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

Answer:

The molarity of the base is 0.2664 M

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