Chapter 16, Problem 112IL

A hydrogen atom in the organic base pyridine, C₅H₅N, can be substituted by various atoms or groups to give XC₅H₄N, where X is an atom such as Cl or a group such as CH₃. The following table gives K_a values for the conjugate acids of a variety of substituted pyridines.

1. (a) Suppose each conjugate acid is dissolved in sufficient water to give a 0.050 M solution. Which solution would have the highest pH? The lowest pH?
2. (b) Which of the substituted pyridines is the strongest Brønsted base? Which is the weakest Brønsted base?

Interpretation Introduction

Interpretation:

The conjugate acid which have the highest $\text{pH}$ and which have the lowest $\text{pH}$ has to be stated.

Concept Introduction:

The $\text{pH}$ is a measure of Hydronium ion concentration. The expression for $\text{pH}$ is,

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{\text{+}}\right]$ (1)

Higher the value of $K_{\text{a}}$, stronger the acid and for an acidic solution $\text{pH}$ value is always less than $7$. That means higher the value of $K_{\text{a}}$ for a stronger acid smaller the value of $\text{pH}$.

Answer to Problem 112IL

The solution containing $4-C{H}_3\left(C_5{H}_5{N}^{+}\right)$ has the highest $\text{pH}$ and the solution containing $4-N{O}_2\left(C_5{H}_5{N}^{+}\right)$ has the lowest $\text{pH}$.

Explanation of Solution

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For Any acid HA,

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$ (2)

An equilibrium constant $\left(K\right)$ with subscript $\text{a}$ indicate that it is an equilibrium constant of an acid in water.

Given:

The $K_{\text{a}}$ value for each conjugate acids are:

$\begin{array}{cc}\text{Group}\left(\text{X}\right)& K_a\text{of}\text{Acid}\left[\text{4}-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)\right]\\ {\text{NO}}_{\text{2}}& 5.9\times {10}^{-2}\\ \text{Cl}& 1.5\times {10}^{-4}\\ \text{H}& 6.8\times {10}^{-6}\\ {\text{CH}}_{\text{3}}& 1\times {10}^{-6}\end{array}$

Initial Concentration of each solution is $0.05\text{M}$.

Set up an ICE table for the reaction of $\text{4}-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ with water at equilibrium,

$\begin{array}{cccccccc}\text{Equilibrium}& \text{4}-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)\left(\text{aq}\right)& +& {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)& \rightleftharpoons & \text{4}-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{4}}\text{N}\right)\left(\text{aq}\right)& +& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)\\ \text{Initial}\left(\text{M}\right)& 0.05& & & & 0& & 0\\ \text{Change}& -x& & & & +x& & +x\\ \text{At}\text{Equilibrium}& 0.05-x& & & & x& & x\end{array}$

Substitute the values in equation (2) to calculate $x$ for each acid,

$K_a=\frac{x^2}{0.05-x}$ (3)

Substitute the value of $K_{\text{a}}$ for $\text{4}-{\text{NO}}_{\text{2}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ in equation (3) to calculate the value of $x$,

$\begin{array}{c}5.9\times {10}^{-2}=\frac{x^2}{0.05-x}\\ x=0.054\end{array}$

The value of $x$ is equal to the concentration of hydronium ion that is $0.054\text{ mol}\cdot {\text{L}}^{-1}$.

Substitute the value of hydronium ion concentration in equation (1) to calculate the value of $\text{pH}$,

$\begin{array}{c}\text{pH}=-\log \left(0.054\right)\\ =1.26\end{array}$

Therefore, the value of $\text{pH}$ for $\text{4}-{\text{NO}}_{\text{2}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ is $1.26$.

Similarly, Substitute the value of $K_{\text{a}}$ for $\text{4}-\text{Cl}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ in equation (3) to calculate the value of $x$,

$\begin{array}{c}1.5\times {10}^{-4}=\frac{x^2}{0.05-x}\\ x=0.0027\end{array}$

The value of $x$ is equal to the concentration of hydronium ion that is $0.0027\text{ mol}\cdot {\text{L}}^{-1}$

Substitute the concentration of hydronium ion in equation (1) to calculate the value of $\text{pH}$,

$\begin{array}{c}\text{pH}=-\log \left(0.0027\right)\\ =2.56\end{array}$

Therefore, the value of $\text{pH}$ for $\text{4}-\text{Cl}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ is $2.56$.

