Question

When double displacement reaction of an aqueous solution of lead (II) nitrate and aqueous solution of potassium sulfate occurs, precipitate form will be?

• A. $Pb{SO}_4$
• B. $KN{O}_3$
• C. $K_{2}Pb$
• D. ${NO}_2$
• E. ${SO}_3$

Answer 1

The correct option is A $Pb{SO}_4$

$Pb\left(N{O}_3{)}_2\right(aq)+K_{2}S{O}_4(aq)\to PbS{O}_4(ppt.)+2KN{O}_3(aq)$
Double displacement reactions may be defined as the chemical reactions in which one component each of both the reacting molecules is exchanged to form the products. During this reaction, the cations and anions of two different compounds switch places, forming two entirely different compounds.
The general equation which represents a double displacement reaction can be written as:
Double displacement reactions generally take place in aqueous solutions in which the ions precipitate and there is an exchange of ions.