Question

A 25.0 mL sample of a 0.0500 M solution of aqueous trimethylamine is titrated with a...

A 25.0 mL sample of a 0.0500 M solution of aqueous trimethylamine is titrated with a 0.0625 M solution of HCl. 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.

Homework Answers

Answer #1

mmol of base = MV = 0.05*25 = 1.25 mmol

a)

mmol of HCl = MV = 10*0.0625 = 0.625

this is a buffer

mmol of base left = 1.25 - 0.625 = 0.625

mmol of conjguate formed = 0 + 0.625

pOH = pKb + log(conjugate/base)

pOH = 4.19 + log(0.625/0.625)

pOH = 4.19

pH = 14-pOH =14-4.19 = 0.81

b)

mmol of HCl = MV = 20*0.0625 = 1.25

this is equivalence point

Vtotal= 25+20 = 45 mL

[BH+] = mmol/V = 1.25/45 = 0.02777

BH+ = B+ + H+

Ka = [B][H+]/[BH+]

(10^-(14-4.19)) = x*x/(0.02777-x)

x = 2.07*10^-6

pH = -log(2.07*10^-6 = 5.68

c)

mmol of HCl = MV = 30*0.0625 = 1.875

mmol of HCl left = 1.875-1.25 = 0.625

[H+] = mmol/V = 0.625/ (25+30) = 0.0113

pH = -log(0.0113) = 1.946

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