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Pad? 14:08 * 71% ncia.wwnorton.com Search Textbook Solutions | Chegg.com Smartwork5 K CH R 19 elisa.gonzalez507@myci.csuci.edu 03/26/18 SCORE A 25.0 mL sample of a 0.3000 M solution of aqueous trimethylamine is titrated with a 0.3750 M solution of HCI. Calculate the pH of the solution after 10.0,20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N 4.19 at 25°C. 1st attempt Part 1 (1 pt) d See Periodic Table Q See Hint pH after 10.0 mL of acid have been added- Part 2 (1 pt) pH after 20.0 mL of acid have been added = Part 3 (1 pt) pH after 30.0 mL of acid have been added- SUBMIT

Explanation / Answer

mmoles of (CH3)3N = 25 x 0.300 = 7.5

pKb = 4.19

part 1)

mmoles of HCl = 10 x 0.3750 = 3.75

this is half - equivalence point :

here : pOH = pKb

pOH = 4.19

pH = 9.81

part 2)

mmoles of HCl = 20 x 0.375 = 7.5

this is equivalence point .here salt remains

salt concentration = 7.5 / 20 + 25 = 0.166 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (4.19 + log 0.167)

pH = 5.29

part 3)

mmoles of HCl = 11.25

here strong acid reamins

[H+] = 0.0682 M

pH = 1.17

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