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aliya0001 [1]
3 years ago
15

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
Chemistry
1 answer:
Degger [83]3 years ago
8 0

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

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