Chapter 14, Problem 34QAP

A buffer is prepared using the butyric acid/butyrate $\left({\text{HC}}_4{\text{H}}_7{\text{O}}_2/{\text{C}}_4{\text{H}}_7{\text{O}}_2{}^{-}\right)$acid-base pair. The ratio of acid to base is 2.2 and Ka for butyric acid is $1.54\times {10}^{-5}$.

(a) What is the pH of this buffer?

(b) Enough strong base is added to convert 15% of butyric acid to the butyrate ion. What is the pH of the resulting solution?

(c) Strong acid is added to the buffer to increase its pH. What must the acid/base ratio be so that the pH increases by exactly one unit (e.g., from 2 to 3) from the answer in (a)?

Interpretation Introduction

(a)

Interpretation:

The butyric acid and butyrate base is used to prepare the buffer solution. The ratio butyric acid and butyrate base is $\frac{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{]}}=2.2$ in a buffer solution. The equilibrium constant of butyric acid, K_a is 1.54×10-5. The pH value of this buffer solution is to be determined.

Concept introduction:

The buffer solution is formed by the mixture of acid and the conjugate base. It is an aqueous solution which maintains the pH of the solution constant for many chemical applications.

The formula of the pH is −

${\text{pH=-log}}_{\text{10}}{\text{[H}}^{\text{+}}\text{]}$

Answer to Problem 34QAP

The pH value of this buffer solution is 4.47.

Explanation of Solution

The chemical equation for the buffer solution can be represented as-

${\text{HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{(aq)}\rightleftharpoons {\text{H}}^{\text{+}}{\text{(aq)+C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{(aq)}$

The equilibrium constant of acid for the above equation can be written as-

$K_a=\frac{{\text{[H}}^{\text{+}}{\text{][C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}$.. ....................(1)

Or, it can be represented as follows:

$\frac{K_a}{[{\text{H}}^{+}]}=\frac{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}$

Given that-

K_a = 1.54×10-5 for ${\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}$

$\frac{[H{C}_4{H}_7{O}_2]}{[C_4{H}_7{O}_2{}^{-}]}=2.2$

Put the above values in equation (1),

$\frac{{\text{K}}_{\text{a}}}{{\text{[H}}^{\text{+}}\text{]}}\text{=}\frac{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}$

$[{\text{H}}^{+}]=1.54\times {10}^{-5}\times 2.2$

$[{\text{H}}^{+}]=3.388\times {10}^{-5}$

The pH value of the solution-

$\text{pH}=-{\log }_{10}[{\text{H}}^{+}]$.. ......…. (2)

$\text{pH}=-{\log }_{10}\left(3.388\times {10}^{-5}\right)$

The pH value of solution = 4.47

Interpretation Introduction

(b)

Interpretation:

The butyric acid is converted to butyrate ion by adding 15% of the strong base. The pH value of this solution is to be determined.

Concept introduction:

The buffer solution is formed by the mixture of acid and the conjugate base. It is an aqueous solution which maintains the pH of the solution constant for many chemical applications.

The formula of the pH is −

${\text{pH=-log}}_{\text{10}}{\text{[H}}^{\text{+}}\text{]}$

Answer to Problem 34QAP

The pH value of this solution is 4.66 when the butyric acid is converted to butyrate ion by adding 15% of the strong base.

Explanation of Solution

The calculated value-

$[{\text{H}}^{+}]=3.388\times {10}^{-5}$

Therefore,

$[{\text{C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_2^{-}]=1.86\times {10}^{-7}=[{\text{H}}^{+}]$

Given that −

$\frac{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{]}}=2.2$

Or,

${\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]=2}{\text{.2×[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{]}$${\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}=2.2\times 3.388\times {10}^{-5}$

${\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}=7.4536\times {10}^{-5}$

The butyric acid is converted to butyrate ion by adding 15% of the strong base.

Given −

${\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}]=7.4536\times {10}^{-5}$

${\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}]=1.86\times {10}^{-7}=[{\text{H}}^{\text{+}}\text{]}$

Therefore,

${{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}_{\text{final}}=\left[7.4536\times {10}^{-5}-7.4536\times {10}^{-5}\times \frac{15}{100}\right]$

${{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}_{\text{final}}=6.3355\times {10}^{-5}$

${{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}_{\text{final}}=\left[1.86\times {10}^{-7}+1.86\times {10}^{-7}\times \frac{15}{100}\right]$

${{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}_{\text{final}}=4.506\times {10}^{-5}$

The value of $[{\text{H}}^{+}]$ is calculated by using the following equation :

${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}$.. ........(1)

Given that-

K_a = 1.54×10-5 for ${\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}$

${{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}_{\text{final}}=6.3355\times {10}^{-5}$

${{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}_{\text{final}}=4.506\times {10}^{-5}$

Put the above values in equation (1),

$[{\text{H}}^{+}]=1.54\times {10}^{-5}\times \frac{6.335\times {10}^{-5}}{4.506\times {10}^{-5}}$

$[{\text{H}}^{+}]=2.16\times {10}^{-5}$

The pH value of the solution-

${\text{pH=-log}}_{\text{10}}{\text{[H}}^{\text{+}}\text{]}$.. ......…. (2)

$\text{pH=}-{\log }_{10}\left(2.16\times {10}^{-5}\right)$

The pH value of solution = 4.66

Interpretation Introduction

(c)

Interpretation:

The pH value of the buffer solution is increased by adding strong acid to this solution. In this case, the ratio of $\frac{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}{}^{\text{-}}\text{]}}$ is to be determined so that pH value increased by one unit.

Concept introduction:

The buffer solution is formed by the mixture of acid and the conjugate base. It is an aqueous solution which maintains the pH of the solution constant for many chemical applications.

The formula of the pH is −

$\text{pH}=-{\log }_{10}[{\text{H}}^{+}]$

Answer to Problem 34QAP

The ratio of $\frac{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]}}{{\text{[HPO}}_{\text{4}}{}^{\text{2-}}\text{]}}$ is 0.22.

Explanation of Solution

The chemical equation for the buffer solution can be represented as-

${\text{HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_2(aq)\rightleftharpoons {\text{H}}^{+}(aq)+{\text{C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_2^{-}(aq)$

The equilibrium constant of acid for the above equation can be written as-

$K_a=\frac{{\text{[H}}^{\text{+}}{\text{][C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}$.. .................... (1)

Or, it can be represented as-

$\frac{{\text{K}}_{\text{a}}}{{\text{[H}}^{\text{+}}\text{]}}\text{=}\frac{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}$

Given that-

K_a = 1.54×10-5 for ${\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}$

Given that pH value is increased by one unit. Initial pH value is 4.47.

The pH value is 5.47. Then the concentration of H+ -

$\text{pH=}-{\log }_{10}[{\text{H}}^{+}]$ ………….. (2)

$5.47=-{\log }_{10}[{\text{H}}^{+}]$

$[H^{+}]=3.388\times {10}^{-6}$

Then the ratio is calculated by using equation (1)-

$\frac{{\text{K}}_{\text{a}}}{{\text{[H}}^{\text{+}}\text{]}}=\frac{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}$ ……………(3)

Given −

K_a = 1.54×10-5

$[H^{+}]=3.388\times {10}^{-6}$

Put the values in equation (3)

$\frac{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}=\frac{1.54\times {10}^{-5}}{3.388\times {10}^{-6}}$

Or,

$\frac{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}=\frac{3.388\times {10}^{-6}}{1.54\times {10}^{-5}}$

Then, the ratio $\frac{{\text{[HC}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{]}}{{\text{[C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}^{\text{-}}\text{]}}=0.22$