Chapter 7, Problem 7.47AP

Interpretation Introduction

Interpretation:

The number of equivalents presents in each ion of $\text{2}\text{.5}\text{mole KCl}$ and $\text{2}\text{.5}{\text{mole CaSO}}_{\text{4}}$ has to be given.

Concept Introduction:

An equivalent is the number of moles of charge, which a mole of ion contributes to a solution.  The number of equivalents per mole of an ion equals the charge on the ion.

Answer to Problem 7.47AP

The number of equivalents present in $\text{2}{\text{.5 mole K}}^{\text{+}}$ is $\text{2}\text{.5}\text{Eq}{\text{K}}^{\text{+}}\text{.}$

The number of equivalents present in $\text{2}{\text{.5 mole Cl}}^{\text{-}}$ is $\text{2}\text{.5}\text{Eq}{\text{Cl}}^{\text{-}}\text{.}$

The number of equivalents present in $\text{2}{\text{.5 mole Ca}}^{\text{2+}}$ is $\text{5}\text{.0 Eq}{\text{Ca}}^{\text{2+}}\text{.}$

The number of equivalents present in $\text{2}{\text{.5 mole SO}}_{\text{4}}{}^{\text{2-}}$ is $\text{5}\text{.0 Eq}{\text{SO}}_{\text{4}}{}^{\text{2-}}\text{.}$

Explanation of Solution

Potassium chloride $\left(\text{KCl}\right)$ consists of ${\text{K}}^{\text{+}}$ and ${\text{Cl}}^{\text{-}}$ ions.

Potassium has $\text{+1}$ charge, there are $\text{1 Eq}$ of ${\text{K}}^{\text{+}}$ ions per mole of ${\text{K}}^{\text{+}}.$  The number of equivalents present is calculated as,

  Number of equivalents=$\text{2}\text{.5}\text{mol}{\text{K}}^{\text{+}}\text{×}\frac{\text{1}\text{Eq}{\text{K}}^{\text{+}}}{\text{1}\text{mole}{\text{K}}^{\text{+}}}\text{=2}\text{.5}\text{Eq}{\text{K}}^{\text{+}}$

The number of equivalents present in $\text{2}{\text{.5 mole K}}^{\text{+}}$ is $\text{2}\text{.5}\text{Eq}{\text{K}}^{\text{+}}\text{.}$

Chlorine has $\text{-1}$ charge, there are $\text{1 Eq}$ of ${\text{Cl}}^{\text{-}}$ ions per mole of ${\text{Cl}}^{\text{-}}\text{.}$  The number of equivalents present is calculated as,

  Number of equivalents=$\text{2}\text{.5}\text{mol}{\text{Cl}}^{\text{-}}\text{×}\frac{\text{1}\text{Eq}{\text{Cl}}^{\text{-}}}{\text{1}\text{mole}{\text{Cl}}^{\text{-}}}\text{=2}\text{.5}{\text{Eq Cl}}^{\text{-}}$

The number of equivalents present in $\text{2}{\text{.5 mole Cl}}^{\text{-}}$ is $\text{2}\text{.5}\text{Eq}{\text{Cl}}^{\text{-}}\text{.}$

Calcium sulphate $\left({\text{CaSO}}_{\text{4}}\right)$ consists of ${\text{Ca}}^{\text{2+}}$ and ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ ions.

Calcium has $\text{+2}$ charge, there are $\text{2 Eq}$ of ${\text{Ca}}^{\text{2+}}$ ions per mole of ${\text{Ca}}^{\text{2+}}.$  The number of equivalents present is calculated as,

  Number of equivalents=$\text{2}\text{.5}\text{mol}{\text{Ca}}^{\text{2+}}\text{×}\frac{\text{2}\text{Eq}{\text{Ca}}^{\text{2+}}}{\text{1}\text{mole}{\text{Ca}}^{\text{2+}}}\text{=5}\text{.0Eq}{\text{Ca}}^{\text{2+}}$

The number of equivalents present in $\text{2}{\text{.5 mole Ca}}^{\text{2+}}$ is $\text{5}\text{.0 Eq}{\text{Ca}}^{\text{2+}}\text{.}$

Sulphate has $\text{-2}$ charge, there are $\text{2 Eq}$ of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ ions per mole of ${\text{SO}}_{\text{4}}{}^{\text{2-}}.$  The number of equivalents present is calculated as,

  Number of equivalents=$\text{2}\text{.5}\text{mol}{\text{SO}}_{\text{4}}{}^{\text{2-}}\text{×}\frac{\text{2}\text{Eq}{\text{SO}}_{\text{4}}{}^{\text{2-}}}{\text{1}\text{mole}{\text{SO}}_{\text{4}}{}^{\text{2-}}}\text{=5}{\text{.0Eq SO}}_{\text{4}}{}^{\text{2-}}$

The number of equivalents present in $\text{2}{\text{.5 mole SO}}_{\text{4}}{}^{\text{2-}}$ is $\text{5}\text{.0 Eq}{\text{SO}}_{\text{4}}{}^{\text{2-}}\text{.}$