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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: magnus on February 13, 2019, 05:27:42 PM

Title: aqueous solutions
Post 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
Title: Re: aqueous solutions
Post by: AWK on February 13, 2019, 06:29:08 PM: 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.
Title: Re: aqueous solutions
Post by: Enthalpy on February 14, 2019, 07:21:52 AM: 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
Title: Re: aqueous solutions
Post by: magnus on February 14, 2019, 12:35:39 PM: Quote from: AWK on February 13, 2019, 06:29:08 PM
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
Title: Re: aqueous solutions
Post by: mjc123 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?
Title: Re: aqueous solutions
Post by: AWK on February 14, 2019, 03:58:10 PM: Quote from: magnus on February 14, 2019, 12:35:39 PM
Quote from: AWK on February 13, 2019, 06:29:08 PM
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?
Quote
[H⁺]=√Ka*C = 6.92*10ˉ³ => pH = -log (6.92*10ˉ³) = 2.15
[OH^-] = cα
Title: Re: aqueous solutions
Post by: magnus on February 16, 2019, 01:35:23 PM: You calculated K_b. How K_a and K_b are related?
Quote
Ka x Kb = [ H3O+] [ OH– ] = Kw
Title: Re: aqueous solutions
Post by: magnus on February 16, 2019, 01:39:35 PM: Quote from: mjc123 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?

More than a salt, it is an organic base...
Title: Re: aqueous solutions
Post by: AWK on February 16, 2019, 05:39:51 PM: Equlibrium writen for this reaction
Quote
RNH₂ + H₂O ⇄ RNH₃^⁺ + OH^-
gives K_b value

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

and you know how calculated K_b
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.
Title: Re: aqueous solutions
Post by: magnus 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
Title: Re: aqueous solutions
Post by: AWK 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 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