is 0.104M which means that 1000mL solution contains 0.104mol of NaOH, so the amount of moles used in the titration will be
[tex] \frac{0.104mol}{1000mL} * 259.4mL = 4.00* 10^{-7} mol[/tex]
The reaction of NaOH and HCl is one to one so the moles of HCl that were titrated will also be [tex]4.00* 10^{-7} mol[/tex]
The Molarity is therefore
[tex] \frac{4.00* 10^{-7} mol}{168.61 mL} * 1000mL = 2.38* 10^{-6} M[/tex]
