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### Topic: aqueous solutions  (Read 1382 times)

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#### magnus

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##### aqueous solutions
« on: February 13, 2019, 05:27:42 PM »
An organic base CH₃(CH₂)₇NH₂ dissolved in water at the concentration of 0.1 M is dissociated by 6.7%. How much is the pH of the solution and the Ka of the
base?
RNH2 + H2O ⇄ RNH3+ + OH-

I have some difficulty solving this exercise

#### AWK

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##### Re: aqueous solutions
« Reply #1 on: February 13, 2019, 06:29:08 PM »
Use Ostwald Dilution Law (not abbreviated). Then calculate Ka from Kb
From these data ( 0.1 M is dissociated by 6.7%) you can easily calculate pOH.
AWK

#### Enthalpy

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##### Re: aqueous solutions
« Reply #2 on: February 14, 2019, 07:21:52 AM »

#### magnus

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##### Re: aqueous solutions
« Reply #3 on: February 14, 2019, 12:35:39 PM »
Use Ostwald Dilution Law (not abbreviated). Then calculate Ka from Kb
From these data ( 0.1 M is dissociated by 6.7%) you can easily calculate pOH.

I thought this:   Ka= α²C/(1-α) =>  0.067² * 0.1 / 1 - 0.067 = 4.811*10ˉ⁴
[H⁺]=√Ka*C = 6.92*10ˉ³    =>    pH = -log (6.92*10ˉ³) = 2.15

#### mjc123

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##### Re: aqueous solutions
« Reply #4 on: February 14, 2019, 12:57:39 PM »
That would work assuming you dissolved a salt RNH3+X-, and RNH3+ dissociated to the extent of 6.7%. But that is not the situation you have, is it? How might you adapt this method to your situation?

#### AWK

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##### Re: aqueous solutions
« Reply #5 on: February 14, 2019, 03:58:10 PM »
Use Ostwald Dilution Law (not abbreviated). Then calculate Ka from Kb
From these data ( 0.1 M is dissociated by 6.7%) you can easily calculate pOH.

I thought this:   Ka= α²C/(1-α) =>  0.067² * 0.1 / 1 - 0.067 = 4.811*10ˉ⁴
You calculated Kb. How Ka and Kb are related?
Quote
[H⁺]=√Ka*C = 6.92*10ˉ³    =>    pH = -log (6.92*10ˉ³) = 2.15
[OH-] = cα
AWK

#### magnus

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##### Re: aqueous solutions
« Reply #6 on: February 16, 2019, 01:35:23 PM »
You calculated Kb. How Ka and Kb are related?
Quote
Ka  x Kb  = [ H3O+] [ OH– ] = Kw

#### magnus

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##### Re: aqueous solutions
« Reply #7 on: February 16, 2019, 01:39:35 PM »
That would work assuming you dissolved a salt RNH3+X-, and RNH3+ dissociated to the extent of 6.7%. But that is not the situation you have, is it? How might you adapt this method to your situation?

More than a salt, it is an organic base...

#### AWK

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##### Re: aqueous solutions
« Reply #8 on: February 16, 2019, 05:39:51 PM »
Equlibrium writen for this reaction
Quote
RNH2 + H2O ⇄ RNH3+ + OH-
gives Kb value

You calculated Kb, not Ka;
Quote
I thought this:   Ka= α²C/(1-α) =>  0.067² * 0.1 / 1 - 0.067 = 4.811*10ˉ⁴

and you know how calculated Kb
Quote
Ka  x Kb = Kw

Why you did not do it?

Calculation of [OH-] from this information is  very simple
Quote
0.1 M is dissociated by 6.7%

Then you can calculate [H3O+], and in the next step pH
or
you can calculate pOH, then pH.
« Last Edit: February 16, 2019, 05:51:06 PM by AWK »
AWK

#### magnus

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##### Re: aqueous solutions
« Reply #9 on: February 20, 2019, 12:26:23 AM »
The results of the test are compatible with the data I have previously marked, which is why I thought it was fine.
pH=2.17, Ka=4.8×10-4
pH=11.83, Ka=5.0×10-10
pH=11.83, Ka=4.8×10-4
pH=2.17, Ka=5.0×10-10

#### AWK

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##### Re: aqueous solutions
« Reply #10 on: February 20, 2019, 03:02:00 AM »
K = α²C/(1-α)
is a general formula for the Ostwald dilution law expressed by degree of dissociation - for amines, it is called Kb; for acids  - Ka and that's why you have to convert Kb to Ka.
.
pH of bases is always greater than 7. Value of 2.17 is called pOH. So the answer in accordance with your question is only: pH=11.83, Ka=4.8×10-4
AWK