Chapter 17, Problem 83A

Interpretation Introduction

Interpretation:

The concentration of cadmium iodate is to be determined when solubility product is given at a particular temperature.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

The solubility product is given as

${\text{K}}_{\text{sp}}\text{ = }\left[{\text{Cd}}^{\text{+2}}\right]\text{ ×}{\left[{\text{IO}}_{\text{3}}{}^{\text{-1}}\right]}^{\text{2}}$

Answer to Problem 83A

The molar concentration of cadmium ion and iodate ion are $\text{1}{\text{.79×10}}^{\text{-3}}\text{mol/L}$ and $\text{3}{\text{.58×10}}^{\text{-3}}\text{mol/L}$, respectively.

Explanation of Solution

Given information : The solubility product of cadmium iodate is $\text{2}{\text{.3×10}}^{\text{-8}}$.

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

$\text{ Cd}{\left({\text{IO}}_{\text{3}}\right)}_{\text{2}}\rightleftharpoons {\text{Cd}}^{\text{2+}}{\text{+ 2IO}}_{\text{3}}{}^{\text{-1}}$

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

$\text{C = +x +2x }$ (Change in concentration or change in amount from initial state to equilibrium)

$\text{E= x 2x }$ (Equilibrium concentration)

The solubility product is given as

$\begin{array}{l}\hfill \\ {\text{ K}}_{\text{sp}}\text{= }\left[{\text{Cd}}^{\text{2+}}\right]\text{ × }{\left[{\text{IO}}_{\text{3}}{}^{\text{1-}}\right]}^{\text{2}}\hfill \\ {\text{ = x × 4x}}^{\text{2}}\hfill \\ {\text{ = 4x}}^{\text{3}}\hfill \end{array}$

$\sqrt[\text{3}]{{\text{(K}}_{\text{sp}}\text{/4 }}\text{= x}$

Or, $\text{x = 1}{\text{.79 ×10}}^{\text{-3}}\text{mol/L}$

The concentration of cadmium ion is $\text{1}{\text{.79 ×10}}^{\text{-3}}\text{mol/L}\text{.}$

And the concentration of iodate ion is $\text{= 2×x = 2×1}{\text{.79 ×10}}^{\text{-3}}{\text{=}}^{}\text{3}{\text{.58 ×10}}^{\text{-3}}\text{mol/L}\text{.}$

Or,

Conclusion

The molar concentration of cadmium ion and iodate ion are $\text{1}{\text{.79×10}}^{\text{-3}}\text{mol/L}$ and $\text{3}{\text{.58×10}}^{\text{-3}}$ mol/L, respectively.