Chapter 2, Problem 2.48P

Which compounds can be deprotonated by ${\text{OH}}^{-}$, so that equilibrium favors the products? Refer to the $\text{p}{K}_{\text{a}}$ table in Appendix A.

(a). $\text{HCOOH}$

(b). ${\text{H}}_{\text{2}}\text{S}$

(c).

(d). ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$

Interpretation Introduction

Interpretation: Among the given compounds, those which are deprotonated by ${\text{OH}}^{-}$ ion are to be identified.

Concept Introduction: According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid.

In a reaction which strongly favors the formation of products, the base used to remove proton from the acid should be stronger than the base formed when the proton is removed. In a reaction which favors the products, equilibrium will favors the formation of the weaker acid or weaker base. Or the $\text{p}{K}_{\text{a}}$ value of the products is higher than the $\text{p}{K}_{\text{a}}$ value of the reactants.

Answer to Problem 2.48P

Correct answer: The correct options are (a) and (b).

Explanation of Solution

Reason for correct answer:

(a) The deprotonation of an acid takes place in the presence of a strong base whose $\text{p}{K}_{\text{a}}$ value is greater than the $\text{p}{K}_{\text{a}}$ value of an acid.

The given compound is formic acid, $\text{HCOOH}$ whose $\text{p}{K}_{\text{a}}$ value is $3.8$. The deprotonation reaction of formic acid with hydroxide ion is shown below.

$\begin{array}{l}\text{HCOOH}+{\text{OH}}^{-}\rightleftharpoons {\text{HCOO}}^{-}+\text{H}-\text{OH}\\ \text{Acid Base Conjugate Conjugate }\\ \text{ base acid }\end{array}$

The $\text{p}{K}_{\text{a}}$ value of ${\text{H}}_{\text{2}}\text{O}$ is $15.7$ which is higher than $\text{p}{K}_{\text{a}}$ value of formic acid. Therefore, equilibrium favors the formation of weak acid with a higher $\text{p}{K}_{\text{a}}$ value. Hence, equilibrium favors the product formation and deprotonation of an acid occurs.

(b) The given compound is hydrogen sulphide, ${\text{H}}_{\text{2}}\text{S}$ whose $\text{p}{K}_{\text{a}}$ value is $7.0$. The deprotonation reaction of ${\text{H}}_{\text{2}}\text{S}$ with hydroxide ion is shown below.

$\begin{array}{l}{\text{H}}_{\text{2}}\text{S}+{\text{OH}}^{-}\rightleftharpoons {\text{HS}}^{-}+\text{ H}-\text{OH}\\ \text{Acid Base Conjugate Conjugate }\\ \text{ base acid }\end{array}$

The $\text{p}{K}_{\text{a}}$ value of ${\text{H}}_{\text{2}}\text{O}$ is $15.7$ which is higher than $\text{p}{K}_{\text{a}}$ value of ${\text{H}}_{\text{2}}\text{S}$. Therefore, equilibrium favors the formation of weak acid with a higher $\text{p}{K}_{\text{a}}$ value. Hence, equilibrium favors the product formation and deprotonation of an acid occurs.

Reason for incorrect options:

(c) The given compound is toluene, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{3}}$ whose $\text{p}{K}_{\text{a}}$ value is $41$. The deprotonation reaction of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{3}}$ with hydroxide ion is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{3}}+{\text{OH}}^{-}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_2^{-}+\text{ H}-\text{OH}\\ \text{Acid Base Conjugate Conjugate }\\ \text{ base acid }\end{array}$

The $\text{p}{K}_{\text{a}}$ value of ${\text{H}}_{\text{2}}\text{O}$ is $15.7$ which is lower than $\text{p}{K}_{\text{a}}$ value of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{3}}$. Equilibrium favors the formation of weak acid with a higher $\text{p}{K}_{\text{a}}$ value but in this case, a stronger acid with less $\text{p}{K}_{\text{a}}$ value is formed. Hence, equilibrium does not favor the product formation and deprotonation of an acid will not occur.

(d) The given compound is methyl amine, ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ whose $\text{p}{K}_{\text{a}}$ value is $40$. The deprotonation reaction of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ with hydroxide ion is shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}+{\text{OH}}^{-}\rightleftharpoons {\text{CH}}_3{\text{NH}}^{-}+\text{ H}-\text{OH}\\ \text{Acid Base Conjugate Conjugate }\\ \text{ base acid }\end{array}$

The $\text{p}{K}_{\text{a}}$ value of ${\text{H}}_{\text{2}}\text{O}$ is $15.7$ which is lower than $\text{p}{K}_{\text{a}}$ value of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$. Equilibrium favors the formation of weak acid with a higher $\text{p}{K}_{\text{a}}$ value but in this case, a stronger acid with less $\text{p}{K}_{\text{a}}$ value is formed. Hence, equilibrium does not favor the product formation and deprotonation of an acid will not occur.

Conclusion

Among the given compounds, formic acid, $\text{HCOOH}$ and hydrogen sulphide, ${\text{H}}_{\text{2}}\text{S}$ gets deprotonated by hydroxide ion. Hence, options (a) and (b) are correct.