Question

Identify all chemical species present in an aqueous solution of \(C{a_3}{\left( {P{O_4}} \right)_2}\)and list these species in decreasing order of their concentrations. (Hint: Remember that the \(PO_4^{3 - }\) ion is a weak base.)

Short answer

Decreasing order of their concentrations

\(\left[ {{\rm{C}}{{\rm{a}}^{3 + }}} \right] > \left[ {{\rm{O}}{{\rm{H}}^ - }} \right] > \left[ {{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}} \right] > \left[ {{{\rm{H}}_2}{\rm{PO}}_4^ - } \right] > \left[ {{\rm{HP}}{{\rm{O}}_4}^{2 - }} \right] > \left[ {{\rm{P}}{{\rm{O}}_4}^{3 - }} \right]\)

Step-by-step solution

Define chemical species

A chemical species is a chemical substance or ensemble composed of chemically identical molecular entities that can explore the same set of molecular energy levels on a characteristic or delineated time scale. These energy levels determine the way the chemical species will interact with others.

Calculating decreasing order of their first and second concentrations

First reaction\( - {\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2}\)is a salt, so it ionizes completely

The highest concentration\( - \left[ {{\rm{C}}{{\rm{a}}^{2 + }}} \right]\)

Second reaction

\({\rm{\; - \;}}{{\rm{K}}_a}{\rm{\;of\;HPO}}_4^{2 - }{\rm{\;is\;}}4.2 \cdot {10^{ - 13}}\)

\({\rm{\; - Hence,\;}}{{\rm{K}}_b}{\rm{\;of\;PO}}_4^{3 - }{\rm{\;is\;}}\frac{1}{{{K_a}}} = \frac{1}{{4.2 - {{10}^{ - 1S}}}} = 2.38 \cdot {10^{12}}\)

The lowest concentration \( - \left[ {{\rm{PO}}_4^{3 - }} \right]\)

 Step 3: Calculating decreasing order of their third and fourth concentrations

Third reaction

\({{\rm{K}}_a}{\rm{\;of\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - {\rm{is\;}}6.2 \cdot {10^{ - 8}}\)

\({\rm{\;Hence,\;}}{{\rm{K}}_b}{\rm{\;of\;HPO}}_4^{3 - }{\rm{\;is\;}}\frac{1}{{{K_a}}} = \frac{1}{{6.2 - {{10}^{ - \infty }}}} = 1.61 \cdot {10^7}\)

Since\({K_b}\)is too large, almost all\({\rm{HP}}{{\rm{O}}_4}^{2 - }\)will react with water to produce\({{\rm{H}}_2}{\rm{PO}}_4^ - \)

The second lowest concentration\( - \left[ {{\rm{HPO}}_4^{2 - }} \right]\)

Fourth reaction

\({{\rm{K}}_a}{\rm{\;of\;}}{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}{\rm{\;is\;}}7.5 \cdot {10^{ - 3}}\)

\({{\rm{K}}_b}{\rm{\;of\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - {\rm{is\;}}\frac{1}{{{K_a}}} = \frac{1}{{7.5 \cdot {{10}^{ - 3}}}} = 1.33 \cdot {10^2}\)

Since\({{\rm{K}}_b}\)is too large, most of\({{\rm{H}}_2}{\rm{P}}{{\rm{O}}_4} - \)will react with water to produce\({{\rm{H}}_3}{\rm{PO}}_4^ - \)

The third lowest concentration -\(\left[ {{{\rm{H}}_2}{\rm{PO}}_4^ - } \right]\)

Decreasing order of their concentrations

\(\left[ {{\rm{C}}{{\rm{a}}^{3 + }}} \right] > \left[ {{\rm{O}}{{\rm{H}}^ - }} \right] > \left[ {{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}} \right] > \left[ {{{\rm{H}}_2}{\rm{PO}}_4^ - } \right] > \left[ {{\rm{HP}}{{\rm{O}}_4}^{2 - }} \right] > \left[ {{\rm{P}}{{\rm{O}}_4}^{3 - }} \right]\)