Chapter 17, Problem 72A

Interpretation Introduction

Interpretation:

The concentration of ${\text{Sc}}^{\text{+3}}$ ion is to be calculated.

Concept introduction:

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

The solubility product is given as

${\text{K}}_{\text{SP}}\text{= }\left[{\text{Sc}}^{\text{+3}}\right]\text{×}{\left[{\text{F}}^{\text{-1}}\right]}^{\text{3}}$

Answer to Problem 72A

The concentration of ${\text{Sc}}^{\text{+3}}$ ion required to form precipitate is $\text{9}{\text{.6×10}}^{\text{-15}}\text{M}$

Explanation of Solution

Given information : The solubility product of scandium fluoride is $\text{4}{\text{.2×10}}^{\text{-18}}$.

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

${\text{ScF}}_{\text{3}}\underset{}{\rightleftharpoons }{\text{Sc}}^{\text{3+}}{\text{+ 3F}}^{\text{-1}}$

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

$\text{C = +x + 3x }$ (Change in concentration or change in amount from initial state to equilibrium)

$\text{ E= x 3x }$ (Equilibrium concentration)

The solubility product is given as

${\text{K}}_{\text{sp}}\text{= }\left[{\text{Sc}}^{\text{3+}}\right]\text{×}{\left[{\text{F}}^{\text{1-}}\right]}^{\text{3}}$

$\begin{array}{l}{\text{K}}_{\text{sp}}\text{= }\left[{\text{Sc}}^{\text{3+}}\right]\text{×}{\left[{\text{F}}^{\text{1-}}\right]}^{\text{3}}\\ \left[{\text{Sc}}^{\text{3+}}\right]\text{=}\frac{{\text{K}}_{\text{sp}}}{{\left[{\text{F}}^{\text{1-}}\right]}^{\text{3}}}\\ \text{=}\frac{\text{4}{\text{.2×10}}^{\text{-18}}}{{\left(\text{0}\text{.076}\right)}^{\text{3}}}\\ \text{=}\frac{\text{4}{\text{.2×10}}^{\text{-14}}}{\text{4}\text{.39}}\\ \text{=9}{\text{.6×10}}^{\text{-15}}\text{M}\end{array}$

Precipitation is occurs when ionic product is greater than solubility product.

The concentration of ${\text{Sc}}^{\text{+3}}$ ion required to form the precipitation form is $\text{9}{\text{.6×10}}^{\text{-15}}\text{M}$

Conclusion

The concentration of ${\text{Sc}}^{\text{+3}}$ ion required to form precipitate is $\text{9}{\text{.6×10}}^{\text{-15}}\text{M}$