Respuesta :
Explanation:
We know that relation between [tex]pK_{a}[/tex] and [tex]pK_{b}[/tex] is as follows.
[tex]pK_{a} + pK_{b}[/tex] = 14
As it is given that [tex]pK_{a}[/tex] is 8.18. Therefore, calculate the value of [tex]pK_{b}[/tex] as follows.
[tex]pK_{a} + pK_{b}[/tex] = 14
[tex]8.18 + pK_{b}[/tex] = 14
[tex]pK_{b}[/tex] = 14 - 8.18
= 5.82
Similarly, as value of pH is given as 7.18. Therefore, value of pOH will be as follows.
pH + pOH = 14
7.18 + pOH = 14
pOH = 6.82
Let us take that B represents the enzyme. Hence, its reaction with proton will be as follows.
[tex]B + H^{+} \rightarrow BH^{+}[/tex] (protonated active enzyme)
Hence, pOH = [tex]pK_{b} + log\frac{[BH^{+}]}{[B]}[/tex]
6.82 = 5.82 + [tex]log_{10} \frac{[BH^{+}]}{[B]}[/tex]
[tex]\frac{[BH^{+}]}{[B]}[/tex] = 10
Therefore, percentage of active enzyme = %[tex][BH^{+}][/tex] = [tex]\frac{10}{10 + 1} \rimes 100[/tex]
%[tex][BH^{+}][/tex] = 90.9%
Thus, we can conclude that 90.9% is the percentage of the enzyme which is active in a buffer at pH 7.18.
