Please help me solve this problem.. thank you. Uric acid dissociates as shown in

Question

Please help me solve this problem..

thank you.

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 4.89.

Uric acid dissociates as shown in the figure at the right with a pKa of 5.80, and can be tredHN 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 4.89 HA What would be the ratio of the unprotonated form, AT, 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

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

4.89= 5.8 + log [A-]/[HA

4.89-5.80= log [A-]/[HA]

[A-]/[HA] =10-0.91 =0.12303

Ka= [A-][H+]/[HA] but pKa= 5.8 Ka =10-5.8 =1.5485*10-6

but [A-] /[HA] =0.12303

1.5485*10-6= 0.12303*[H+]

[H+] = 1.5485*10-6 /0.12303 =1.2882*10-5

[therfore [A-] = 1.2882*10-5 and [HA] =1.2882*10-5/0.12303=0.000107408 M

moles of HA undissociated= 0.000107408/ (1.2882*10-5+0.000107408)=89%

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