Chapter 13, Problem 61QAP

Phthalic acid H₂C₈H₄O₄, is a diprotic acid. It is used to make phenolphthalein indicator. $K_{a1}=0.0012$, and $K_{a2}=3.9\times {10}^{-6}$. Calculate the pH of a 2.9 M solution of phthalic acid. Estimate [HC₈H₄O₄^-] and (C₈H₄O₄^2-].

Interpretation Introduction

Interpretation:

The pH of solution, the concentration of ${\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)$ and that of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)$ needs to be calculated.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is concentration of conjugate base and $\left[\text{HB}\right]$ is the concentration of weak acid.

The pH of the solution is calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion.

Answer to Problem 61QAP

The pH of solution is 1.23, the concentration of ${\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)$ is $0.059\text{ M}$ and that of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)$ is $3.9\times {10}^{-6}\text{ M}$.

Explanation of Solution

The molecular formula of phthalic acid is ${\text{H}}_{\text{2}}{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}\text{O}$. The first and second ionization constant is $0.0012$ and $3.9\times {10}^{-6}$ respectively.

The molarity of solution is 2.9 M.

The concentration of each species can be calculated using the ICE table as follows:

$\begin{array}{l}{\text{ H}}_{\text{2}}{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}\text{O}\left(aq\right)\rightleftharpoons {\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ I2.9\text{ - -}\\ C-xxx\\ E2.9-xxx\end{array}$

Now, the second ionization reaction is represented as follows:

$\begin{array}{l}{\text{ HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)\rightleftharpoons {\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ Ix\text{ - }x\\ C-yyy\\ Ex-yyx+y\end{array}$

According to above two ICE tables, the equilibrium concentration of ${\text{H}}_{\text{2}}{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}\text{O}\left(aq\right)$, ${\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)$, ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)$ and ${\text{H}}^{+}\left(aq\right)$ is $2.9-x$, $x-y$, y and $x+y$ respectively.

The expression for dissociation constant is as follows:

${\text{K}}_{a1}=\frac{\left[{\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{H}}_2{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}\text{O}\left(aq\right)\right]}$

Putting the values,

$0.0012=\frac{\left(x-y\right)\left(x+y\right)}{\left(2.9-x\right)}$.. ..... (1)

Similarly, the expression for the second ionization constant is as follows:

${\text{K}}_{a2}=\frac{\left[{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)\right]}$

Putting the values,

$3.9\times {10}^{-6}=\frac{y\left(x+y\right)}{\left(x-y\right)}$.. ..... (2)

Since, the second dissociation constant is very low thus; the value of y can be neglected from the denominator in the equation (2)

$3.9\times {10}^{-6}=\frac{y\left(x\right)}{\left(x\right)}=y$

Thus, the value of y is $3.9\times {10}^{-6}$.

From the ICE table, it is concentration of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)$, thus,

$\left[{\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)\right]=3.9\times {10}^{-6}\text{ M}$

Now, in the equation (1),

$0.0012=\frac{\left(x-y\right)\left(x+y\right)}{\left(2.9-x\right)}$

Calculate the ratio of acid dissociation constant and concentration of acid; if it is less than 0.05 then the value of x from the denominator can be neglected.

$\frac{0.0012}{2.9}=4.14\times {10}^{-4}$

The calculated value is less than 0.05 thus,

$0.0012=\frac{\left(x-y\right)\left(x+y\right)}{\left(2.9\right)}$

Or,

$0.0012=\frac{x^2-y^2}{2.9}$

Putting the value of y in the above equation,

$0.0012=\frac{x^2-{\left(3.9\times {10}^{-6}\right)}^2}{2.9}=\frac{x^2-\left(1.521\times {10}^{-11}\right)}{\left(2.9\right)}$

Or,

$\begin{array}{l}{x}^2=\left(0.0012\right)\left(2.9\right)+\left(1.521\times {10}^{-11}\right)=3.48\times {10}^{-3}\\ x=\sqrt{3.48\times {10}^{-3}}=0.059\end{array}$

Thus, the concentration of ${\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)$ will be:

$\left[{\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)\right]=x-y$

Putting the values,

$\left[{\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)\right]=0.059-\left(3.9\times {10}^{-6}\right)=0.059\text{ M}$

The concentration of ${\text{H}}^{+}\left(aq\right)$ will be:

$\left[{\text{H}}^{+}\left(aq\right)\right]=x+y$

Putting the values,

$\left[{\text{H}}^{+}\left(aq\right)\right]=0.059+\left(3.9\times {10}^{-6}\right)=0.059\text{ M}$

The pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Putting the values,

$\text{pH}=-\log \left(0.059\right)=1.23$

Therefore, the pH of solution is 1.23, the concentration of ${\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)$ is $0.059\text{ M}$ and that of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)$ is $3.9\times {10}^{-6}\text{ M}$.

Conclusion

Therefore, the pH of solution is 1.23, the concentration of ${\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{-}\left(aq\right)$ is $0.059\text{ M}$ and that of ${\text{C}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}^{2-}\left(aq\right)$ is $3.9\times {10}^{-6}\text{ M}$.