Chapter 17.3, Problem 24PP

Interpretation Introduction

Interpretation:

The solubility product is to be calculated.

Concept introduction:

The solubility product is the product of concentration of its constituent ions with each ion concentration term raised to the power equal to number of ions produced or dissociated.

Answer to Problem 24PP

The solubility product of $\text{AgCl}$ is $\text{1}{\text{.68×10}}^{\text{-5}}$.

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

$\text{AgCl}\rightleftharpoons {\text{Ag}}^{\text{+}}{\text{+Cl}}^{-}$

I $\text{= 0 0 }$ (initial concentration)

C $\text{= +x +x }$ (change in concentration or change in amount from initial state to equilibrium)

E $\text{= x x }$ (equilibrium concentration)

Given information:

Solubility $\text{=1}{\text{.86×10}}^{\text{-4}}\text{g/100}$

$\begin{array}{l}\\ =\frac{1.86\times {10}^{-4}\times 1000}{100}\\ =1.86\times {10}^{-3}\\ \end{array}$

Molar mass of silver iodide = $\text{143}\text{.32 g/mol}$

$\begin{array}{l}\text{=}\frac{\text{1}{\text{.86×10}}^{\text{-3}}}{\text{143}\text{.32}}\\ \text{=0}{\text{.012978×10}}^{\text{-3}}\text{mol/L}\\ \text{=1}{\text{.297×10}}^{\text{-5}}\text{mol/L}\end{array}$

SO, The solubility of silver ion and chlorine ion are equal whose value is $\text{=1}{\text{.297×10}}^{\text{-5}}\text{mol/L}$.

The solubility product is calculated as,

$\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\left[{\text{Ag}}^{\text{+}}\right]\text{×}\left[{\text{Cl}}^{\text{-}}\right]\\ {\text{K}}_{\text{sp}}\text{=}\left(\text{1}{\text{.297×10}}^{\text{-5}}\right)\text{×}\left(\text{1}{\text{.297×10}}^{\text{-5}}\right)\\ {\text{K}}_{\text{sp}}\text{=1}{\text{.68×10}}^{\text{-10}}\end{array}$

The solubility product of $\text{AgCl}$ is $\text{1}{\text{.68×10}}^{\text{-10}}$.

Conclusion

The solubility product of $\text{AgCl}$ is $\text{1}{\text{.68×10}}^{\text{-5}}$.