Chapter 22, Problem 22.80P

(a)

Interpretation Introduction

Interpretation:

The affect of acid rain on the leaching of phosphate into ground water from terrestrial phosphate rock has to be given.  The solubility of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ in pure water with $\text{pH 7}\text{.0}$ has to be calculated.

Explanation of Solution

The solubility of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ can be calculated from its ${\text{K}}_{\text{sp}}$.  It is $\text{1}{\text{.2×10}}^{\text{-29}}$.

The phosphate ion is derived from a weak acid.  the $\text{pH}$ of the solution impacts the acid-base equilibrium.

The solubility of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ in pure water can be calculated by using the equilibrium expression and concentration of ${\text{H}}^{\text{+}}$.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\rightleftarrows {\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2PO}}_{\text{4}}^{\text{3-}}{}_{\text{(aq)}}\\ \text{Initial}\text{-}\text{0}\text{0}\\ \text{Change}\text{-x}\text{+3x}\text{+2x}\end{array}}\\ \text{Equilibrium}\text{-}\text{3x}\text{2x}\end{array}$

The equilibrium constant can be calculated as

  $\begin{array}{l}{\text{K}}_{\text{sp}}{\text{= [Ca}}^{\text{2+}}{\text{]}}^{\text{3}}{{\text{[PO}}_{\text{4}}^{\text{3-}}\text{]}}^{\text{2}}\\ \text{1}{\text{.2×10}}^{\text{-29}}{\text{ = (3x)}}^{\text{3}}{\text{(2x)}}^{\text{2}}{\text{ = 108x}}^{\text{5}}\\ \text{x = 6}{\text{.4439×10}}^{\text{-7}}\text{M}\end{array}$

The solubility of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ in pure water is $\text{6}{\text{.4×10}}^{\text{-7}}\text{M}$.

(b)

Interpretation Introduction

Interpretation:

The affect of acid rain on the leaching of phosphate into ground water from terrestrial phosphate rock has to be given.  The solubility of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ in moderately acidic rain water with $\text{pH 4}\text{.5}$ has to be calculated.

Explanation of Solution

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ at $\text{pH 4}\text{.5}$ is

  ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = 10}}^{\text{-4}\text{.5}}\text{ = 3}{\text{.162×10}}^{\text{-5}}\text{M}$

The phosphate ion can gain one ${\text{H}}^{\text{+}}$ and forms ${\text{HPO}}_{\text{4}}^{\text{2-}}$. On further gain of one more ${\text{H}}^{\text{+}}$, it forms ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$, on addition of last ${\text{H}}^{\text{+}}$, it forms ${\text{H}}_3{\text{PO}}_{\text{4}}^{}$.  The ${\text{K}}_{\text{a}}$ values for phosphate ions are ${\text{K}}_{{\text{a}}_{\text{1}}}\text{=7}{\text{.2×10}}^{\text{-3}}\text{,}{\text{K}}_{{\text{a}}_{\text{2}}}\text{=6}{\text{.3×10}}^{\text{-8}}\text{and}{\text{K}}_{{\text{a}}_{\text{3}}}\text{=4}{\text{.2×10}}^{\text{-13}}$.

  ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}{\text{+ H}}_{\text{3}}{\text{O}}^{\text{+}}$

The equilibrium constant can be calculated as

  $\begin{array}{l}{\text{K}}_{{\text{a}}_{\text{1}}}\text{= }\frac{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}}\\ \frac{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}}\text{=}\frac{{\text{K}}_{{\text{a}}_{\text{1}}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \frac{\text{7}{\text{.2×10}}^{\text{-3}}}{\text{3}{\text{.162×10}}^{\text{-5}}}\text{=227}\text{.70}\end{array}$

  ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{HPO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}{\text{+ H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium constant can be calculated as

  $\begin{array}{l}{\text{K}}_{{\text{a}}_{\text{2}}}\text{= }\frac{{\text{[HPO}}_{\text{4}}^{\text{2-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}}\\ \frac{{\text{[HPO}}_{\text{4}}^{\text{2-}}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}}\text{=}\frac{{\text{K}}_{{\text{a}}_{\text{2}}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \frac{\text{6}{\text{.3×10}}^{\text{-8}}}{\text{3}{\text{.162×10}}^{\text{-5}}}\text{=1}{\text{.9924×10}}^{\text{-3}}\end{array}$

  ${\text{HPO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{PO}}_{\text{4}}^{\text{3-}}{}_{\text{(aq)}}{\text{+ H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium constant can be calculated as

