Question

Complete and balance the following acid-base equations:

Answer 1

Answer:

a) $HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

b) $H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

c) $Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

Explanation:

a)

when $HClO_4$ is added to LiOH, lithium chlorate and water is formed.

$HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous $H_2SO_4$ is added to NaOH, $Na_2SO_4$ and $H_2O$ is formed. It is a neutrilization reaction.

$H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

No. of Na atom in left hand side = 1

No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

Now, it can been seen that all the atoms are balanced.

c) When $Ba(OH)2$ reacts with HF, baroum fluoride and water is formed.

$Ba(OH)2 + HF\rightarrow BaF_2 + H_2O$

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

$Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O$

In order to balance H and O on either side, multiply H2O by 2.

$Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$