Q21.145 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: Evaluate the value of K sp

Silver is oxidized during the production of AgCl.

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

 $∆{\text{G}}^{\text{o}}\text{ = - RTlnK}$ 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: ${\text{K}}_{\text{sp}}{\text{ = 1.90××10}}^{\text{ - 4}}$