  $\begin{array}{l}{\text{K}}_{{\text{a}}_{\text{3}}}\text{= }\frac{{\text{[PO}}_{\text{4}}^{\text{3-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HPO}}_{\text{4}}^{\text{2-}}\text{]}}\\ \frac{{\text{[PO}}_{\text{4}}^{\text{3-}}\text{]}}{{\text{[HPO}}_{\text{4}}^{\text{2-}}\text{]}}\text{=}\frac{{\text{K}}_{{\text{a}}_{\text{3}}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \frac{\text{4}{\text{.2×10}}^{\text{-13}}}{\text{3}{\text{.162×10}}^{\text{-5}}}\text{=1}{\text{.32827×10}}^{\text{-8}}\end{array}$

From the above calculations, the concentration of ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$ is at least $\text{1000}$ times more than the concentration of any other species.  It is the dominant species and from this the value of ${\text{[PO}}_{\text{4}}^{\text{3-}}\text{]}$ is calculated.

  $\begin{array}{l}\frac{{\text{[PO}}_{\text{4}}^{\text{3-}}\text{]}}{{\text{[HPO}}_{\text{4}}^{\text{2-}}\text{]}}\text{ = 1}{\text{.32827×10}}^{\text{-8}}\text{and}\frac{{\text{[HPO}}_{\text{4}}^{\text{2-}}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}}\text{ = 1}{\text{.9924×10}}^{\text{-3}}\\ {\text{[PO}}_{\text{4}}^{\text{3-}}\text{] = (1}{\text{.32827×10}}^{\text{-8}}{\text{)[HPO}}_{\text{4}}^{\text{2-}}\text{]}\text{and}{\text{[HPO}}_{\text{4}}^{\text{2-}}\text{] = (1}{\text{.9924×10}}^{\text{-3}}{\text{)[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}\\ {\text{[PO}}_{\text{4}}^{\text{3-}}\text{] = (1}{\text{.32827×10}}^{\text{-8}}{\text{)[HPO}}_{\text{4}}^{\text{2-}}\text{] = (1}{\text{.32827×10}}^{\text{-8}}\text{)(1}{\text{.9924×10}}^{\text{-3}}{\text{)[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}\\ {\text{[PO}}_{\text{4}}^{\text{3-}}\text{] = (2}{\text{.6464×10}}^{\text{-11}}{\text{)[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}\end{array}$

Substitute this in the ${\text{K}}_{\text{sp}}$ expression,

  $\begin{array}{l}{\text{K}}_{\text{sp}}{\text{ = [Ca}}^{\text{2+}}{\text{]}}^{\text{3}}{{\text{[PO}}_{\text{4}}^{\text{3-}}\text{]}}^{\text{2}}\text{ = 1}{\text{.2×10}}^{\text{-29}}\\ {\text{K}}_{\text{sp}}{\text{ = [Ca}}^{\text{2+}}{\text{]}}^{\text{3}}{\text{{(2}{\text{.6464×10}}^{\text{-11}}{\text{)[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}}}^{\text{2}}\end{array}$

Rearrange,

  ${\text{K}}_{\text{sp}}\text{/(2}{\text{.6464×10}}^{\text{-11}}{\text{)}}^{\text{2}}{\text{ = [Ca}}^{\text{2+}}{\text{]}}^{\text{3}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}}^{\text{2}}$

The concentration of calcium ions is represented as $\text{3x}$ and the concentration of dihydrogen phosphate ion as $\text{2x}$.

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{/(2}{\text{.6464×10}}^{\text{-11}}{\text{)}}^{\text{2}}{\text{ = [3x]}}^{\text{3}}{\text{[2x]}}^{\text{2}}\\ \frac{\text{(1}{\text{.2×10}}^{\text{-29}}\text{)}}{{\text{(2}{\text{.6464×10}}^{\text{-11}}\text{)}}^{\text{2}}}{\text{ = 108x}}^{\text{5}}\\ \text{x = 1}{\text{.0967×10}}^{\text{-2}}\text{ = 1}{\text{.1×10}}^{\text{-2}}\text{M}\end{array}$

Acid rains increases the leaching of phosphate into ground water from terrestrial phosphate rock due to the protonation of ${\text{PO}}_{\text{4}}^{\text{3-}}{\text{ to HPO}}_{\text{4}}^{\text{2-}}\text{and}{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$.

From the calculations of (a) and (b), the solubility increases from pure water $\text{6}{\text{.4×10}}^{\text{-7}}\text{M}$ to acidic rain water $\text{1}{\text{.1×10}}^{\text{-2}}\text{M}$.