uric acid dissociates as shown in the figure at the right with a pKa of 5.80, an
ID: 983707 • Letter: U
Question
uric acid dissociates as shown in the figure at the right with a pKa of 5.80, and can be treated HN N 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 5.20. HA the protonated form, HA, for uric acid in What would be the ratio of the unprotonated form, this urine sample? Number HA What would be the fraction of the uric acid in the neutral, protonated form in this sample? Number fraction as HA Check Answer Next ExExplanation / Answer
Henderson–Hasselbalch equation can be written as
HA-à A-+ H+
Ka= [A-] [H+]/[HA] (1)
Given PH= 5.2, -log [H+] =5.2 [ H+] =10^(-5.2)=6.31*10-6M =[A-]
Pka= 5.8 ka= 10^(-5.8)= 1.59*10-6
From (1) 1.59*10-6= 6.31*10-6 [A-]/[HA]
[A-]/[HA]= 1.59/6.31=0.252
[HA] =[A-]/0.252 =6.31*10-6/0.252 =25.04*10-6
[A-]+[HA]+ H+] = 6.31*10-6+6.31*10-6 +25.04*10-6= 37.66*10-6 ( at Equilibrium)
[HA/{[A-] + [HA]+[H+] }= 25.04*10-6/37.66*10-6= 0.66
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