Chapter 17, Problem 89A

Interpretation Introduction

Interpretation:

The greater molar solubility of calcium phosphate or iron phosphate 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

${\text{K}}_{\text{sp}}\text{ = }{\left[{\text{Ca}}^{\text{+2}}\right]}^{\text{3}}\text{×}{\left[{\text{PO}}_{\text{4}}{}^{\text{-3}}\right]}^{\text{2}}$

And

${\text{K}}_{\text{sp}}\text{= }\left[{\text{Fe}}^{\text{+3}}\right]\text{×}{\left[{\text{PO}}_{\text{4}}{}^{\text{-3}}\right]}^{}$

Answer to Problem 89A

The molar solubility of calcium phosphate is $\text{0}{\text{.644× 10}}^{\text{-6}}$ mol/L which is greater than that of iron phosphate whose value is ${\text{1×10}}^{\text{-11}}$ mol/L. the molar solubility of calcium phosphate and iron phosphate in g/L are $\text{199}{\text{.75×10}}^{\text{-6}}$ and $\text{150}{\text{.82×10}}^{\text{-11}}$ g/L respectively.

Explanation of Solution

Given information : The solubility product of calcium phosphate and iron phosphate are $\text{1}{\text{.2×10}}^{\text{-29}}$ and $\text{1}{\text{.0×10}}^{\text{-22}}$ respectively.

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

${\text{Ca}}_{\text{3(}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}\rightleftharpoons {\text{3Ca}}^{\text{2+}}{\text{+ 2PO}}_{\text{4}}{}^{\text{-3}}$

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

$\text{C = +3x +2x }$ (Change in concentration or change in amount from initial state to equilibrium)

$\text{ E= 3x 2x }$ (Equilibrium concentration)

The solubility product is given as

$\begin{array}{l}{\text{ K}}_{\text{sp}}\text{= }{\left[{\text{Ca}}^{\text{2+}}\right]}^{\text{3}}\text{× }{\left[{\text{PO}}_{\text{4}}{}^{\text{3-}}\right]}^{\text{2}}\hfill \\ \text{ = }{\left(\text{3x}\right)}^{\text{3}}\text{×}{\left(\text{2x}\right)}^{\text{2}}\hfill \\ {\text{ = 108x}}^{\text{5}}\hfill \end{array}$

$\sqrt[\text{5}]{\left({\text{K}}_{\text{sp}}\text{/108}\right)}{\text{= x}}^{}$

Or, $\begin{array}{l}\text{}\text{x =}\sqrt[\text{5}]{\left(\text{1}{\text{.2x10}}^{\text{-29}}\text{/ 108}\right)}\hfill \\ {}^{}\text{ =}\sqrt[\text{5}]{\text{0}{\text{.0111 ×10}}^{\text{-29}}}\text{mol/L}\hfill \\ \text{ =}\sqrt[\text{5}]{\text{0}{\text{.111 ×10}}^{\text{-30}}}\text{mol/L}\hfill \\ \text{ =0}{\text{.6443×10}}^{\text{-6}}\text{mol/L}\hfill \end{array}$

Or, The molar solubility is $\left(\text{0}{\text{.6443×10}}^{\text{-6}}\text{mol/L}\right)$

$\text{Fe}\left({\text{PO}}_{\text{4}}\right)\rightleftharpoons {\text{ Fe}}^{\text{3+}}{\text{+ PO}}_{\text{4}}{}^{\text{-3}}$

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

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

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

The solubility product is given as

$\begin{array}{l}{\text{ K}}_{\text{sp}}\text{= }\left[{\text{Fe}}^{\text{2+}}\right]\text{ × }{\left[{\text{PO}}_{\text{4}}{}^{\text{3-}}\right]}^{}\hfill \\ \text{ = }\left(\text{x}\right)\text{ ×}{\left(\text{x}\right)}^{}\hfill \\ {\text{ = x}}^{\text{2}}\hfill \end{array}$

$\sqrt[\text{2}]{\left(\text{Ksp}\right)}\text{ = x}$

Or, $\begin{array}{l}\text{x =}\sqrt[\text{2}]{\left(\text{1}\text{.0}\times {\text{10}}^{\text{-22}}\right)}\hfill \\ {}^{}\text{ = 1}\times {\text{10}}^{\text{-11}}\text{mol/L}\hfill \end{array}$

Or, The molar solubility of iron phosphate is $\left({\text{1×10}}^{\text{-11}}\text{mol/L}\right)$

The molar solubility of iron phosphate in g/L is $\text{150}{\text{.82 ×10}}^{\text{-11}}$ g/L

Hence, on comparison we find that the molar solubility of calcium phosphate is greater than iron phosphate.

Conclusion

The molar solubility of calcium phosphate is $\text{0}{\text{.644× 10}}^{\text{-6}}\text{mol/L}$ which is greater than that of iron phosphate whose value is ${\text{1×10}}^{\text{-11}}\text{mol/L}$. The molar solubility of calcium phosphate and iron phosphate in g/L are $\text{199}{\text{.75×10}}^{\text{-6}}$ and $\text{150}{\text{.82×10}}^{\text{-11}}$ g/L respectively.