Chapter 14, Problem 43E

Consider a solution that contains both C₅H₅N and C₅H₅NHNO₃. Calculate the ratio [C₅H₅N]/[C₅H₅NH⁺] if the solution has the following pH values:

a. pH = 4.50

b. pH = 5.00

c. pH = 5.23

d. pH = 5.50

Interpretation Introduction

Interpretation:

Different $\text{pH}$ values of solution of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ and ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}$ is given. The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is to be calculated.

Concept introduction:

The $\text{pOH}$ value is the measure of ${\text{OH}}^{-}$ ions. The relation between $\text{pOH}$ and $\text{p}{K}_{\text{b}}$ is given by Henderson-Hassel Bach equation. According to this equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Answer to Problem 43E

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ for $\text{pH}=\text{4}\text{.50}$ is $\underset{\_}{0.18}$.

Explanation of Solution

Explanation

To determine $\text{pOH}$

The $\text{pOH}$ value of a solution of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ and ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}$ is $\underset{\_}{9.5}$.

The $\text{pH}$ value of solution is $\text{4}\text{.50}$.

The relation between $\text{pH}$ and $\text{pOH}$ is,

$\text{pH}+\text{pOH}=\text{14}$

Where,

• $\text{pH}$ is a measure of concentration of ${\text{H}}^{\text{+}}$.
• $\text{pOH}$ is a measure of concentration of ${\text{OH}}^{-}$.

Substitute the value of $\text{pOH}$ in above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{4}\text{.50}+\text{pOH}=\text{14}\\ \text{pOH}=\underset{\_}{9.5}\end{array}$

To determine the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]$

The value of $K_{\text{b}}$ for ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is $\text{1}\text{.7}\times {\text{10}}^{-\text{9}}$.

The $\text{pOH}$ value of solution is $\text{9}\text{.5}$.

The $\text{pOH}$ is calculated using the Henderson-Hassel Bach equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Where,

• $\text{pOH}$ is the measure of ${\text{OH}}^{-}$ ions.
• $\text{p}{K}_{\text{b}}$ is the measure of basic strength.

The formula of $\text{p}{K}_{\text{b}}$ is,

$\text{p}{K}_{\text{b}}=-\log K_{\text{b}}$

Where,

• $K_{\text{b}}$ is base equilibrium constant.

Substitute the value of $\text{p}{K}_{\text{b}}$ in the above equation.

$\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Substitute all the given values in the above equation.

$\begin{array}{c}\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}\\ \text{9}\text{.5}=-\log \left(\text{1}\text{.7}\times {\text{10}}^{-\text{9}}\right)+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \text{9}\text{.5}=\text{8}\text{.769}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\end{array}$

Simplify the above equation.

$\begin{array}{c}\text{9}\text{.5}=\text{8}\text{.769}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\text{0}\text{.731}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\underset{\_}{5.3826}\end{array}$

To find the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is,

$\begin{array}{c}\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\text{5}\text{.3826}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}=\frac{\text{1}}{\text{5}\text{.3826}}\\ =\underset{\_}{0.18}\end{array}$

Interpretation Introduction

Interpretation:

Different $\text{pH}$ values of solution of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ and ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}$ is given. The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is to be calculated.

Concept introduction:

The $\text{pOH}$ value is the measure of ${\text{OH}}^{-}$ ions. The relation between $\text{pOH}$ and $\text{p}{K}_{\text{b}}$ is given by Henderson-Hassel Bach equation. According to this equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Answer to Problem 43E

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ for $\text{pH}=\text{5}\text{.00}$ is $\underset{\_}{0.59}$.

Explanation of Solution

Explanation

To determine the $\text{pOH}$value

The $\text{pH}$ value of solution is $\text{5}\text{.00}$.

The relation between $\text{pH}$ and $\text{pOH}$ is,

$\text{pH}+\text{pOH}=\text{14}$

Where,

• $\text{pH}$ is a measure of concentration of ${\text{H}}^{\text{+}}$.
• $\text{pOH}$ is a measure of concentration of ${\text{OH}}^{-}$.

Substitute the value of $\text{pOH}$ in above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{5}\text{.00}+\text{pOH}=\text{14}\\ \text{pOH}=\underset{\_}{9.0}\end{array}$

To determine the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]$

The value of $K_{\text{b}}$ for ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is $\text{1}\text{.7}\times {\text{10}}^{-\text{9}}$.

The $\text{pOH}$ value of solution is $\text{9}\text{.0}$.

The $\text{pOH}$ is calculated using the Henderson-Hassel Bach equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Where,

• $\text{pOH}$ is the measure of ${\text{OH}}^{-}$ ions.
• $\text{p}{K}_{\text{b}}$ is the measure of basic strength.

The formula of $\text{p}{K}_{\text{b}}$ is,

$\text{p}{K}_{\text{b}}=-\log K_{\text{b}}$

Where,

• $K_{\text{b}}$ is base equilibrium constant.

Substitute the value of $\text{p}{K}_{\text{b}}$ in the above equation.

$\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Substitute all the given values in the above equation.

$\begin{array}{c}\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}\\ \text{9}\text{.0}=-\log \left(\text{1}\text{.7}\times {\text{10}}^{-\text{9}}\right)+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \text{9}\text{.0}=\text{8}\text{.769}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\end{array}$

Simplify the above equation.

$\begin{array}{c}\text{9}\text{.0}=\text{8}\text{.769}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\text{0}\text{.231}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\underset{\_}{1.7021}\end{array}$

To find the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is,

$\begin{array}{c}\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\text{1}\text{.7021}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}=\frac{\text{1}}{\text{1}\text{.7021}}\\ =\underset{\_}{0.59}\end{array}$

Interpretation Introduction

Interpretation:

Different $\text{pH}$ values of solution of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ and ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}$ is given. The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is to be calculated.

