Question

A student does a titration with 14.80 mL of 0.225M H2SO4 and 25.00 mL of unknown LiOH solution. What is the molarity of the base Pls hurry

Answer 1

The molarity of the base is 0.2664 M

Step-by-step explanation:

Firstly, we write the complete and balanced titration reaction.

`H_(2)SO_(4(aq))` +
`2 LiOH_((aq))` →
`Li_(2)SO_(4(aq))` +
`2H_(2)O_((l))`

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;

`V_(a)` = 14.80 mL ,
`V_(b)` = 25.00 mL ,
`C_(a)` = 0.225M and
`C_(b)` = ?

`n_(a)` = 1 ,
`n_(b)` = 2

We use the relation;

`C_(a)`
`V_(a)`/
`n_(a)` =
`C_(b)`
`V_(b)`/
`n_(b)`

Plugging the values, we have ;

(0.225 × 14.80)/1 = (
`C_(b)` × 25.00)/2

`C_(b)` = (2 × 0.225 × 14.80)/25

`C_(b)` = 0.2664 M