Consider the titration of 50.00 mL of 0.100 M trimethylamine ( (CH3)3N, Kb = 6.25 x...

Consider the titration of 50.00 mL of 0.100 M trimethylamine ( (CH3)3N, Kb = 6.25 x 10-5) with 0.100 M HI. Calculate the pH at the initial point.

Expert Solution

(CH3)3N + H2O -------------------------> (CH3)3NH+ + OH-

0.100 0 0

0.100 -x x x

Kb = x^2 / 0.1-x

6.25 x 10^-5 =x^2 / 0.1-x

x^2 + 6.25 x 10^-5 x - 6.25 x 10^-6 = 0

x = 2.47 x 10^-3

[OH-] = 2.47 x 10^-3 M

pOH = -log [OH-]

pOH = -log (2.47 x 10^-3)

pOH = 2.61

pH +pOH = 14

pH = 11.39

queen_honey_blossom answered 1 month ago

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