Question

The $p{K_a}$ of butyric acid ${\text{HBut}}$ is 4.7. How do you calculate ${K_b}$ for the butyrate ion ${\text{Bu}}{{\text{t}}^ - }$?

Answer 1

Hint:The dissociation constant is the measure of the strength of the acid and bases. The higher the value of the dissociation constant stronger is the acid or base and vice versa. It is represented as \[{{\text{K}}_{\text{a}}}\] in the case of acids and ${{\text{K}}_{\text{b}}}$ in the case of bases.
In the case of weak acids or bases, the value of the dissociation constant is less.
Here, we have to first determine the and ${\text{p}}{{\text{K}}_{\text{w}}}$ using ${\text{p}}{{\text{K}}_{\text{a}}}$ of acid then calculate the ${{\text{K}}_{\text{b}}}$.

Complete answer:
The butyric acid is dissociated as follows:
${\text{HBut}} \to {{\text{H}}^ + }\, + {\text{Bu}}{{\text{t}}^ - }$
Here, we can see that the acid ${\text{HBut}}$ produces ${\text{Bu}}{{\text{t}}^ - }$ which is the conjugate base.
Here, we have to first calculate ${\text{p}}{{\text{K}}_{\text{b}}}$ as follows:
The ${\text{p}}{{\text{K}}_{\text{w}}}$is given as follows:
${\text{p}}{{\text{K}}_{\text{w}}} = {\text{p}}{{\text{K}}_{\text{a}}}\left( {{\text{species}}} \right) + {\text{p}}{{\text{K}}_{\text{b}}}\left( {{\text{conjugate}}\,{\text{base}}} \right)$
Rearrange the above equation for ${\text{p}}{{\text{K}}_{\text{b}}}$.
${\text{p}}{{\text{K}}_{\text{b}}}{\text{ = p}}{{\text{K}}_{\text{w}}} - {\text{p}}{{\text{K}}_{\text{a}}}$
At ${25^^\circ }\,{\text{C}}$ ${\text{p}}{{\text{K}}_{\text{w}}}$ is 14 and substitute the value of ${\text{p}}{{\text{K}}_{\text{a}}}$ the 4.7 in the rearranged equation for ${\text{p}}{{\text{K}}_{\text{b}}}$.
${\text{p}}{{\text{K}}_{\text{b}}}{\text{ = 14}} - 4.7$
${\text{p}}{{\text{K}}_{\text{b}}}{\text{ = }}9.3$
The ${\text{p}}{{\text{K}}_{\text{b}}}$ is given as negative logarithm of ${{\text{K}}_{\text{b}}}$.
${\text{p}}{{\text{K}}_{\text{b}}} = - \log {{\text{K}}_{\text{b}}}$
$\log {{\text{K}}_{\text{b}}} = - {\text{p}}{{\text{K}}_{\text{b}}}$
Here, substitute the value of the ${\text{p}}{{\text{K}}_{\text{b}}}$ calculated.
$\log {{\text{K}}_{\text{b}}} = - {\text{9}}{\text{.3}}$
${{\text{K}}_{\text{b}}} = {10^{ - {\text{9}}{\text{.3}}}}$

Thus, the value of the ${K_b}$ for the butyrate ion is ${10^{ - {\text{9}}{\text{.3}}}}$.

Note:There is the various definition of the acids and bases. As per the Arrhenius concept, acid is a proton donor in an aqueous solution while the base is the hydroxide ion donor in an aqueous solution.