Chapter 20, Problem 91A

Interpretation Introduction

Interpretation:

the solubility of silver iodide is to be calculated.

Concept introduction:

The solubility product of a solute is the product of solubility of its constituent ions.

Answer to Problem 91A

The solubility of silver iodide is 5.916x10-9 mol/L.

Explanation of Solution

Let the amount of silver iodide dissolved at equilibrium be x mol/L.

Reaction for dissocation of AgI:

$AgI\to A{g}^{+}+I^{-}$

Solubility product:

$\begin{array}{l}{K}_{sp}=[A{g}^{+}][I^{-}]\\ K_{sp}=x\ast x\\ 3.5\times {10}^{-17}=x^2\\ x=\sqrt{3.5\times {10}^{17}}\\ x=5.916\times {10}^{-9}\end{array}$

Conclusion

The solubility of silver iodide was found to be 5.916x10-9 mol/L using its solubility product.