Q19.69P - Chemistry: Molecular Nature Of Matter And Change

Question

Write the ion-product expressions for

$\left(\text{a}\right)\mathbf{\text{ lead(II)iodide; }}\left(\text{b}\right)\mathbf{\text{ strontium sulfate; }}\left(\text{c}\right)\mathbf{\text{ cadmium sulfide.}}$

Step-by-Step Solution

Answer

The ion-product expression of the compounds is,

$$\begin{gathered} \left(\text{a}\right)\text{ Lead(II) iodide: Ksp = }\left[{\text{Pb}}^{\text{2 + }}\right]{\left[{\text{I}}^{\text{ - }}\right]}^{\text{2}} \\ \left(\text{b}\right)\text{ Strontium sulfate: Ksp = }\left[{\text{Sr}}^{\text{2 + }}\right]\left[{\text{SO}}_{\text{4}}^{\text{2 - }}\right]. \\ \left(\text{c}\right)\text{ Cadmium sulfide: Ksp = }\left[{\text{Cd}}^{\text{2 + }}\right]\left[{\text{S}}^{\text{2 - }}\right] \end{gathered}$$

Concept Introduction

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

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

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 Lead Iodide

Let us obtain the ion-product expression for lead (II) iodide.

The formula for lead (II) iodide is ${\mathrm{PbI}}_2$.

${\mathrm{PbI}}_2\left(\mathrm{s}\right)\iff {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)$

We squared $\left[{\text{I}}^{\text{ - }}\right]$ because we have 2 mols of in the equation.

Therefore, the ion-product expression for lead $\left(\mathrm{II}\right)$ iodide is $\text{Ksp = }\left[{\text{Pb}}^{\text{2 + }}\right]{\left[{\text{I}}^{\text{ - }}\right]}^{\text{2}}$.

Obtaining ion-product expression for Strontium Sulfate

Let us obtain the ion-product expression for strontium sulphate.

The formula for silver cyanide is $SrS{O}_4$.

${\mathrm{SrSO}}_4\left(\mathrm{s}\right)\iff {\mathrm{Sr}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$

$\text{Ksp = }\left[{\text{Sr}}^{\text{2 + }}\right]\left[{\text{SO}}_{\text{4}}^{\text{2 - }}\right]$

Therefore, the ion-product expression for strontium sulphate is$\left[{\text{Sr}}^{\text{2 + }}\right]\left[{\text{SO}}_{\text{4}}^{\text{2 - }}\right]$.

Obtaining ion-product expression for Cadmium Sulfide

Let us obtain the ion-product expression for cadmium sulfide.

The formula for cadmium sulfide is $CdS$.

$$\begin{gathered} \mathrm{CdS}\left(\mathrm{s}\right)\iff {\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right) \\ \mathrm{Ksp}=\left[{\mathrm{Cd}}^{2+}\right]\left[{\mathrm{S}}^{2-}\right] \end{gathered}$$

Therefore, the ion-product expression for cadmium sulfide is $\left[C{d}^{2+}\right]\left[S^{2-}\right]$.