PPR - Endocrine University of Oklahoma saplinglearning.com saplinglearning.com O

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PPR - Endocrine University of Oklahoma saplinglearning.com saplinglearning.com O saplingleaming.com/ibiscms/mod/ibi5Nie w.php?id-3871 623 Sapling Learning Andrea Brisby Jump to... macmillan leaming 010/27/2017 11:55 PM3.6/510/24/201 7 01:21 PM 4e e Print Calculator Periodic Table # Attempts Score Gradebook Assignn Question 14 of 14 Available l Map Due Date: 2 Sapling Learning eanng A 1.37 L buffer solution consists of 0.298 M propanoic acid and 0.174 M sodium propanoate. Calculate the pH of the solution following the addition of 0.064 moles of HCI. Assume that any contribution of the HCI to the volume of the solution is negligible. The Ka of propanoic acid is 1.34 x10 Grade Cat Descriptio 5 Number Solutions You can You can You have There is 6 8 9 eTextbo Help Wi 2011-2017 Sapling Learning, Inc. about uscareersprivacy policy terms of use contact us help 9:47 PM Type here to search @ ND4») 10242017 ES

Explanation / Answer

mol of HCl added = 0.064 mol

C2H5COO- will react with H+ to form C2H5COOH

Before Reaction:

mol of C2H5COO- = 0.174 M *1.37 L

mol of C2H5COO- = 0.2384 mol

mol of C2H5COOH = 0.298 M *1.37 L

mol of C2H5COOH = 0.4083 mol

after reaction,

mol of C2H5COO- = mol present initially - mol added

mol of C2H5COO- = (0.2384 - 0.064) mol

mol of C2H5COO- = 0.1744 mol

mol of C2H5COOH = mol present initially + mol added

mol of C2H5COOH = (0.4083 + 0.064) mol

mol of C2H5COOH = 0.4723 mol

Ka = 1.34*10^-5

pKa = - log (Ka)

= - log(1.34*10^-5)

= 4.8729

since volume is both in numerator and denominator, we can use mol instead of concentration

we have below equation to be used

pH = pKa + log {[conjugate base]/[acid]}

= 4.8729+ log {0.1744/0.4723}

= 4.44

Answer: 4.44

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