Chapter 12, Problem 12STP

Interpretation Introduction

Interpretation:

Correct option must be chosen regarding the net ionic reaction when potassium chromate and lead acetate are allowed to react.

Concept introduction:

Potassium chromate and lead acetate react to produce lead chromate and potassium acetate.

In the net ionic equation all the common ions to reactant side and product side (spectator ions) are cancelled.

Answer to Problem 12STP

Correct answer: Option B is correct.

Explanation of Solution

Reason for correct option: Option C is correct because in the reaction of potassium chromate and lead acetate there will be formation of lead chromate and potassium acetate. Potassium and acetate ions are cancelled as these are spectator ions.

${\text{Pb}}^{\text{2+}}{\text{(aq)+CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}\to {\text{PbCrO}}_{\text{4}}\text{(s)}\text{.}$

Conclusion

Reasons for incorrect options: Option A is incorrect as the reaction is not showing the formation of lead chromate.

Option B is incorrect as one lead ion must reacts with one chromate ion as both have valency 2.

Option D is incorrect as it is not showing the formation of lead chromate.

Option E is incorrect as it is not showing chromate ion as divalent anion.