c acid dissociates as shown in the figure at HN N the right with a pKa of 5.80,

Question

c acid dissociates as shown in the figure at HN N the right with a pKa of 5.80, and can be treated HN N as a 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 5.03 HA What would be the ratio of the unprotonated form, A to the protonated form, HA, for uric acid in 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

Explanation / Answer

we have to use the formula

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

5.03 = 5.80 + log [A-]/ [HA]

log [A-]/ [HA] = 5.03 - 5.80

[A-]/ [HA] = 10-0.77 = 0.1698 ---------> 1

from the problem it is clear that

[HA] + [A-] = 5.03 -----------> 2

from the equation 1 [A-] = 0.1689 x [HA]

put this value in equation 2

[HA] +   0.1689 x [HA] = 5.03

[HA] x 1.1689 = 5.03

[HA] = 5.03/1.1689 = 4.3031

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