CHEM 1406-221HY Intr x CHEM 1406-221HY-intro x C Chegg Study Guided Sox CHEM 140

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CHEM 1406-221HY Intr x CHEM 1406-221HY-intro x C Chegg Study Guided Sox CHEM 1406-221HY Intro X G pling php?id 238494. CHEM 1406-221 HY Introductory Chemistry I-G Johnson Activities and Due Dates My Assig 4/13/2016 02:00 PM 56.8/100 @4/13/2016 o4:50 PM Attempts score Print Calculator Periodic Table Question 20 of 20 Uric acid dissociates as shown in the figure at the right with a pKa of 5.80, and can be treated HN 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 6.02 HA What would be the ratio of the unprotonated form, A, to the protonated form, HA, for uric acid in this urine sample? Previous 3 Give Up & View Solution Check Answer Next Exit 8 Hint ht D 2011 2016 Sapling Learnin Resources Assignment Information Available From: 3/23/2016 06:00 PM 4/13/2016 02:00 PM Due Date: Late submissions: Allowed with 109 of the points possible deducted per day 04/23/2016 02:00 PM Points Possible: Grade Category: Homework Description: Homework Late You can check your answers. You can view solutions when you complete or give up on any question You have five attempts per question. in your question for each incorrect attempt at that For multiple-choice questions, the penalty depends on the number of choices available. O Help With This Topic O Web Help & Videos O Technical Support and Bug Reports contact us he O Logout l Help

Explanation / Answer

pKa = 5.8

pH = 6.02

apply buffer equation

pH = pKa + log(A-/HA)

6.02 = 5.8 + log(Ratio)

ratio = 10^(6.02 -5.8 ) = 1.6595

ratio = 1.6595

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