Chapter 16, Problem 16.47QP

Phthalic acid, H₂C₈H₄O₄, is a diprotic acid used in the synthesis of phenolphthalein indicator. K_a1 = 1.2 × 10⁻³, and K_a2 = 3.9 × 10⁻⁶. a Calculate the hydronium-ion concentration of a 0.015 M solution. b What is the concentration of the C₈H₄O₄²⁻ ion in the solution?

Interpretation Introduction

Interpretation:

The concentration of hydronium ion and concentration of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{\text{2-}}$ ion of a 0.015 M phthalic acid solution has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$ is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$ is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as $K_{a1}and{K}_{a2}$

For triprotic acids, the successive ionization constants are designated as $K_{a1},K_{a2}and{K}_{a3}$

To Calculate: The concentration of hydronium ion and concentration of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{\text{2-}}$ ion of a 0.015 M phthalic acid solution

Answer to Problem 16.47QP

• The concentration of hydroxide ion of the given phthalic acid solution is $3.7\times 10^{-3}M$
• The concentration of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{\text{2-}}$ ion in the given phthalic acid solution is $3.9\times 10^{-6}M$

Explanation of Solution

Given data:

Phthalic acid ${\text{(H}}_{\text{2}}{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}})$  is a diprotic acid

The concentration of the phthalic acid solution = 0.015 M

The value of ${\text{K}}_{\text{a1}}$ = $1.2\times {10}^{-3}$

The value of ${\text{K}}_{\text{a2}}$ = $3.9\times {10}^{-6}$

Ionizations of Phthalic acid:

Phthalic acid is a diprotic acid

The conjugate base resulting from the first ionization is the acid for the second ionization, and its starting concentration is the equilibrium concentration from the first ionization.

Let us represent phthalic acid ${\text{(H}}_{\text{2}}{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}})$ as ${\text{H}}_{\text{2}}\text{Ph}$ and its first ionization conjugate anion ${\text{(HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{-})$ as ${\text{HPh}}^{\text{-}}$ and second ionization conjugate anion ${\text{(C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{-})$ as ${\text{Ph}}^{\text{2-}}$

The ionizations of phthalic acid is as follows,

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{Ph}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ HPh}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{HPh}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ Ph}}^{\text{2-}}{}_{\text{(aq)}}\end{array}$

The ionization constant for first ionization is: ${\text{K}}_{\text{a1}}$ = $1.2\times {10}^{-3}$

The ionization constant for second ionization is: ${\text{K}}_{\text{a2}}$ = $3.9\times {10}^{-6}$

First ionization:

Construct an equilibrium table for first ionization.

Let x be the unknown in the first ionization.

	${\text{H}}_{\text{2}}{\text{Ph}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ HPh}}^{\text{-}}{}_{\text{(aq)}}$
Initial $(M)$	0.015 $-x$ 0.015-x	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

Write the equilibrium expression and substitute the equilibrium concentrations into it.

$\begin{array}{l}{\text{ K}}_{\text{a1}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][HPh}}^{\text{-}}\text{]}}{{\text{[H}}_{\text{2}}\text{Ph]}}\\ 1.2\times {10}^{-3}=\frac{x^2}{0.015-x}\\ x\text{ cannot be ignored}\text{. So, solve }x\text{ using quadratic equation}\\ x^2+(0.0012x)-1.80\times {10}^{-5}=0\\ x=\frac{-0.0012\pm \sqrt{{(0.0012)}^2-4(-1.80\times {10}^{-5})}}{2}\\ =\frac{-0.0012\pm 0.008569}{2}\\ \text{On considering positive root, we get}\\ x\text{=3}\text{.7}\times {\text{10}}^{\text{-3}}\text{M}\end{array}$

Thus,

The concentration of the hydrogen ion $[{\text{H}}^{\text{+}}\text{]}$ is $\text{3}\text{.7}\times {\text{10}}^{\text{-3}}\text{M}$

The concentration of the ${\text{(HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{-})$ ion is $\text{3}\text{.7}\times {\text{10}}^{\text{-3}}\text{M}$

Second ionization and concentration of   ${\text{(C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{-})$:

Construct an equilibrium table for second ionization.

Let y be the unknown in the second ionization.

The equilibrium concentration of the ${\text{(HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{-})$ ion after the first ionization becomes the starting concentration for the second ionization.

Additionally, the equilibrium concentration of ${\text{H}}^{+}$ is the starting concentration for the second ionization.

	${\text{HPh}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ Ph}}^{\text{2-}}{}_{\text{(aq)}}$
Initial $(M)$	$\text{3}\text{.7}\times {\text{10}}^{\text{-3}}$ $-y$ $\text{3}\text{.7}\times {\text{10}}^{\text{-3}}-y$	$0.00$	$0.00$
Change $(M)$		$+y$	$+y$
Equilibrium $(M)$		$\text{3}\text{.7}\times {\text{10}}^{\text{-3}}+y$	$y$

$\begin{array}{l}{\text{ K}}_{\text{a2}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][Ph}}^{\text{2-}}\text{]}}{{\text{[HPh}}^{\text{-}}\text{]}}\\ 3.9\times {10}^{-6}=\frac{(\text{3}\text{.7}\times {\text{10}}^{\text{-3}}+y)(y)}{\text{3}\text{.7}\times {\text{10}}^{\text{-3}}-y}\end{array}$

Assuming that y is very small and applying approximations, $\text{3}\text{.7}\times {\text{10}}^{\text{-3}}+y$ and $\text{3}\text{.7}\times {\text{10}}^{\text{-3}}-y\approx \text{3}\text{.7}\times {\text{10}}^{\text{-3}}$ gives,

$\begin{array}{l}3.9\times {10}^{-6}=\frac{(\text{3}\text{.7}\times {\text{10}}^{\text{-3}})(y)}{\text{3}\text{.7}\times {\text{10}}^{\text{-3}}}\\ y=3.9\times {10}^{-6}\text{ M}\end{array}$

Therefore, the concentration of ${\text{[C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{-}]$ is $3.9\times {10}^{-6}\text{ M}$

Conclusion

The concentration of the hydronium ion is calculated as $\text{3}\text{.7}\times {\text{10}}^{\text{-3}}\text{M}$

The concentration of ${\text{[C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}^{-}]$ is calculated as $3.9\times {10}^{-6}\text{ M}$