Question

When we mix the solutions of lead (II) nitrate and potassium iodide.

(i) What is the colour of the precipitate formed? Can you name the compound precipitated?

(ii) Write the balanced chemical equation for this reaction.

(iii) Is this also a double displacement reaction?

Answer 1

Hint: We will determine the product of the reaction and check their solubility. The compound which will be insoluble in water will form a precipitate. By writing the chemical formula of each compound we will write the chemical equation. We will determine if the two compounds are exchanging their parts.

Complete answer:

(i) The colour of the precipitate formed and the name of the compound is as follows:

The chemical reaction takes place between lead (II) nitrate and potassium iodide, so potassium nitrate and lead iodide form.

Nitrates are soluble in water, so the product potassium nitrate will be soluble. The lead halides are insoluble in water, so lead iodide will be precipitated.

So, the name of the compound precipitated is lead iodide and it is a yellow colour precipitate.

(ii) The balanced chemical equation for this reaction is as follows:

The chemical formula of lead (II) nitrate is  $ {\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}} $  and potassium iodide  $ {\text{KI}} $ .

 $ {\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{(aq)}}\, + \,{\text{KI}}\,{\text{(aq)}}\, \to \,{\text{Pb}}{{\text{I}}_2}{\text{(s)}}\, + {\text{KN}}{{\text{O}}_{\text{3}}}\,{\text{(aq)}} $ 

On the left side of the equation, two nitrate ions are present, so add two on the left side, in front of potassium nitrate.

 $ {\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{(aq)}}\, + \,{\text{KI}}\,{\text{(aq)}}\, \to \,{\text{Pb}}{{\text{I}}_2}{\text{(s)}}\, + 2\,{\text{KN}}{{\text{O}}_{\text{3}}}\,{\text{(aq)}} $ 

Now on the left side, two iodide ions are present, so add two on the right side, in front of potassium iodide.

 $ {\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{(aq)}}\, + \,2\,{\text{KI}}\,{\text{(aq)}}\, \to \,{\text{Pb}}{{\text{I}}_2}{\text{(s)}}\, + {\text{2KN}}{{\text{O}}_{\text{3}}}\,{\text{(aq)}} $ 

So, the balanced chemical equation is,  $ {\text{Pb}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{(aq)}}\, + \,2\,{\text{KI}}\,{\text{(aq)}}\, \to \,{\text{Pb}}{{\text{I}}_2}{\text{(s)}}\, + {\text{2KN}}{{\text{O}}_{\text{3}}}\,{\text{(aq)}} $ 

(iii) We will determine the type of reaction as follows:

In a double displacement reaction, two compounds exchange their parts to form two new compounds. The reaction takes place between lead (II) nitrate and potassium iodide, the lead and potassium exchange their anions to give lead iodide and potassium nitrate.

So, this is a double displacement reaction.

Note:

All nitrates are soluble. Halides are also soluble only the halides of lead, mercury, and solver are insoluble. Lead iodide is a yellow colour crystalline solid. It becomes red-orange on heating. Soluble compounds are donated by (aq) in the equation whereas the insoluble which form precipitate are donated by (s). The reaction in which one element replaces the other element from the compound is known as a single displacement reaction. For example, the reaction between zinc and copper sulphates, the reaction gives zinc sulphate and copper.