Question

The${\mathbf{K}}_a$ of phenylacetic acid is$\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{-5}$, and the${\mathbf{pK}}_a$of propionic acid is$\mathbf{4}\mathbf{.}\mathbf{87}$.

(a)Calculate the${\mathbf{pK}}_a$of phenylacetic acid and the${\mathbf{K}}_a$of propionic acid.

(b) Which one of these is the stronger acid? Calculate how much stronger an acid it is.

(c) Predict whether the following equilibrium will favor the reactants or the products.

Short answer

(a)$p{K}_a$of phenylacetic acid is found to be$4.28$.

$K_a$of propionic acid is found to be$1.35\times {10}^{-5}$

(b) Phenylacetic acid is stronger acid than propionic acid.

Phenylacetic acid is $0.59p{K}_a$units stronger than propionic acid.

(c) Equilibrium will favor the reactants.

Step-by-step solution

Conjugate acid-base pair

The remaining part of acidafter losing a proton${\text{(H}}^{\text{+}}\text{)}$will have a tendency to accept a proton${\text{(H}}^{\text{+}}\text{)}$. Hence, it will behave as a base. These pairs of substances that differ from one another by a proton${\text{(H}}^{\text{+}}\text{)}$are known as conjugate acid-base pairs. Consider a general example of an acid:

$\text{HA\hspace{0.17em}(acid)}\rightleftharpoons {\text{H}}^{+}\text{(proton)\hspace{0.17em}}+{\text{A}}^{-}\text{\hspace{0.17em}(conjugate\hspace{0.17em}\hspace{0.17em}base)}$

Kaand  pKa, Kband pKb

$K_a$is known as the acid dissociation constant and its value indicates the relative strength of the acid.$p{K}_a$is defined as the negative logarithm (base$10$) of$K_a$. Mathematically ,$p{K}_a=-\log K_a$.$p{K}_a$value of strong acids is usually zero or even negative while thevalue of weaker acids are greater than$4$ .

$K_b$is known as the base dissociation constant and its value indicates the relative strength of the base. $p{K}_b$is defined as the negative logarithm (base$10$ ) of $K_b$.

Mathematically, $p{K}_b=-\log K_b$

Equilibrium positions of acid-base reactions

The formation of weaker acid and weaker base is favored by the acid-base equilibrium. Larger $p{K}_a$means weaker is the acid while larger $p{K}_b$means weaker is the base. Both the weaker acid and weaker base exist as either reactants or products and they are always present on the same side of the equation.

Calculation, explanation and prediction of the equilibrium

(a) As per given data,

$K_a$of phenylacetic acid

$p{K}_a$of propionic acid

Find the $p{K}_a$of phenylacetic acid from its given $K_a$by using the relation.

$p{K}_a=-\log K_a$

Now,

$\begin{array}{c}p{K}_a=-\log K_a\\ =-\log (5.2\times {10}^{-5})\\ =-(\log 5.2-5\log 10)\\ =-(0.72-5)\\ =4.28\end{array}$

$p{K}_a$of phenylacetic acid is found to be $4.28$.

Find the $K_a$of propionic acid from its given $p{K}_a$by using the relation $K_a={10}^{-p{K}_a}$

Now,

$\begin{array}{c}{K}_a={10}^{-p{K}_a}\\ ={10}^{-4.87}\\ =1.35\times {10}^{-5}\end{array}$

$K_a$of propionic acid is found to be$1.35\times {10}^{-5}$

(b)$p{K}_a$of propionic acid$=4.87$

$p{K}_a$of phenylacetic acid$=4.28$

It can be seen that$p{K}_a$of propionic acid is greater than$p{K}_a$of phenylacetic acid.

The strength of an acid or base is given by its$p{K}_a$value. The smaller$p{K}_a$value means stronger is the acid while the larger$p{K}_a$value means stronger is the base. Therefore, phenylacetic acid is stronger acid than propionic acid.

Now,

$\begin{array}{c}\text{Difference\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}\hspace{0.17em}}p{K}_a=4.87-4.28\\ =0.59\end{array}$

Phenylacetic acid is$0.59p{K}_a$units stronger than propionic acid.

(c)

The weaker acid and weaker base are both present in the reactant side, hence the equilibrium will favor the reactants.