Answer:
The molarity of the base is 0.2664 M
Explanation:
Firstly, we write the complete and balanced titration reaction.
[tex]H_{2}SO_{4(aq)}[/tex] + [tex]2 LiOH_{(aq)}[/tex] → [tex]Li_{2}SO_{4(aq)}[/tex] + [tex]2H_{2}O_{(l)}[/tex]
From the reaction, we can identify that 1 mole of the acid reacted with 2 moles of the alkali (base) to yield salt and water only
We identify the following also from the question;
[tex]V_{a}[/tex] = 14.80 mL , [tex]V_{b}[/tex] = 25.00 mL , [tex]C_{a}[/tex] = 0.225M and [tex]C_{b}[/tex] = ?
[tex]n_{a}[/tex] = 1 , [tex]n_{b}[/tex] = 2
We use the relation;
[tex]C_{a}[/tex][tex]V_{a}[/tex]/[tex]n_{a}[/tex] = [tex]C_{b}[/tex] [tex]V_{b}[/tex]/ [tex]n_{b}[/tex]
Plugging the values, we have ;
(0.225 × 14.80)/1 = ( [tex]C_{b}[/tex] × 25.00)/2
[tex]C_{b}[/tex] = (2 × 0.225 × 14.80)/25
[tex]C_{b}[/tex] = 0.2664 M
