Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: magnus 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
-
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.
-
Water can't dissolve 0.1M of octylamine
https://www.researchgate.net/publication/258071172_IUPAC-NIST_Solubility_Data_Series_96_Amines_with_Water_Part_1_C4-C6_Aliphatic_Amines chap 2.19
or
https://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=DE&language=de&productNumber=O5802&brand=ALDRICH&PageToGoToURL=https%3A%2F%2Fwww.sigmaaldrich.com%2Fcatalog%2Fproduct%2Faldrich%2Fo5802%3Flang%3Dde
-
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
-
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?
-
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?
[H⁺]=√Ka*C = 6.92*10ˉ³ => pH = -log (6.92*10ˉ³) = 2.15
[OH-] = cα
-
You calculated Kb. How Ka and Kb are related?
Ka x Kb = [ H3O+] [ OH– ] = Kw
-
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...
-
Equlibrium writen for this reaction
RNH2 + H2O ⇄ RNH3+ + OH-
gives Kb value
You calculated Kb, not Ka; I thought this: Ka= α²C/(1-α) => 0.067² * 0.1 / 1 - 0.067 = 4.811*10ˉ⁴
and you know how calculated Kb
Ka x Kb = Kw
Why you did not do it?
Calculation of [OH-] from this information is very simple
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.
-
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
-
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