Question
In: Chemistry
Consider a 0.64 M solution of dimethylamine, (CH3)2NH (Kb = 5.4×10-4). Mark the major species in...
Consider a 0.64 M solution of dimethylamine, (CH3)2NH (Kb = 5.4×10-4).
Mark the major species in the solution.
_______TrueFalse (CH3)2NH
_______TrueFalse
(CH3)2NH2+
_______TrueFalse H2O
_______TrueFalse H+
_______TrueFalse OH-
Complete the following table (in terms of "x", the amount of dimethylamine which reacts). Minus signs must be included, omit positive signs and omit molarity units (they are assumed).
| (CH3)2NH | (CH3)2NH2+ | OH- | |
|---|---|---|---|
| Initial | |||
| Change | |||
| Equilibrium | 0.64 - x |
Determine the equilibrium concentration of
(CH3)2NH2+
M
Calculate the pH of the solution.
pH =
Solutions
Expert Solution
(CH3)2NH ----------------> (CH3)2NH2+ + OH-
0.64 0 0 ---------> initial
- x +x +x ---------> change
0.64 - x x x ---------> equilibrium
Kb = [(CH3)2NH2+][OH-] / [(CH32NH)]
5.4 x 10^-4 = x^2 / 0.64 - x
x = 0.0183
equilibrium concentration of (CH3)2NH2+ = 0.018 M
[OH-] = 0.0183 M
pOH = - log (0.0183) = 1.74
pH = 12.26
Major species :
OH- , (CH3)2NH2+ , (CH3)2NH
queen_honey_blossom answered 2 years agoRelated Solutions
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(b) After the addition of 14.0 mL
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Consider the titration of 40.0 mL of 0.0600 M (CH3)2NH (a weak base; Kb = 0.000540)...
Consider the titration of 40.0 mL of 0.0600 M
(CH3)2NH (a weak base; Kb =
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A 24.3 mL sample of 0.349 M dimethylamine, (CH3)2NH, is titrated with 0.275 M hydrochloric acid....
A 24.3 mL sample of 0.349 M
dimethylamine,
(CH3)2NH, is titrated with
0.275 M hydrochloric acid.
(1) Before the addition of any
hydrochloric acid, the pH is _____
(2) After adding 13.6 mL of
hydrochloric acid, the pH is _____
(3) At the titration midpoint, the pH is
_____
(4) At the equivalence point, the pH is
_____
(5) After adding 45.9 mL of
hydrochloric acid, the pH is _____
Equilibrium constants are found here:
https://docs.google.com/document/d/1cNVOoEOVMqPSx8QdL8mc3zd8yFTwBx8VWd2MiDCDA1o/edit
1)A 21.7 mL sample of 0.267 M dimethylamine, (CH3)2NH, is titrated with 0.241 M perchloric acid....
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Consider the titration of 65 mL of 1.75 M diethylamine, (CH3CH2)2NH (Kb = 1.3×10-3) with 0.25...
Consider the titration of 65 mL of 1.75 M diethylamine,
(CH3CH2)2NH (Kb = 1.3×10-3) with 0.25 M HCl. How many mL of HCl
will need to be added to the diethylamine solution to reach a pH of
11.11?
The answer is NOT 455 mL or 45.5 mL
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