Chapter 13, Problem 59QAP

Consider a 0.45 M solution of ascorbic acid, ${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}$, one form of Vitamin C.

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\left(aq\right)\rightleftharpoons {\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)K_{\text{a}}=7.9\times {10}^{-5}\\ {\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)K_{\text{a}}=1.6\times {10}^{-12}\end{array}$

Calculate the pH of this solution and estimate $\left[{\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\right]$ and $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{\text{2}-}\right]$.

Interpretation Introduction

Interpretation:

The pH of solution, concentration of ${\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)$ and ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)$ needs to be determined.

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 59QAP

The pH of solution is 2.22, the concentration of ${\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)$ is $6.0\times {10}^{-3}\text{ M}$ and that of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)$ is $1.6\times {10}^{-12}$.

Explanation of Solution

The given acid is ascorbic acid ${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}$. The ionization reactions are as follows:

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\left(aq\right)\rightleftharpoons {\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right){\text{ K}}_a{}_1=7.9\times {10}^{-5}\\ {\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)+{\text{H}}^{+}\left(aq\right){\text{ K}}_{a2}=1.6\times {10}^{-12}\end{array}$

The initial concentration of ${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}$ is 0.45 M. Consider the first ionization, the ICE table is as follows:

$\begin{array}{l}{\text{ H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\left(aq\right)\rightleftharpoons {\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ I0.45\text{ - -}\\ C-xxx\\ E0.45-xxx\end{array}$

Now, the second ionization reaction is represented as follows:

$\begin{array}{l}{\text{ HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ Ix\text{ - }x\\ C-yyy\\ Ex-yyx+y\end{array}$

According to above two ICE tables, the equilibrium concentration of ${\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\left(aq\right)$, ${\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)$, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)$ and ${\text{H}}^{+}\left(aq\right)$ is $0.45-x$, $x-y$, y and $x+y$ respectively.

The expression for dissociation constant is as follows:

${\text{K}}_{a1}=\frac{\left[{\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\right]}$

Putting the values,

$7.9\times {10}^{-5}=\frac{\left(x-y\right)\left(x+y\right)}{\left(0.45-x\right)}$.. ..... (1)

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

${\text{K}}_{a2}=\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)\right]}$

Putting the values,

$1.6\times {10}^{-12}=\frac{y\left(x+y\right)}{\left(x-y\right)}$.. ..... (2)

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

$1.6\times {10}^{-12}=\frac{y\left(x\right)}{\left(x\right)}=y$

Thus, the value of y is $1.6\times {10}^{-12}$.

From the ICE table, it is concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)$, $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)\right]$ thus,

$\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)\right]=1.6\times {10}^{-12}\text{ M}$

Now, in the equation (1),

$7.9\times {10}^{-5}=\frac{\left(x-y\right)\left(x+y\right)}{\left(0.45-x\right)}$

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.

$\frac{7.9\times {10}^{-5}}{0.45}=1.76\times {10}^{-4}$

The calculated value is less than 0.05 thus,

$7.9\times {10}^{-5}=\frac{\left(x-y\right)\left(x+y\right)}{\left(0.45\right)}$

Or,

$7.9\times {10}^{-5}=\frac{x^2-y^2}{\left(0.45\right)}$

Putting the value of y in the above equation,

$7.9\times {10}^{-5}=\frac{x^2-{\left(1.6\times {10}^{-12}\right)}^2}{\left(0.45\right)}=\frac{x^2-\left(2.25\times {10}^{-24}\right)}{\left(0.45\right)}$

Or,

$\begin{array}{l}{x}^2=\left(7.9\times {10}^{-5}\right)\left(0.45\right)+\left(2.25\times {10}^{-24}\right)=3.56\times {10}^{-5}\\ x=\sqrt{3.56\times {10}^{-5}}=6\times {10}^{-3}\end{array}$

Thus, the concentration of ${\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)$ will be:

$\left[{\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)\right]=x-y$

Putting the values,

$\left[{\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)\right]=\left(6\times {10}^{-3}\right)-\left(1.6\times {10}^{-12}\right)=6.0\times {10}^{-3}\text{ M}$

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

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

Putting the values,

$\left[{\text{H}}^{+}\left(aq\right)\right]=\left(6\times {10}^{-3}\right)+\left(1.6\times {10}^{-12}\right)=6.0\times {10}^{-3}\text{ M}$

The pH of the solution can be calculated as follows:

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

Putting the values,

$\text{pH}=-\log \left(6.0\times {10}^{-3}\right)=2.22$

Therefore, the pH of solution is 2.22, the concentration of ${\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)$ is $6.0\times {10}^{-3}\text{ M}$ and that of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)$ is $1.6\times {10}^{-12}$.

Conclusion

Therefore, the pH of solution is 2.22, the concentration of ${\text{HC}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{-}\left(aq\right)$ is $6.0\times {10}^{-3}\text{ M}$ and that of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}{}^{2-}\left(aq\right)$ is $1.6\times {10}^{-12}$.