Uric acid dissociates as shown in the figure at the right with a pKa of 5.80, an

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 5.27.

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

[A-]/[HA]=_______________

B) What would be the fraction of the uric acid in the neutral, protonated form in this sample?

fraction of HA=____________

Explanation / Answer

a)

pH = pKa + log [A-]/[HA]
5.27 = 5.8 + log [A-]/[HA]
log [A-]/[HA] = - 0.53
[A-]/[HA] = 0.295 : 1

b)
[HA/A--]= 1/0.295= 3.388

Fraction of A- in sample = 0.295 / (0.295+3.388) = 0.08

Fraction of HA in sample = 3.388 / (0.295+3.388) = 0.92

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