Question

Calculate K_sp for iron(II) sulfide given the following data:
$$\begin{gathered} \mathit{F}\mathit{e}\mathit{S}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{F}\mathit{e}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{S}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{ }\mathit{V} \\ \mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{F}\mathit{e}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{44}\mathbf{ }\mathit{V} \end{gathered}$$

Short answer

Thevalue of K_sp for iron (II) sulphide is$5\times {10}^{-20}.$

Step-by-step solution

Equilibrium equation of Iron (II) sulfide and half-reactions

The equilibrium equation can be denoted as:

$FeS\left(s\right)\to F{e}^{2+}\left(aq\right)+S^{2-}\left(aq\right)---\left(1\right)$

The half-reactions can be given as:

$$\begin{gathered} FeS\left(s\right)+2e^{-}\to F{e}^{2+}\left(aq\right)+S^{2-}\left(aq\right) E^o=-1.01 V---\left(2\right) \\ F{e}^{2+}\left(aq\right)+2e^{-}\to FeS\left(s\right) E^o=-0.44 V--\left(3\right) \end{gathered}$$

Calculating the standard potential for the equation

On adding equation (2) and (3),

$FeS\left(s\right)\to F{e}^{2+}\left(aq\right)+S^{2-}\left(aq\right)$

On calculating the standard potential,

$$\begin{gathered} E^o=(-1.01 V)+(0.44 V) \\ =-0.57 V \\ \log K_{sp}=\frac{n{E^o}_{cell}}{0.0591} \\ \log K_{sp}=\frac{2\times (-0.57)}{0.0591} \\ \log K_{sp}=-19.3 \\ {K}_{sp}=5\times {10}^{-20} \end{gathered}$$