12/1/2017 12:55 AM 62.1/ 10011 /22/2017 02:18 AM Gradebook Assignment Informatio

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12/1/2017 12:55 AM 62.1/ 10011 /22/2017 02:18 AM Gradebook Assignment Information een Print Calculator Periodic Table Available From: Due Date: Points Possible: Grade Category: 10/4/2017 11:55 PM 12/1/2017 12:55 AM 100 Graded Question 38 of 39 Sapling Learning Calculate the volume (in mL) of a 0.460 M KOH solution that should be added to 4.500 g of HEPES (MW#238.306 g/mol, pKa = 7.56) to give a pH of 7.22. Description: Number Policies: Homework KOH volume-D mL You can check your answers. You can view solutions when you complete or give up on any question. You can keep trying to answer each question until you get it right or give up. You lose 5% of the points available to each answer in your question for each incorrect attempt at that answer O eTextbook O Help With This Topi O Web Help & Videos O Previous Give Up & View Solution O Check Answer O NextExit | Technical Support and Bug Re Hint

Explanation / Answer

Since HEPES is a weak acid, you can use the Henderson-Hasselbach equation:

pH = pKa + log([X-]/[HX])

Hence, the ratio of the two is:

[X-]/[HX] = 10^(7.22-7.56) = 0.457

or

[X-] = 0.457*[HX]

Now consider the following reaction:

HX + KOH ---> NaX + H2O

Moles of HEPES = 4.5 / 238.306 = 0.01888

the equilibrium amount of HX is (0.01888 - x) moles and the equilibrium amount of X- is x; therefore,

x = 0.457(0.01888 - x)
1.457x = 0.008631 moles
x = 0.00592 moles

this amount is exactly the amount of KOH that was used to neutralize HEPES, thus the volume of KOH(aq) was:

(0.00592 moles) / (0.460 M KOH) = 0.0128780 L or 12.88 mL

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