Chapter 8, Problem 21A

Interpretation Introduction

Interpretation: The ionization of RbOH and CsOH in water needs to be explained.

Concept Introduction: Acids are the substances which tend to give ${\text{H}}^{\text{+}}$ ions in their solution whereas bases give ${\text{OH}}^{\text{-}}$ ions.

The acid-base reaction is the reaction of acid and base to form salt and water. Here salt is formed from the cation of base and anion of acid.

Answer to Problem 21A

  $\begin{array}{l}{\text{RbOH}}_{\text{(aq)}}\stackrel{{\text{H}}_{\text{2}}\text{O(l) }}{\to }{\text{Rb}}_{\text{(aq)}}^{\text{+}}{\text{+OH}}_{\text{(aq)}}^{\text{-}}\\ {\text{CsOH}}_{\text{(aq)}}\stackrel{{\text{H}}_{\text{2}}\text{O(l) }}{\to }{\text{Cs}}_{\text{(aq)}}^{\text{+}}{\text{+OH}}_{\text{(aq)}}^{\text{-}}\end{array}$

Explanation of Solution

Given information: RbOH and CsOH are strong bases.

On the basis of strength of bases; they can be further classified as strong and weak base. A strong base can be ionized completely in its solution whereas a weak base will ionized partially ionized in its solution. The ionization of RbOH and CsOH can be written as:

  $\begin{array}{l}{\text{RbOH}}_{\text{(aq)}}\stackrel{{\text{H}}_{\text{2}}\text{O(l) }}{\to }{\text{Rb}}_{\text{(aq)}}^{\text{+}}{\text{+OH}}_{\text{(aq)}}^{\text{-}}\\ {\text{CsOH}}_{\text{(aq)}}\stackrel{{\text{H}}_{\text{2}}\text{O(l) }}{\to }{\text{Cs}}_{\text{(aq)}}^{\text{+}}{\text{+OH}}_{\text{(aq)}}^{\text{-}}\end{array}$

Conclusion

  $\begin{array}{l}{\text{RbOH}}_{\text{(aq)}}\stackrel{{\text{H}}_{\text{2}}\text{O(l) }}{\to }{\text{Rb}}_{\text{(aq)}}^{\text{+}}{\text{+OH}}_{\text{(aq)}}^{\text{-}}\\ {\text{CsOH}}_{\text{(aq)}}\stackrel{{\text{H}}_{\text{2}}\text{O(l) }}{\to }{\text{Cs}}_{\text{(aq)}}^{\text{+}}{\text{+OH}}_{\text{(aq)}}^{\text{-}}\end{array}$