Chapter 20, Problem 20.35AP

Interpretation Introduction

Interpretation:

The validation corresponding to the fact that $0.1M$ solution of acetic acid $\left(\text{p}{K}_a=4.76\right)$ turns litmus red, and that a $0.1M$ solution of phenol $\left(\text{p}{K}_a=9.95\right)$ does not is to be stated.

Concept introduction:

The $\text{pH}$ of a solution is represented as the negative logarithm of the concentration of proton.

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

This scale is defined from $1$ to $14$. It helps to determine whether the given solution is acidic, basic or neutral. The solutions with $\text{pH}<\text{7}$ are acidic in nature, with $\text{pH}=\text{7}$ are neutral in nature and with $\text{pH}>\text{7}$ are basic in nature.

Answer to Problem 20.35AP

The value of $\text{pH}$ for $0.1M$ solution of acetic acid is below $3$ that is $2.89$ and the value of $\text{pH}$ for $0.1M$ solution of phenol is above $3$ that is $5.48$. So, $0.1M$ solution of acetic acid $\left(\text{p}{K}_a=4.76\right)$ turns litmus red, and that a $0.1M$ solution of phenol $\left(\text{p}{K}_a=9.95\right)$ does not.

Explanation of Solution

The given $\text{p}{K}_a$ value $0.1M$ solution of acetic acid is $4.76$.

So, the value of $K_a$ for $0.1M$ solution of acetic acid is ${10}^{-4.76}=0.0000174$.

The ionization table for acetic acid for initial and final concentration is given below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\to {\text{CH}}_{\text{3}}{\text{CO}}_2^{-}+{\text{H}}^{+}\\ \text{Initial}\text{concentration}\text{0}\text{.1}0\text{}0\\ \text{Final concentration}0.1-xxx\end{array}$

The expression to calculate the value of $K_a$ is given below.

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{CH}}_{\text{3}}{\text{CO}}_2^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\right]}$

Substitute the values of respective concentrations and $K_a$ in the above expression.

$\begin{array}{rll}0.0000174& =& \frac{\left[x\right]\left[x\right]}{\left[\text{0}\text{.1}-x\right]}\\ 0& =& x^2+0.0000174x-0.00000174\\ x& =& 0.0013\end{array}$

The value of $\text{pH}$ solution is represented as follows.

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$ … (1)

Substitute the value of ${\text{H}}^{+}$ as $0.0013$ for $0.1M$ solution of acetic acid in equation (1).

$\begin{array}{rll}\text{pH}& =& -\log \left[0.0013\right]\\ & =& 2.89\end{array}$

Therefore, the value of $\text{pH}$ for $0.1M$ solution of acetic acid is $2.89$ that is below the $\text{pH}$ value $3$ and turns the litmus paper red.

The given $\text{p}{K}_a$ value $0.1M$ solution of phenol is $9.95$.

So, the value of $K_a$ for $0.1M$ solution of phenol is ${10}^{-9.95}=1.1\times {10}^{-10}$.

The ionization table for phenol for initial and final concentration is given below.

$\begin{array}{l}\text{PhOH}\to {\text{H}}^{+}+{\text{PhO}}^{-}\\ \text{Initial}\text{concentration}\text{0}\text{.1}0\text{}0\\ \text{Final concentration}0.1-xxx\end{array}$

The expression to calculate the value of $K_a$ is given below.

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

Substitute the values of respective concentrations and $K_a$ in the above expression.

$\begin{array}{rll}1.1\times {10}^{-10}& =& \frac{\left[x\right]\left[x\right]}{\left[\text{0}\text{.1}-x\right]}\\ 0& =& x^2+1.1\times {10}^{-10}x-1.1\times {10}^{-11}\\ x& =& 3.3\times {10}^{-6}\end{array}$

Substitute the value of ${\text{H}}^{+}$ as $3.3\times {10}^{-6}$ for $0.1M$ solution of phenol in equation (1).

$\begin{array}{rll}\text{pH}& =& -\log \left[3.3\times {10}^{-6}\right]\\ & =& 5.48\end{array}$

Therefore, the value of $\text{pH}$ for $0.1M$ solution of phenol is $5.48$ that is above the $\text{pH}$ value $3$ and does not turns the litmus paper red.

Conclusion

The $0.1M$ solution of acetic acid $\left(\text{p}{K}_a=4.76\right)$ turns litmus red, and that a $0.1M$ solution of phenol $\left(\text{p}{K}_a=9.95\right)$ does not because acetic acid has its $\text{pH}$ value below $3$ whereas, phenol has its $\text{pH}$ value above $3$.