Concept introduction:

The $\text{pOH}$ value is the measure of ${\text{OH}}^{-}$ ions. The relation between $\text{pOH}$ and $\text{p}{K}_{\text{b}}$ is given by Henderson-Hassel Bach equation. According to this equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Answer to Problem 43E

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ for $\text{pH}=\text{5}\text{.23}$ is $\underset{\_}{1}$.

Explanation of Solution

Explanation

To determine the $\text{pOH}$

The $\text{pOH}$ value of a solution of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ and ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}$ is $\underset{\_}{8.77}$.

The $\text{pH}$ value of solution is $\text{5}\text{.23}$.

The relation between $\text{pH}$ and $\text{pOH}$ is,

$\text{pH}+\text{pOH}=\text{14}$

Where,

• $\text{pH}$ is a measure of concentration of ${\text{H}}^{\text{+}}$.
• $\text{pOH}$ is a measure of concentration of ${\text{OH}}^{-}$.

Substitute the value of $\text{pOH}$ in above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{5}\text{.23}+\text{pOH}=\text{14}\\ \text{pOH}=\underset{\_}{8.77}\end{array}$

To determine the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]$

The value of $K_{\text{b}}$ for ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is $\text{1}\text{.7}\times {\text{10}}^{-\text{9}}$.

The $\text{pOH}$ value of solution is $\text{8}\text{.77}$.

The $\text{pOH}$ is calculated using the Henderson-Hassel Bach equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Where,

• $\text{pOH}$ is the measure of ${\text{OH}}^{-}$ ions.
• $\text{p}{K}_{\text{b}}$ is the measure of basic strength.

The formula of $\text{p}{K}_{\text{b}}$ is,

$\text{p}{K}_{\text{b}}=-\log K_{\text{b}}$

Where,

• $K_{\text{b}}$ is base equilibrium constant.

Substitute the value of $\text{p}{K}_{\text{b}}$ in the above equation.

$\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Substitute all the given values in the above equation.

$\begin{array}{c}\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}\\ \text{8}\text{.77}=-\log \left(\text{1}\text{.7}\times {\text{10}}^{-\text{9}}\right)+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \text{8}\text{.77}=\text{8}\text{.77}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\end{array}$

Simplify the above equation.

$\begin{array}{c}\text{8}\text{.77}=\text{8}\text{.77}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\text{0}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\underset{\_}{1}\end{array}$

To determine the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is,

$\begin{array}{c}\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\text{1}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}=\frac{\text{1}}{\text{1}}\\ =\underset{\_}{1}\end{array}$

Interpretation Introduction

Interpretation:

Different $\text{pH}$ values of solution of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ and ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}$ is given. The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is to be calculated.

Concept introduction:

The $\text{pOH}$ value is the measure of ${\text{OH}}^{-}$ ions. The relation between $\text{pOH}$ and $\text{p}{K}_{\text{b}}$ is given by Henderson-Hassel Bach equation. According to this equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Answer to Problem 43E

Answer:

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ for $\text{pH}=\text{5}\text{.50}$ is $\underset{\_}{1.86}$.

Explanation of Solution

Explanation

To determine the $\text{pOH}$

The $\text{pH}$ value of solution is $\text{5}\text{.50}$.

The relation between $\text{pH}$ and $\text{pOH}$ is,

$\text{pH}+\text{pOH}=\text{14}$

Where,

• $\text{pH}$ is a measure of concentration of ${\text{H}}^{\text{+}}$.
• $\text{pOH}$ is a measure of concentration of ${\text{OH}}^{-}$.

Substitute the value of $\text{pOH}$ in above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{5}\text{.50}+\text{pOH}=\text{14}\\ \text{pOH}=\underset{\_}{8.50}\end{array}$

To determine the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]$

The value of $K_{\text{b}}$ for ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is $\text{1}\text{.7}\times {\text{10}}^{-\text{9}}$.

The $\text{pOH}$ value of solution is $\text{8}\text{.50}$.

The $\text{pOH}$ is calculated using the Henderson-Hassel Bach equation,

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Where,

• $\text{pOH}$ is the measure of ${\text{OH}}^{-}$ ions.
• $\text{p}{K}_{\text{b}}$ is the measure of basic strength.

The formula of $\text{p}{K}_{\text{b}}$ is,

$\text{p}{K}_{\text{b}}=-\log K_{\text{b}}$

Where,

• $K_{\text{b}}$ is base equilibrium constant.

Substitute the value of $\text{p}{K}_{\text{b}}$ in the above equation.

$\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}$

Substitute all the given values in the above equation.

$\begin{array}{c}\text{pOH}=-\log K_{\text{b}}+\log \frac{\left[\text{Conjugate}\text{acid}\right]}{\left[\text{Weak}\text{base}\right]}\\ \text{8}\text{.50}=-\log \left(\text{1}\text{.7}\times {\text{10}}^{-\text{9}}\right)+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \text{8}\text{.50}=\text{8}\text{.769}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\end{array}$

Simplify the above equation.

$\begin{array}{c}\text{8}\text{.50}=\text{8}\text{.769}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}\\ \log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=-\text{0}\text{.269}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\underset{\_}{0.538}\end{array}$

To determine the ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$

The ratio of $\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]/\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]$ is,

$\begin{array}{c}\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\text{0}\text{.538}\\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]}=\frac{\text{1}}{\text{0}\text{.538}}\\ =\underset{\_}{1.86}\end{array}$