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

Expert verified

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

01

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.

02

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

03

 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

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

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Most popular questions from this chapter

Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.

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\(\begin{array}{*{20}{c}}{CoS(s) \rightleftharpoons C{o^{2 + }}(aq) + {S^{2 - }}(aq)\quad \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{K_{sp}} = 4.5 \times 1{0^{ - 27}}} \\ {{H_2}S(aq) + 2{H_2}O(l) \rightleftharpoons 2{H_3}{O^ + }(aq) + {S^{2 - }}(aq)\quad \;\;\;\;\;\;\;\;\;K = 1.0 \times 1{0^{ - 26}}} \end{array}\;\;\)

Question: What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases, it may be necessary to control the \(pH\).(Hint: Consider the \({K_{sp}}\)values given in

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(b) \(S{O_4}^{2 - }\;and\;C{l^ - }\)

(c) \(H{g^{2 + }}\;and\;C{o^{2 + }}\)

(d) \(Z{n^{2 + }}\;and\;S{r^{2 + }}\)

(e) \(B{a^{2 + }}\;and\;M{g^{2 + }}\)

(f) \(CO_3^{2 - }\;and\;O{H^ - }\)

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