Q19.66P - Chemistry: Molecular Nature Of Matter And Change

Question

Write the ion-product expressions for  silver carbonate;

 barium fluoride;  copper (II) sulphide.

Step-by-Step Solution

Answer

The ion-product expression of the compounds is,

 Silver carbonate: $\text{Ksp = }{\left[{\text{Ag}}^{\text{ + }}\right]}^{\text{2}}\left[{\text{CO}}_{\text{3}}^{\text{2}\text{ -}}\right]$

 Barium fluoride:  $\text{Ksp = }\left[{\text{Ba}}^{\text{2 + }}\right]{\left[{\text{F}}^{\text{ - }}\right]}^{\text{2}}$

 Copper (II) sulphide: $\text{Ksp = }\left[{\text{Cu}}^{\text{2 + }}\right]\left[{\text{s}}^{\text{2 - }}\right]$ 

Concept Introduction

The $\mathbf{\text{Qsp}}$-ion-product expression is obtained by multiplying the concentrations of ions generated by the dissolution of a chemical. When a solution is saturated, the $\mathbf{\text{Qsp}}$  value is referred to as the $\mathbf{\text{Ksp}}$ value (solubility-product constant).

${\mathbf{\text{MX}}}_{\mathbf{2}}\mathbf{>}\mathbf{>}{\mathbf{M}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{X}}^{\mathbf{-}}$

We do not include the solid and liquid states in the  equations-

$\mathbf{\text{Ksp = }}\left[{\text{M}}^{\text{2 + }}\right]{\left[{\text{X}}^{\text{ - }}\right]}^{\mathbf{\text{2}}}$

Obtaining ion-product expression for Silver Carbonate

 Let us obtain the ion-product expression for barium fluoride.

The formula for barium fluoride is $Ba{F}_2$ .

$Ba{F}_2\left(s\right)\iff B{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)$ 

 $\text{Ksp = }\left[{\text{Ba}}^{\text{2 + }}\right]{\left[{\text{F}}^{\text{ - }}\right]}^{\text{2}}$

We squared  $\left[{\text{F}}^{-}\right]$ because we have two mols of ${\text{F}}^{-}$.

Therefore, the ion-product expression for barium fluoride is  $\text{Ksp = }\left[{\text{Ba}}^{\text{2 + }}\right]{\left[{\text{F}}^{\text{ - }}\right]}^{\text{2}}$

Obtaining ion-product expression for copper Barium Fluoride

Let us obtain the ion-product expression for copper $\left(\mathrm{II}\right)$sulphide.

The formula for copper $\left(\mathrm{II}\right)$sulphide is $\mathrm{CuS}$.

$CuS\left(s\right)\iff C{u}^{2+}\left(aq\right)+S^{2-}\left(aq\right)$

width="116" height="18" style="max-width: none; vertical-align: -4px;" $\text{Ksp = }\left[{\text{Cu}}^{\text{2 + }}\right]\left[{\text{S}}^{\text{2 - }}\right]$

Therefore, the ion-product expression for copper $\left(\mathrm{II}\right)$sulphide is $\text{Ksp = }\left[{\text{Cu}}^{\text{2 + }}\right]\left[{\text{S}}^{\text{2 - }}\right]$.