[magnus]

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]

Use Ostwald Dilution Law (not abbreviated). Then calculate K_a from K_b
From these data ( 0.1 M is dissociated by 6.7%) you can easily calculate pOH.

[Enthalpy]

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

[magnus]

Use Ostwald Dilution Law (not abbreviated). Then calculate K_a from K_b
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]

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]

Use Ostwald Dilution Law (not abbreviated). Then calculate K_a from K_b
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 K_b. How K_a and K_b are related?

[H⁺]=√Ka*C = 6.92*10ˉ³    =>    pH = -log (6.92*10ˉ³) = 2.15

[OH^-] = cα

[magnus]

You calculated K_b. How K_a and K_b are related?

Ka  x Kb  = [ H3O+] [ OH– ] = Kw

[magnus]

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]

Equlibrium writen for this reaction

RNH₂ + H₂O ⇄ RNH₃⁺ + OH^-

gives K_b value

You calculated K_b, not K_a;

I thought this:   Ka= α²C/(1-α) =>  0.067² * 0.1 / 1 - 0.067 = 4.811*10ˉ⁴

and you know how calculated K_b

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.

[magnus]

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]

 K = α²C/(1-α) 
is a general formula for the Ostwald dilution law expressed by degree of dissociation - for amines, it is called K_b; for acids  - K_a and that's why you have to convert K_b 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