If the conc. of HCl is 1.5 M, then 1000ml of HCl contain 1.5 mol
let 20ml of HCl (used in titration)contain x
⇒ x = (20 ml × 1.5 mol) ÷1000ml
x = 0.03 mol
∴ moles of HCl used to dissolve the LiOH was 0.03 mol
The reaction is characterised by the equation: HCl + LiOH → LiCl + H₂O
Based on the equation, the mole ratio of HCl : LiOH is 1 : 1
∴ if moles of HCl reacting = 0.03 mol
then moles of LiOH = 0.03mol
Now, since 25ml of LiOH reacted with the HCl,
then it implies that 25ml of LiOH contain 0.03 mol
∴ Let 25ml of LiOH contain 0.03mol
and 1000ml of LiOH contain y
⇒ y = (1000 ml × 0.03 mol) ÷ 25ml
y = 1.2 mol
∴ the concentration of the base (LiOH) in this titration is 1.2 M (OPTION A)
