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Question: The \(p{K_a}\) of butyric acid \({\text{HBut}}\) is 4.7. How do you calculate \({K_b}\) for the buty...

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

Explanation

Solution

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 Ka{{\text{K}}_{\text{a}}} in the case of acids and Kb{{\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 pKw{\text{p}}{{\text{K}}_{\text{w}}} using pKa{\text{p}}{{\text{K}}_{\text{a}}} of acid then calculate the Kb{{\text{K}}_{\text{b}}}.

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

Thus, the value of the Kb{K_b} for the butyrate ion is 10−9.3{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.