Question

A student prepared 750 mL of a stock solution of 0.1 M NaOH from solid NaOH and deionized water. To standardize the NaOH solution it was titrated against 20.00 mL aliquots* of 0.1134 M HCl to a light pink endpoint using phenolphthalein as an indicator. The average titre for 3 titrations was 22.78 mL of NaOH.
How many moles of NaOH were used to titrate each aliquot of the HCl?

Answer 1

C(NaOH)V(NaOH)=C(HCl)V(HCl)=n(NaOH)

n(NaOH)=0.1134*20.00*0.001=2.268*10^-3 mol NaOH