Chapter 8, Problem 15A

Interpretation Introduction

Interpretation :

The balanced net ionic equations for the precipitation of silver (I), lead (II) and mercury (II) as chlorides must be given.

Concept Introduction :

Silver (I), lead (II) and mercury (II) chlorides have very low solubility products and hence these can be easily precipitated as corresponding chloride salts.

Answer to Problem 15A

  ${\text{Ag}}^{\text{+}}{\text{(aq)+Cl}}^{\text{-}}\text{(aq)}\to \text{AgCl(s)}\text{.}$

  ${\text{Pb}}^{\text{2+}}{\text{(aq)+2Cl}}^{\text{-}}\text{(aq)}\to {\text{PbCl}}_{\text{2}}\text{(s)}\text{.}$

  ${\text{Hg}}^{\text{2+}}{\text{(aq)+2Cl}}^{\text{-}}\text{(aq)}\to {\text{HgCl}}_{\text{2}}\text{(s)}\text{.}$

Explanation of Solution

Silver ion is precipitated as silver chloride after reaction with chloride ion from HCl.

Lead ion is precipitated as lead chloride after reacting with two chloride ions from HCl.

Mercury ion is precipitated as mercury chloride after reacting with two chloride ions from HCl.

Spectator ions are removed from both the side of chemical equations.