Chapter 7, Problem 114E

(a)

Interpretation Introduction

Interpretation: The pH of 0.10 M solution of ${\text{Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}{}^{\text{3+}}$ needs to be determined.

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

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

  $\begin{array}{l}{\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

(b)

Interpretation Introduction

Interpretation: The pH of iron (ii) nitrate and iron (III) nitrate for 1.0 M solution needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

  $\begin{array}{l}{\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$