Chapter 16, Problem 16.90QA

Interpretation Introduction

To find:

The $pH$ of $1.00 M Cu{\left(N{O}_3\right)}_2$.

Answer to Problem 16.90QA

Solution:

$pH$ of $1.00 M Cu{\left(N{O}_3\right)}_2$ is $3.76$

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 $Cu{\left(N{O}_3\right)}_2$ dissolves in water, the ${Cu}^{2+}$ ions react with water to give a hydrated cupric ion $Cu{\left(H_{2}O\right)}_6^{2+}$.

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 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 a weak acid.

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

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

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

3) Given:

i) $\left[Cu{\left(N{O}_3\right)}_2\right]=1.00 M$

ii) For ${Cu}^{2+} \left(aq\right)$, $K_a=3\times {10}^{-8}$

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

4) Calculation:

$Cu{\left(N{O}_3\right)}_2$ is a salt which shows complete dissociation in an aqueous medium.

$Cu{\left(N{O}_3\right)}_2\left(aq\right)\to {Cu}^{2+} \left(aq\right)+2N{O}_3^{-} \left(aq\right)$

The formula for hydrated metal ion can be written as,$Cu{\left(H_{2}O\right)}_6^{2+}$. The hydrated ${Cu}^{2+}$ 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 ${Cu}^{2+}$ ion is left with one fewer $H_{2}O$ ligand and a bond between the complex and $O{H}^{-}$ ion.

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

Now draw the RICE table incorporating the concentrations.

	$\left[Cu{\left(H_{2}O\right)}_6^{2+}\right] M$	$\left[Cu{\left(H_{2}O\right)}_5{\left(OH\right)}^{+}\right] M$	$\left[H_3{O}^{+}\right] M$
Initial (I)	$1.00$	$0$	$0$
Change (C)	$-x$	$+x$	$+x$
Equilibrium (E)	$1.00-x$	$x$	$x$

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

$3\times {10}^{-8}= \frac{\left[x\right]\left[x\right]}{\left[1.00-x\right]}$

Now we make the simplifying assumption that $x$ is much smaller than $1.00 M$. Therefore, we can ignore the $x$ term in the equilibrium value of  $\left[Cu{\left(H_{2}O\right)}_6^{2+}\right]$ and use the simplified value in the expression.

$3\times {10}^{-8}= \frac{\left[x\right]\left[x\right]}{\left[1.00\right]}$

${\left[x\right]}^2=\left(3\times {10}^{-8}\right)\left(1.00\right)=3\times {10}^{-8}$

$\left[x\right]=\left[H_3{O}^{+}\right]=1.732 \times {10}^{-4}$ M

Calculating the $pH$:

$pH= -\log \left[H_3{O}^{+}\right]= -\mathrm{l}\mathrm{o}\mathrm{g}(1.732 \times {10}^{-4})$

$pH=3.76$

$pH$ of $1.00 M Cu{\left(N{O}_3\right)}_2$ is $3.76$

Conclusion:

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