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 5.88.

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

I am aware that one should use the henderson-hasslebalch equation to solve this but I am stuck after this step:

.08=log[A-]/[HA]   I dont how how to solve for or divide by log.... Please explain each step with words so I understand. Thanks

Explanation / Answer

For a monoprotic acid,

HA <==> A- + H+

Let x amount be dissociated then,

[A-] = [H+] = x M

[HA] = [HA-x] M

pH = 5.88

pKa = 5.80

Feed values,

Ka = [A-][H+]/[HA]

Ka = (x)(x)/(HA-x)

pH = 5.88 = -log[H+]

[H+] = 1.32 x 10^-6 = x

pKa = 5.80 = -log[Ka]

Ka = 1.58 x 10^-6

Feed values,

1.58 x 10^-6 = (1.32 x 10^-6)^2/(HA-1.32 x 10^-6)

1.58 x 10^-6[HA] - 2.09 x 10^-12 = 1.74 x 10^-12

[HA] = 2.42 x 10^-6 M

[A-] = 1.32 x 10^-6 M

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