Question

When the solutions of lead (II) nitrate and potassium iodide are mixed:
(a) What is the name and color of the precipitate formed?
(b) Write the balanced chemical equation for this reaction?

Answer 1

We will first write the correct chemical formula for the reactants and find out their oxidation number as to what product will be formed. After knowing about the product we will check the solubility of the product to see which one is precipitated.

Complete step by step answer:
The chemical formula of lead (II) nitrate ${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$
The chemical formula of potassium iodide ${\text{KI}}$
The oxidation number of Pb in ${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ is + 2}}$
The oxidation number of ${\text{N}}{{\text{O}}_{\text{3}}}{\text{ in Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ is - 1}}$
The oxidation number of ${\text{K in KI is + 1}}$
The oxidation number of ${\text{I in KI is - 1}}$
Since it is a precipitation reaction so on observing the oxidation number of the ions in the reactant, we can easily confirm the product. So, the equation is
${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ + KI }} \to {\text{KN}}{{\text{O}}_{\text{3}}}{\text{ + Pb}}{{\text{I}}_{\text{2}}}$
Now we will balance the equation as the reactant side contain ${\text{2 molecules of N}}{{\text{O}}_{\text{3}}}$ and the product side contain only one molecule so we multiply it by $2$. So full balanced equation is
${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ + 2KI }} \to {\text{ 2KN}}{{\text{O}}_{\text{3}}}{\text{ + Pb}}{{\text{I}}_{\text{2}}}$
We know that all the compounds of potassium are soluble in water so the product ${\text{KN}}{{\text{O}}_{\text{3}}}$is dissolved in water and the product ${\text{Pb}}{{\text{I}}_{\text{2}}}$ is the precipitate here.
The lead nitrate is a white salt and has colorless aqueous solution and same is for potassium iodide it is a white salt and has colorless aqueous solution. The precipitate ${\text{Pb}}{{\text{I}}_{\text{2}}}$ formed is bright yellow in color.

Note: The above reaction is a type of precipitation reaction. In these types of reactions, the products formed have different solubilities so that we can separate them, one will be soluble and other will be insoluble in medium and we can separate it further.
We can always write a well-balanced chemical reaction by calculating the oxidation number of ions present in the reactant. We can write products' chemical composition first and then analyze the number of molecules of each ion present on either side of the reaction and multiplying them by necessary multiple.