Question

Uric acid dissociates as shown in the figure at the right with a pKa of 5.80, and can be treated as a simple monoprotic acid in this pH range. Its solubility in urine depends on the relative amount of protonated and unprotonated forms. A urine sample was found to have a pH of 6.80.

What would be the ratio of the unprotonated form, A–, to the protonated form, HA, for uric acid in this urine sample?

Answer 1

Given that pH of urine = 6.8

             pKa of uric acid = 5.8

From Henderson-Hasselbalch's equation,

pH = pKa + log [A-]/HA]

pH = pKa + log [unprotonated form] / [protonated form]

6.8 = 5.8 + log [unprotonated form] / [protonated form]

log [unprotonated form] / [protonated form] = 1

[unprotonated form] / [protonated form] = 10¹ = 10

Hence, ratio = 10 : 1

Therefore,

the ratio of the unprotonated form, A–, to the protonated form, HA, for uric acid = 10 : 1