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Chrome File Edit View History Bookmarks People Window New Tab ? RwW14 Ahd-Base Mixing and ) e Calculate The PH Of Each Of·× C O www.webassign.net/web/Student/Assignment-Re CH 201 Syllabus an... &My; Pack Portal DAEC Extra Credit 17817743 D TOPHAT i Moodle D 20 Textbook D Dymamic Perlodic T.. buses D Dik Gentys Motst N Nettix Bo p 4. O 824 points | Previous Answers Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.73-M HCI to 560. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative. a) water pH before mixing-7.00 pH after mixing 1.32 pH change--5.68 b) 0.161 M C2Hy02 pH before mixing - 8.96 pH after miking 8.9 pH change c) 0.161 M HC2H302 pH before mixing- pH after mixing PH change d) a buffer solution that is 0.161 M in each C2Hyo21 and HC2H302 pH before mixing- pH after mixing- pH change = Submit Answer Save Progress 5 -(24 points

Explanation / Answer

HCl added = 2.73 M x 10 ml = 27.3 mmol

a) H2O

initial pH before addition = 7

final pH after addition = -log(27.3 mmol/570 ml) = 1.320

pH change = 1.32 - 7= -5.68

b) initial pH

C2H3O2- + H2O <==> HC2H3O2 + OH-

5.55 x 10^-10 = x^2/0.161

x = [OH-] = 9.46 x 10^-6 M

pOH = -log[OH-] = 5.02

pH = 14 - pOH = 8.98

afer addition of HCl

initial C2H3O2- = 0.161 M x 560 ml = 90.16 mmol

moles HC2H3O2 formed = 27.3 mmol

moles C2H3O2- remained = 62.86 mmol

pH = pKa + log(C2H3O2-/(HC2H3O2)

      = 4.75 + log(62.86/27.3)

      = 5.11

change in pH = -3.87

c) initial pH

HC2H3O2 + H2O <==> C2H3O2- + H3O+

1.8 x 10^-5 = x^2/0.161

x = [H3O+] = 1.70 x 10^-3 M

pH = -log[H3O+] = 2.77

after adding HCl

final pH = -log(2.73) = 0.44

change in pH = -2.33

d) buffer

initial pH = pKa = 4.75

after adding HCl

final HC2H3O2 = 0.161 M x 560 ml + 27.3 mmol = 117.46 mmol

final C2H3O2 = 0.161 M x 560 ml - 27.3 mmol = 62.86 mmol

pH = 4.75 + log(62.86/117.46) = 4.48

change in pH = -0.27

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