Chapter 19, Problem 87AE

Interpretation Introduction

Interpretation: The pH of $\text{0}\text{.10 M}$ solution of ${\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}^{\text{3+}}$ needs to be calculated.

Concept Introduction: pH stands for "potential of hydrogen", it is a measure of alkalinity or acidity of water-soluble substances. pH number ranges from 0 to 14.

• pH is a logarithmic scale which specify the acidity or basicity of the solution.

  $pH=-\log [H^{+}]$

  $\begin{array}{l}\text{pOH = -log }\left[{\text{OH}}^{\text{-}}\right]\hfill \\ \text{pH + pOH = 14}\hfill \end{array}$

• Neutral solution- A solution is neutral if $pH=7$ .
• Acidic solution- A solution is acidic if $pH<7$ .
• Basic solution- A solution is basic if $pH>7$ .

Answer to Problem 87AE

The pH of $\text{0}\text{.10 M}$ solution of ${\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}^{\text{3+}}$ is $\text{1}\text{.6}$ .

Explanation of Solution

Given:

  ${\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}^{\text{3+}}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\rightleftharpoons {\text{ Fe(H}}_{\text{2}}{\text{O)}}_{\text{5}}{\text{(OH)}}^{\text{2+}}{\text{(aq) + H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) K}}_a\text{ = 6}\text{.0}\times {\text{10}}^{-3}$

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:

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

So, for the reaction:

  ${\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}^{\text{3+}}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\rightleftharpoons {\text{ Fe(H}}_{\text{2}}{\text{O)}}_{\text{5}}{\text{(OH)}}^{\text{2+}}{\text{(aq) + H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}$

The expression for the equilibrium constant is:

  ${\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{Fe(H}}_2{\text{O)}}_5{\text{(OH)}}^{2+}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}{{\left[{\text{Fe(H}}_2{\text{O)}}_6\right]}^{3+}}$

Since, the products are in equimolar ratio so, $\left[{\text{Fe(H}}_2{\text{O)}}_5{\text{(OH)}}^{2+}\right]\text{ = }\left[{\text{H}}_3{\text{O}}^{+}\right]$ .

Now, the expression for the equilibrium constant can be rewritten as:

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{ = }\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}^{\text{2}}}{{\left[{\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}\right]}^{\text{3+}}}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }\sqrt{{\text{K}}_{\text{a}}\left[{\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}^{\text{3+}}\right]}\end{array}$

Substituting the values:

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{ = 6}{\text{.0 × 10}}^{\text{-3}}\\ {\left[{\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}\right]}^{\text{3+}}\text{ = 0}\text{.10 M}\end{array}$

  $\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]\text{ = }\sqrt{\text{(6}\text{.0 }\times {\text{ 10}}^{-3})\times {\text{ (10}}^{-1})}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]\text{ = }\sqrt{6.0\times {\text{ 10}}^{-4}}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]\text{ = 2}\text{.45}\times {\text{10}}^{-2}\text{ M}\end{array}$

Now, the pH of the solution is calculated as:

  $\begin{array}{l}\text{pH = -log}\left[{\text{H}}_3{\text{O}}^{+}\right]\\ \text{pH }=\text{ -log}\left[\text{2}\text{.45}\times {\text{10}}^{-2}\right]\\ \text{pH = 1}\text{.6}\end{array}$

Hence, the pH of $\text{0}\text{.10 M}$ solution of ${\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}^{\text{3+}}$ is $\text{1}\text{.6}$ .

Interpretation Introduction

Interpretation: Whether 1.0 M solution of iron(II) nitrate have greater or lower pH compared to 1.0 M solution of iron(III) nitrate needs to be explained.

Concept Introduction: pH stands for "potential of hydrogen", it is a measure of alkalinity or acidity of water-soluble substances. pH number ranges from 0 to 14.

• pH is a logarithmic scale which specify the acidity or basicity of the solution.

  $pH=-\log [H^{+}]$

Answer to Problem 87AE

1.0 M solution of iron(II) nitrate have greater pH compared to 1.0 M solution of iron(III) nitrate.

Explanation of Solution

The balanced reaction of iron(II) nitrate with water is as follows:

  ${\text{Fe(NO}}_3{)}_2{\text{ + 2H}}_2\text{O }\to {\text{ Fe(OH)}}_2{\text{ + 2HNO}}_3$

From the above balanced reaction it can be observed that 1 mole of iron(II) nitrate forms 2 moles of nitric acid, ${\text{HNO}}_3$ that will give 2 moles of ${\text{H}}^{+}$ ions.

Thebalanced reaction of iron(III) nitrate with water is as follows:

  ${\text{Fe(NO}}_3{)}_3{\text{ + 3H}}_2\text{O }\to {\text{ Fe(OH)}}_3{\text{ + 3HNO}}_3$

From the above balanced reaction it can be observed that 1 mole of iron(III) nitrate forms 3 moles of nitric acid, ${\text{HNO}}_3$ that will give 3 moles of ${\text{H}}^{+}$ ions.

The relation between pH and concentration of ${\text{H}}^{+}$ ions is inverse that is higher the value of pH, lower will be the concentration of ${\text{H}}^{+}$ ions. Since, the concentration of ${\text{H}}^{+}$ ions in 1.0 M iron(III) nitrate is higher than in 1.0 M iron(II) nitrate so, the pH of 1.0 M iron(III) nitrate is lower than 1.0 M iron(II) nitrate.