Question

Use Appendix D to calculate the${\mathbf{K}}_{\mathbf{sp}}$of AgCl.

Short answer

The value is obtained as: ${\text{K}}_{\text{sp}}{\text{=1.90×10}}^{\text{- 4}}$

Step-by-step solution

Define chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

Evaluate the value of  Ksp

Silver is oxidized during the production of AgCl.

$$\begin{gathered} {\text{Reaction: Ag}}_{\text{(s)}}{\text{+ Cl}}_{\text{(aq)}}^{\text{-}}\to {\text{AgCl}}_{\text{(s)}}{\text{+e}}^{\text{-}} \\ {E}_{\text{half - cell}}^0\text{= - 0.22} \\ ∆{\text{G}}^{\text{o}}{\text{= - nFE}}_{Cell}^0 \\ {\text{= - 1e}}^{\text{-}}\text{×96,500}\frac{\text{C}}{{\text{mole}}^{\text{-}}}\text{×-0.22V} \\ =21,230J \end{gathered}$$

$∆{\text{G}}^{\text{o}}\text{= - RTlnK}$in Action the${\text{K}}_{\text{sp}}$for AgCl may be calculated as:

$\begin{array}{rcl}& & \text{21,230J = - RTlnK}\\ & & \text{lnK =}\frac{\text{- 21,230J}}{\text{RT}}\\ & & {\text{K}}_{\text{sp}}\text{= e}\frac{\text{- 21,230}}{\text{RT}}\\ & & \text{= e}\frac{{\displaystyle \text{- 21,230}}}{{\displaystyle \text{298 K×8.314}\frac{J}{{\displaystyle \text{mol K}}}}}\\ & & {\text{= 1.90×10}}^{\text{- 4}}\end{array}$

Therefore, the value is:${\text{K}}_{\text{sp}}{\text{= 1.90××10}}^{\text{- 4}}$