Chapter 17, Problem 133CP

Interpretation Introduction

Interpretation: The expected osmotic pressure of the given solution needs to be calculated.

Concept introduction: The excess pressure applied to a solution in order to prevent the passage of solvent to the solution is said to be osmotic pressure. The formula to determine the osmotic pressure, $\pi$ is:

  $\pi \text{ = iMRT}$

Where $\text{i}$ is van’t Hoff’s factor, $\text{M}$ is molar concentration, $\text{R}$ is ideal gas constant and $\text{T}$ is temperature in K.

Answer to Problem 133CP

The expected osmotic pressure of the given solution is 6.11 atm.

Explanation of Solution

Given:

0.05 mole of ${\text{Fe}}_2{{\text{(SO}}_4)}_3$ is dissolved in water to make a solution of 1.0 L. The ions present in solution is ${\text{SO}}_{\text{4}}^{\text{2-}}{\text{ (aq) and Fe(H}}_{\text{2}}{\text{O)}}_6^{\text{3+}}\text{(aq)}$ ions.

The ${\text{Fe(H}}_{\text{2}}{\text{O)}}_6^{\text{3+}}$ behaves as an acid as:

  ${\text{Fe(H}}_2{\text{O)}}_6^{3+}\rightleftharpoons {\text{ Fe(OH)(H}}_2{\text{O)}}_5^{2+}{\text{ + H}}^{+}$

Converting temperature from degree Celsius to Kelvin as

  $\text{25}+273\text{ = 298 K}$

The dissociation reaction of ${\text{Fe}}_2{({\text{SO}}_4)}_3$ is:

  ${\text{Fe}}_2{({\text{SO}}_4)}_3(\text{aq) }\to {\text{ 2Fe}}^{3+}({\text{aq) + 3SO}}_4^{2-}(\text{aq)}$

From the above reaction it is observed that the number of ions formed are 5. So, $\text{i = 5}$ .

Now, the osmotic pressure is calculated using formula:

  $\pi \text{ = iMRT}$

Substituting the values:

  $\begin{array}{l}\pi \text{ = 5(0}\text{.05 mol/L)(0}\text{.08206 Latm}\cdot {\text{K}}^{-1}{\text{mol}}^{-1})(298\text{ K)}\\ \pi \text{ = 6}\text{.11 atm}\end{array}$

Hence, the expected osmotic pressure of the given solution is 6.11 atm.

Interpretation Introduction

Interpretation: The ${\text{K}}_{\text{a}}$ for the dissociation reaction of ${\text{Fe(H}}_2{\text{O)}}_6^{3+}$ needs to be calculated.

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 133CP

The ${\text{K}}_{\text{a}}$ for the dissociation reaction of ${\text{Fe(H}}_2{\text{O)}}_6^{3+}$ is $\text{8}\text{.3}\times {\text{10}}^{-3}$ .

Explanation of Solution

The contribution of each ion in the osmotic pressure of the solution of ${\text{Fe}}_2{({\text{SO}}_4)}_3$ is determined as follows:

The dissociation reaction of ${\text{Fe}}_2{({\text{SO}}_4)}_3$ is:

  ${\text{Fe}}_2{({\text{SO}}_4)}_3(\text{aq) }\to {\text{ 2Fe}}^{3+}({\text{aq) + 3SO}}_4^{2-}(\text{aq)}$

From the above reaction the contribution of each ion in the osmotic pressure is calculated as:

  $\begin{array}{l}{\text{Fe}}^{3+}\text{ = }\frac{2}{5}\times \text{ 6}\text{.11 atm}\\ {\text{Fe}}^{3+}\text{ = 2}\text{.44 atm}\end{array}$

  $\begin{array}{l}{\text{SO}}_4^{2-}\text{ = }\frac{3}{5}\times \text{ 6}\text{.11 atm}\\ {\text{SO}}_4^{2-}\text{ = 3}\text{.67 atm}\end{array}$

Since, the actual osmotic pressure is 6.73 and the ion that remain same in both the dissociation is ${\text{SO}}_4^{2-}$ so, the contribution of ${\text{SO}}_4^{2-}$ ions in total osmotic pressure is the same.

The contribution of ${\text{Fe(H}}_2{\text{O)}}_6^{3+}\text{ = 6}\text{.73 - 3}\text{.67 = 3}\text{.06 atm}$ .

The initial concentration of ${\text{Fe(H}}_2{\text{O)}}_6^{3+}$ is:

  ${\text{Fe(H}}_2{\text{O)}}_6^{3+}\text{ = 2(0}\text{.0500 M) = 0}\text{.1000 M}$

Creating ICE table for weak acid, ${\text{Fe(H}}_2{\text{O)}}_6^{3+}$ as:

  $\begin{array}{l}{\text{ Fe(H}}_2{\text{O)}}_6^{3+}\to {\text{ H}}^{+}{\text{ + Fe(OH)(H}}_2{\text{O)}}_5^{2+}\\ \text{Initial: 0}\text{.100 M 0 0}\\ \text{Change: -x +x +x}\\ \text{Equilibrium: 0}\text{.100 - x x x}\end{array}$

Now, calculating the total ion concentration as:

  $\begin{array}{l}\frac{\pi }{\text{RT}}\text{= 0}\text{.100 M - x + x + x}\\ \frac{\pi }{\text{RT}}\text{ = 0}\text{.100 M + x}\end{array}$

Substituting the values:

  $\begin{array}{l}\text{x = }\frac{\text{3}\text{.06 atm}}{\text{0}\text{.08206 Latm}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}(298\text{ K)}}\text{ - 0}\text{.100 M}\\ \text{x = 0}\text{.025 M}\end{array}$

Now, the value of acid dissociation constant is calculated using formula:

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

So, the expression for acid dissociation constant for the above reaction is:

  ${\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{H}}^{+}\right]\left[{\text{Fe(OH)(H}}_2{\text{O)}}_5^{2+}\right]}{\left[{\text{Fe(H}}_2{\text{O)}}_6^{3+}\right]}$

Substituting the values:

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{ = }\frac{{\text{x}}^2}{\text{0}\text{.100 - x}}\\ {\text{K}}_{\text{a}}\text{ = }\frac{{(0.025)}^2}{\text{(0}\text{.100 - 0}\text{.025)}}\\ {\text{K}}_{\text{a}}\text{ = }\frac{0.000625}{\text{0}\text{.075}}\\ {\text{K}}_{\text{a}}\text{ = 8}\text{.3}\times {\text{10}}^{-3}\end{array}$

Thus, the ${\text{K}}_{\text{a}}$ for the dissociation reaction of ${\text{Fe(H}}_2{\text{O)}}_6^{3+}$ is $\text{8}\text{.3}\times {\text{10}}^{-3}$ .