Q22.85CP - Chemistry: Molecular Nature Of Matter And Change

Question

How does acid rain affect the leaching of phosphate into groundwater from terrestrial phosphate rock? Calculate the solubility of ${\mathbf{\text{Ca}}}_{\mathbf{\text{3}}}{\mathbf{\text{(PO}}}_{\mathbf{\text{4}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$ in each of the following: (a) Pure water, pH  (Assume that $\mathbf{\text{P}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{3 - }}}$ does not react with water.) (b) Moderately acidic rainwater, pH  $\mathbf{\text{4.5}}$(Hint: Assume that all the phosphate exists in the form that predominates at this pH.)

Step-by-Step Solution

Answer

a) The calcium phosphate has a solubility of $6.4439\times {10}^{-7}$M.

b) The solubility of acidic rainwater is:$1.0969\times {10}^{-2}M$ .

Step 1: Define Elements

An element is a pure material made up entirely of atoms with the same number of protons in their nuclei, as defined by chemistry. Chemical elements, unlike chemical compounds, cannot be broken down into smaller substances by chemical reactions.

Step 2: Calculation of solubility in pure water

When phosphate is protonated to generate hydrogen phosphate (${\text{HPO}}_{\text{4}}^{\text{2 - }}$) and dihydrogen phosphate (${\text{HPO}}_{\text{4}}^{\text{2 - }}$), acid rain increases the leaching of phosphates, PO, into groundwater (${\text{PO}}_{\text{4}}^{\text{3 - }}$).

a) ${\text{Ca}}_{\text{3}}{\text{(PO}}_{\text{4}}{\text{)}}_{\text{2}}$ is a soluble terrestrial phosphate rock, and the solubility of salt may be calculated using the appropriate solubility constant${\text{K}}_{\text{sp}}$ .

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

 ${\text{Ca}}_{\text{3}}{\text{(PO}}_{\text{4}}{\text{)}}_{\text{2}}$has a solubility product constant of $1.2\times {10}^{-29}$ in pure water (pH ) according to Appendix C, whereas the solubility product constant may be represented as:

$$\begin{gathered} {\text{K}}_{\text{sp}}\text{ = }\frac{{{\text{[Ca}}^{\text{2 + }}\text{]}}^{\text{3}}{{\text{[PO}}_{\text{4}}^{\text{3 - }}\text{]}}^{\text{2}}}{{\text{[Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}} \\ {\text{=1.2×10}}^{-29} \end{gathered}$$

As calcium phosphate is a solid, its concentration is one, hence it may be left out of the equation:

$$\begin{gathered} {\text{K}}_{\text{sp}}{\text{ = [Ca}}^{\text{2 + }}{\text{]}}^{\text{3}}{\text{[PO}}_{\text{4}}^{\text{3 - }}{\text{]}}^{\text{2}} \\ {\text{=1.2×10}}^{-29} \end{gathered}$$

If we assume that the starting concentration of  was one and that the ions concentration was , then after the reaction, some  concentration was acquired by  ions and some  concentration was obtained by  ions at equilibrium, as shown in Table below:

As a result, the equilibrium${\text{K}}_{\text{sp}}$may be expressed as

$$\begin{gathered} {\text{K}}_{\text{sp}}{\text{ = [3x]}}^{\text{3}}{\text{[2x]}}^{\text{2}} \\ {\text{=1.2×10}}^{-29} \end{gathered}$$

Then solving the x equation as:

$$\begin{gathered} 27\times x^3\times 4\times x^2=1.2\times {10}^{-29} \\ 108\times x^5=1.2\times {10}^{-29} \\ x=\sqrt{\frac{1.2\times {10}^{-29}}{108}} \\ x=6.4439\times {10}^{-7} \end{gathered}$$

Therefore, the calcium phosphate has a solubility of $6.4439\times {10}^{-7}$M at pH .

Step 3: Calculation of solubility in acidic rainwater

b) The pH of the surroundings has a significant influence on the equilibrium since phosphate is a conjugate base of a weak phosphoric acid.

Each time a ${\text{H}}^{\text{ + }}$ ion is gained from phosphate:

${\text{PO}}_{\text{4}}^{\text{3 - }}\to {\text{HPO}}_{\text{4}}^{\text{2 - }}\to {\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

In light of the ${\text{K}}_{\text{a}}$${\text{K}}_{\text{a}}$ values in Appendix C,