Chapter 16, Problem 16.89QA

Interpretation Introduction

To find:

The $pH$ of $0.100 M Fe{\left(N{O}_3\right)}_3$.

Answer to Problem 16.89QA

Solution:

$pH$ of $0.100 M Fe{\left(N{O}_3\right)}_3$ is 1.80

Explanation of Solution

1) Concept:

To determine the $pH$ of the solutions of a variety of metal ions, we will find that these ions act as weak acids in a solution. When $Fe{\left(N{O}_3\right)}_3$ dissolves in water, the ${Fe}^{3+}$ ions react with water to give a hydrated ferric ion,$Fe{\left(H_{2}O\right)}_6^{3+}$. A shift in electron density occurs in hydrated metal ions having a generic formula $M{\left(H_{2}O\right)}_6^{n+}$ when $n\ge 2$. The electrons in the $O-H$ bonds of the water molecules surrounding the metal ions are attracted to the positively charged ions. The resulting distortion in electron density increases the likelihood of one of these $O-H$ bonds ionizing and donating a $H^{+}$ ion to a neighboring molecule of water.

2) Formula:

i) $K_a=\frac{\left[A^{-}\right]\left[H_3{O}^{+}\right]}{\left[HA\right]}$

Where, $K_a$= acid ionization constant of weak acid.

$\left[A^{-}\right]$= concentration of conjugate base

$\left[HA\right]$= concentration of acid

ii) $pH= -\log \left[H_3{O}^{+}\right]$

iii) Quadratic equation, $a{x}^2+bx+c=0$

iv) $x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

3) Given:

i) $\left[Fe{\left(N{O}_3\right)}_3\right]=0.100 M$

ii) For ${Fe}^{3+} \left(aq\right)$, $K_a=3\times {10}^{-3}$

The $K_a$ value is taken from table A5.2 of $K_a$ values of hydrated metal ions mentioned in the book.

4) Calculation:

$Fe{\left(N{O}_3\right)}_3$ is a salt which shows complete dissociation in an aqueous medium.

$Fe{\left(N{O}_3\right)}_3\left(aq\right)\to {Fe}^{3+} \left(aq\right)+3N{O}_3^{-} \left(aq\right)$

The formula for hydrated metal ion can be written as,$Fe{\left(H_{2}O\right)}_6^{3+}$. The hydrated ${Fe}^{3+}$ cation draws electron density away from the water molecules of its inner coordination sphere, which makes it possible for one or more of these molecules to donate a $H^{+}$ ion to a water molecule outside the sphere. The hydrated ${Fe}^{3+}$ ion is left with one fewer $H_{2}O$ ligand and a bond between the complex and $O{H}^{-}$ ion.

$Fe{\left(H_{2}O\right)}_6^{3+} \left(aq\right)+H_{2}O \left(l\right)\rightleftharpoons Fe{\left(H_{2}O\right)}_5{\left(OH\right)}^{2+} \left(aq\right)+H_3{O}^{+} \left(aq\right)$

Now draw the RICE table incorporating the concentrations.

	$\left[Fe{\left(H_{2}O\right)}_6^{3+}\right] M$	$\left[Fe{\left(H_{2}O\right)}_5{\left(OH\right)}^{2+}\right] M$	$\left[H_3{O}^{+}\right] M$
Initial (I)	$0.100$	$0$	$0$
Change (C)	$-x$	$+x$	$+x$
Equilibrium (E)	$0.100-x$	$x$	$x$

$K_a=\frac{\left[Fe{\left(H_{2}O\right)}_5{\left(OH\right)}^{2+}\right]\left[H_3{O}^{+}\right]}{\left[Fe{\left(H_{2}O\right)}_6^{3+}\right]}$

$3\times {10}^{-3}= \frac{\left[x\right]\left[x\right]}{\left[0.100-x\right]}$

$3\times {10}^{-4}-\left(3\times {10}^{-3}\right)x=x^2$

$x^2+ \left(3\times {10}^{-3}\right)x-3\times {10}^{-4}=0$

Therefore,  $a=1, b= 3\times {10}^{-3}$ and $c= -3\times {10}^{-4}$.

$x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x= \frac{-\left(0.003\right)\pm \sqrt{{\left(0.003\right)}^2-\left(4\left(1\right)\left(-3\times {10}^{-4}\right)\right)}}{2\times 1 }$

$x= \frac{-0.003 \pm \sqrt{0.001209}}{2}$

$x= 0.01588$ and $x= -0.01888$

The negative value of $x$ has no physical meaning because it gives us a negative $\left[H_3{O}^{+}\right]$ value. Therefore,

$\left[x\right]=\left[H_3{O}^{+}\right]=0.01588$ M

Calculating the $pH$:

$pH= -\log \left[H_3{O}^{+}\right]= -\mathrm{l}\mathrm{o}\mathrm{g}\left(0.01588\right)$

$pH=1.80$

$pH$ of $0.100 M Fe{\left(N{O}_3\right)}_3$ is 1.80

Conclusion:

With the help of the RICE table, the $pH$ of a hydrated metal ion is calculated.