Question

Enter a balanced equation for the reaction between aqueous lead(II) nitrite and aqueous potassium bromide to form solid lead(II) bromide and aqueous potassium nitrite. Express your answer as a chemical equation. Identify all of the phases in your answer.

Answer 1

Answer:

$Pb(NO_{2})_{2}_{(aq)} + 2KBr_{(aq)}----->PbBr_{2}_{(s)} + 2 KNO_{2}_{(aq)}$

Explanation:

The charges are as follows:

$Pb^{+2}$ since oxidation number of lead is 2 as indicated lead (II)nitrite. The number in parentheses indicates the oxidation number.

$NO_{2}^{-1}$ bears the charge -1.

As lead bears 2 positive charges so we need two nitrite ions to form lead (II) nitrite.

$Br^{-1}$ , it is a halogen and bears the charge -1.

$K^{+1}$ , it is an alkali metal.

Since Bromide bears -1 charge and potassium bears +1 charge so we need only 1 ion of bromide and 1 ion of potassium to form potassium bromide.

Similarly, to form lead bromide we need 2 ions of bromide and 1 ion of lead, and to form potassium nitrite we need 1 nitrite ion and 1 potassium ion.

The final balanced equation is as follows, the various states of reactants and products are also written properly.

$Pb(NO_{2})_{2}_{(aq)} + 2KBr_{(aq)}----->PbBr_{2}_{(s)} + 2 KNO_{2}_{(aq)}$