Substitute the value of $K_{\text{a}}$ for $\text{4}-\text{H}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ in equation (3) to calculate the value of $x$,

$\begin{array}{c}6.8\times {10}^{-6}=\frac{x^2}{0.05-x}\\ x=0.00058\end{array}$

The value of $x$ is equal to the concentration of hydronium ion that is $0.00058\text{ mol}\cdot {\text{L}}^{-1}$.

Substitute the concentration of hydronium ion in equation (1) to calculate the value of $\text{pH}$,

$\begin{array}{c}\text{pH}=-\log \left(0.00058\right)\\ =3.23\end{array}$

Therefore, the value of $\text{pH}$ for $\text{4}-\text{H}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ is $3.23$.

Substitute the value of $K_{\text{a}}$ for $\text{4}-{\text{CH}}_{\text{3}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ in equation (3) to calculate the value of $x$,

$\begin{array}{c}1\times {10}^{-6}=\frac{x^2}{0.05-x}\\ x=0.00022\end{array}$

The value of $x$ is equal to the concentration of hydronium ion that is $0.00022\text{ mol}\cdot {\text{L}}^{-1}$.

Substitute the concentration of hydronium ion in equation (1) to calculate the value of $\text{pH}$

$\begin{array}{c}\text{pH}=-\log \left(0.00022\right)\\ =3.65\end{array}$

Therefore, the value of $\text{pH}$ for $\text{4}-{\text{CH}}_{\text{3}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ is $3.65$.

The solution containing $4-C{H}_3\left(C_5{H}_5{N}^{+}\right)$ has the highest $\text{pH}$ and the solution containing $4-N{O}_2\left(C_5{H}_5{N}^{+}\right)$ has the lowest $\text{pH}$.

Interpretation Introduction

Interpretation:

Strongest $bronsted$ base and weakest $bronsted$ base is to be stated in the given substituted pyridines.

Concept Introduction:

A conjugate acid-base pair contains two compounds that differ only by a hydrogen ion and a charge of $+1$. Thus, $4-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ and $\text{4}-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_4\text{N}\right)$ are conjugate acid-base pair. In this pair, $\text{4}-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_4\text{N}\right)$ is a conjugate base of an acid $\text{4}-\text{X}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$.

The stronger the acid, the weaker its conjugate base and vice-verca. That is, the larger the values of $K_{\text{a}}$, the smaller the values of $K_{\text{b}}$.

Answer to Problem 112IL

The strongest bronsted base is $4-C{H}_3\left(C_5{H}_{5}N\right)$ and the weakest bronsted base is $4-N{O}_2\left(C_5{H}_{5}N\right)$.

Explanation of Solution

The dissociation of $4-{\text{CH}}_{\text{3}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ with water is written as follows,

  $4-{\text{CH}}_{\text{3}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons 4-{\text{CH}}_{\text{3}}\left({\text{C}}_{\text{5}}{\text{H}}_4\text{N}\right)\left(\text{aq}\right)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{l}\right)$

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[4-{\text{CH}}_{\text{3}}\left({\text{C}}_{\text{5}}{\text{H}}_4\text{N}\right)\right]}{\left[4-{\text{CH}}_{\text{3}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)\right]}$

A small value of $K_{\text{a}}=1\times {10}^{-6}$ indicates that the acid is weakest among all of these four, therefore it’s conjugate base is strongest.

The dissociation of $4-{\text{NO}}_{\text{2}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)$ with water is written as follows,

    $4-{\text{NO}}_{\text{2}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)+4-{\text{NO}}_{\text{2}}\left({\text{C}}_{\text{5}}{\text{H}}_4\text{N}\right)\left(\text{aq}\right)$

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[4-{\text{NO}}_{\text{2}}\left({\text{C}}_{\text{5}}{\text{H}}_4\text{N}\right)\right]}{\left[4-{\text{NO}}_{\text{2}}\left({\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{+}}\right)\right]}$

A large value of $K_{\text{a}}=5.9\times {10}^{-2}$ indicates that the acid is strongest among all of these four, therefore it’s conjugate base is weakest.

The strongest bronsted base is $4-C{H}_3\left(C_5{H}_{5}N\right)$ and the weakest bronsted base is $4-N{O}_2\left(C_5{H}_{5}N\right)$.