Chapter 18, Problem 125CP

Interpretation Introduction

Interpretation: The concentration of ${\text{Mg}}^{\text{2+}}$ needs to be calculated in a solution that contains 50 ppm ${\text{Mg}}^{\text{2+}}$ after 40 g of ${\text{Na}}_{\text{5}}{\text{P}}_{\text{3}}{\text{O}}_{\text{10}}$ is added to a solution of 1.0 L.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 125CP

The concentration of ${\text{Mg}}^{\text{2+}}$ is $\text{4}\text{.8}\times {\text{10}}^{-11}\text{ M}$ .

Explanation of Solution

Given:

  $\text{pK = -8}\text{.60}$ for the formation of ${\text{MgP}}_3{\text{O}}_{\text{10}}^{-3}$ .

The reaction for the formation of ${\text{MgP}}_3{\text{O}}_{\text{10}}^{-3}$ is:

  ${\text{Mg}}^{2\text{+}}{\text{ + P}}_{\text{3}}{\text{O}}_{\text{10}}^{-5}\rightleftharpoons {\text{ MgP}}_3{\text{O}}_{\text{10}}^{-3}$ , $\text{pK = -8}\text{.60}$

The value of formation constant is calculated as follows:

  $\begin{array}{l}{\text{K = 10}}^{-\text{pK}}\\ \text{K }={\text{ 10}}^{-(-8.6)}\\ \text{K = 4}\text{.0 }\times {\text{ 10}}^8\end{array}$

From the large value of $\text{K}$ , it can be assumed that the reaction undergoes completion.

For obtaining ICE table, the concentration of the reactants involved in the formation of ${\text{MgP}}_3{\text{O}}_{\text{10}}^{-3}$ is calculated using formula:

  $\text{Molarity = }\frac{\frac{\text{mass of the solute}}{\text{Molar mass of the solute}}}{\text{Volume of solution (L)}}$

Substituting the values:

Molar mass of Mg 24.3 g/mol

  $\begin{array}{l}{\text{[Mg}}^{2+}]\text{ = }\frac{\text{50 mg }\times \frac{{10}^{-3}\text{ g}}{1\text{ mg}}}{\text{L}}\times \frac{1\text{ mol}}{\text{24}\text{.3 g}}\\ {\text{[Mg}}^{2+}]\text{ = }\frac{\text{50 }\times {10}^{-3}\text{ g }}{\text{L}}\times \frac{1\text{ mol}}{\text{24}\text{.3 g}}\\ {\text{[Mg}}^{2+}]=\text{ 2}\text{.06}\times {\text{ 10}}^{-3}\text{ M}\simeq \text{2}\text{.1 }\times {\text{ 10}}^{-3}\text{ M}\end{array}$

Molar mass of ${\text{Na}}_{\text{5}}{\text{P}}_{\text{3}}{\text{O}}_{\text{10}}$ 367.9 g/mol.

The mole ratio of ${\text{P}}_3{\text{O}}_{10}^{5-}:{\text{Na}}_5{\text{P}}_3{\text{O}}_{10}$ is 1:1. So,

  $\begin{array}{l}{\text{[P}}_3{\text{O}}_{10}^{5-}\text{] = }\frac{\text{40}{\text{.g Na}}_5{\text{P}}_3{\text{O}}_{10}}{\text{L}}\times \frac{\text{1 mol}}{\text{367}\text{.9 g}}\times \frac{{\text{1 mol P}}_3{\text{O}}_{10}^{5-}}{{\text{1 mol Na}}_5{\text{P}}_3{\text{O}}_{10}}\\ {\text{[P}}_3{\text{O}}_{10}^{5-}\text{] = 0}\text{.109 M}\simeq \text{0}\text{.11 M}\end{array}$

Thus, the limiting reagent is ${\text{Mg}}^{2+}$ (as it is present in lesser molarity).

So, the ICE table for the formation of ${\text{MgP}}_3{\text{O}}_{\text{10}}^{-3}$ is:

  $\begin{array}{l}{\text{ Mg}}^{2+}{\text{ + P}}_3{\text{O}}_{\text{10}}^{5-}\rightleftharpoons {\text{ MgP}}_3{\text{O}}_{10}^{3-}\\ \text{Before: 2}\text{.1 }\times {\text{ 10}}^{-3}\text{ M 0}\text{.11 M 0}\\ \text{Change: -2}\text{.1 }\times {\text{ 10}}^{-3}\text{ -2}\text{.1 }\times {\text{ 10}}^{-3}\text{ 2}\text{.1 }\times {\text{ 10}}^{-3}\\ \text{After: 0 0}\text{.11 2}\text{.1 }\times {\text{ 10}}^{-3}\end{array}$

Solving for the backward reaction:

  $\begin{array}{l}{\text{ Mg}}^{2+}{\text{ + P}}_3{\text{O}}_{\text{10}}^{-5}\rightleftharpoons {\text{ MgP}}_3{\text{O}}_{10}^{3-}\\ \text{Before: }0\text{ 0}\text{.11 2}\text{.1 }\times {\text{ 10}}^{-3}\\ \text{Change: }+\text{x }+\text{x -x}\\ \text{After: x 0}\text{.11 - x 2}\text{.1 }\times {\text{ 10}}^{-3}\text{ - x }\end{array}$

The expression for the formation constant is:

  $\text{K = }\frac{\left[{\text{MgP}}_3{\text{O}}_{10}^{3-}\right]}{\left[{\text{Mg}}^{2+}\right]\left[{\text{P}}_3{\text{O}}_{10}^{5-}\right]}$

Substituting the values:

  $4.0\times {\text{ 10}}^8\text{ = }\frac{\text{2}\text{.1 }\times {\text{ 10}}^{-3}\text{ - x}}{\text{x(0}\text{.11 + x)}}$

Since $\text{2}\text{.1 }\times {\text{10}}^{-3}\text{ >> x}$ and $\text{0}\text{.11>>x}$ so,

  $\begin{array}{l}4.0\times {\text{10}}^8\text{ = }\frac{\text{2}\text{.1}\times {\text{10}}^{-3}}{\text{x(0}\text{.11)}}\\ \text{x(0}\text{.11)= }\frac{\text{2}\text{.1}\times {\text{10}}^{-3}}{4.0\times {\text{10}}^8}\\ \text{x = }\frac{\text{5}\text{.25}\times {\text{10}}^{-12}}{\text{0}\text{.11}}\\ \text{x = 4}\text{.77}\times {\text{10}}^{-11}\text{ M}\simeq \text{ 4}\text{.8}\times {\text{10}}^{-11}\text{ M}\end{array}$

Hence, the concentration of ${\text{Mg}}^{\text{2+}}$ is $\text{4}\text{.8}\times {\text{10}}^{-11}\text{ M}$ .

Conclusion

  $\text{4}\text{.8}\times {\text{10}}^{-11}\text{ M}$ is the concentration of ${\text{Mg}}^{\text{2+}}$ .