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 hour ago

Determine the pH during the titration of 36.3 mL of 0.281 M trimethylamine ((CH3)3N, Kb =...

Determine the pH during the titration of 36.3 mL of 0.281 M trimethylamine ((CH3)3N, Kb = 6.3×10-5) by 0.281 M HBr at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem. (a) Before the addition of any HBr ? (b) After the addition of 15.7 mL of HBr ? (c) At the titration midpoint ? (d) At the equivalence point ? (e) After adding 57.7 mL...

Determine the pH during the titration of 36.0 mL of 0.235 M trimethylamine ((CH3)3N, Kb =...

Determine the pH during the titration of 36.0 mL of 0.235 M trimethylamine ((CH3)3N, Kb = 6.3×10-5) by 0.235 M HBr at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem. a.) Before the addition of any HBr b.) After the addition of 15.3 mL of HBr c.) At the titration midpoint d.) At the equivalence point e.) After adding 56.2 mL of HBr

Consider the titration of 40.0 mL of 0.0600 M (CH3)3N (a weak base; Kb = 6.40e-05)...

Consider the titration of 40.0 mL of 0.0600 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HNO3. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 6.0 mL pH = (c) 12.0 mL pH = (d) 18.0 mL pH = (e) 24.0 mL pH = (f) 26.4 mL pH =

Consider the titration of 30.0 mL of 0.0700 M (CH3)3N (a weak base; Kb = 6.40e-05)...

Consider the titration of 30.0 mL of 0.0700 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HCl. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = _________ (b) 5.3 mL pH = __________ (c) 10.5 mL pH = ____________ (d) 15.8 mL pH = _____________ (e) 21.0 mL pH =___________ (f) 39.9 mL pH = __________

Consider the titration of 70.0 mL of 0.0300 M (CH3)3N (a weak base; Kb = 6.40e-05)...

Consider the titration of 70.0 mL of 0.0300 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HBr. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = _____ (b) 5.3 mL pH = _____ (c) 10.5 mL pH = _____ (d) 15.8 mL pH = _____ (e) 21.0 mL pH = _____ (f) 27.3 mL pH = _____

Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.1000...

Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.1000 M HNO3(aq) after 19.4 mL of the acid have been added. Kb of trimethylamine = 6.5 x 10-5 Correct Answer: 8.30 ± 0.02

Consider the titration of 60.0 mL of 0.100 M NH3 Kb= 1.8 x 10^-5 with 0.150...

Consider the titration of 60.0 mL of 0.100 M NH3 Kb= 1.8 x 10^-5 with 0.150 M HCl. Calculate the pH after the following volumes of titrant have been added: 40.5 mL & 60.0 mL

1). A 27.1 mL sample of 0.342 M trimethylamine, (CH3)3N, is titrated with 0.336 M hydrochloric...

1). A 27.1 mL sample of 0.342 M trimethylamine, (CH3)3N, is titrated with 0.336 M hydrochloric acid. The pH before the addition of any hydrochloric acid is _____ 2). A 20.8 mL sample of 0.305 M dimethylamine, (CH3)2NH, is titrated with 0.255 M hydroiodic acid. At the equivalence point, the pH is ______ 3). A 26.2 mL sample of 0.251 M diethylamine, (C2H5)2NH, is titrated with 0.215 M hydrobromic acid. After adding 42.8 mL of hydrobromic acid, the pH is...

A 25.6 mL sample of 0.219 M trimethylamine, (CH3)3N, is titrated with 0.373 M hydrochloric acid....

A 25.6 mL sample of 0.219 M trimethylamine, (CH3)3N, is titrated with 0.373 M hydrochloric acid. After adding 5.62 mL of hydrochloric acid, the pH is .

caluclate the pH in the titration of 50.00 mL of 0.200 M HNO3 by 0.100 M...

caluclate the pH in the titration of 50.00 mL of 0.200 M HNO3 by 0.100 M NaOH after the addition to the acid of solutions of (a) 0 mL NaOH (b) 10.00 ml NaOH (c) 100.00 ml Na OH (d) 150 mL NaOH

Question