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Story Bookmarks Tools Window Help x G calculator Google search x dsmartwork/view.php?lda675592&isStudenta1; Search Question 3 of 9 (1 point) A of a Msolution of aqueous trimethylamine is titrated with a 0.175 M solution of HCI. Calculate the of the solution after 10.0, 20.0,and acid have been added, poko of (CH33N 4.19 at 25 C. pH after 10.0 mL of acid have been added: pH after 20.0 mL of acid have been added: pH after 30.0 mL of acid have been added Check Answer Hint Exit Activity Progress: 1 2 Activity Grade: 9 Points Possible

Explanation / Answer

pKa of (CH3)3N = 4.19

millimoles of trimethylamine = 25 x 0.140 = 3.5

a) after the addition of 10 mL HCl

millimoles of HCl = 10 x 0.175 = 1.75

it is half equivalence point . so

pOH = pKb

pOH = 4.19

pH +pOH =14

pH = 9.81

b) after the addition of 20.0 mL HCl

millimoles of acid = 20 x 0.175 = 3.5

it is equivalence point only salt is formed

salt millimoles = 3.5

salt concentration = millimoles / total volume = 3.5 / (25+20) = 0.078 M

salt is from strong acid weak base so pH <7

pH = 7 -1/2 [pKb + logC]

pH = 7 - 1/2 [4.19 + log 0.078]

pH = 5.46

c)

b) after the addition of 30.0 mL HCl

millimoles of acid = 30 x 0.175 = 5.25

here strong acid reamins.

[HCl] = 5.25 - 3.5 / (30 +25) = 0.0318 M

pH = -log [H+] = -log (0.0318)

pH = 1.50

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