Find the fraction of dissociation, of an aqueous 0 400Mtriethylamine ...

Let be the dissociation of the weak basetrimethylamine BOH B OH initial co ....

Find the fraction of dissociation, Î± of an aqueous 0.400M triethylamine [(CH2CH3)N] solution.

4.2 Ratings (763 Votes)

Let Î± be the dissociation of the weak base,trimethylamine
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â BOH <---> B ⁺ + OH^-

initial conc.Â Â Â Â Â Â Â Â Â Â Â cÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â 0

changeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â -cÎ±Â Â Â Â Â Â Â Â Â Â Â +cÎ±Â Â Â Â Â +cÎ±

Equb. conc.Â Â Â Â Â Â Â Â c(1-Î±)Â Â Â Â Â Â Â cÎ±Â Â Â Â Â cÎ±

Dissociation constant, K_b = (cÎ± x cÎ±) / ( c(1-Î±)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = c Î±² / (1-Î±)

In the case of weak bases Î± is very small so 1-Î± is taken as 1

So K_b = cÎ±²

==> Î± = âˆš ( K_b / c )

Given K_b = 6.46x10^-5

Â Â Â Â Â Â Â Â Â c = concentration = 0.400 M

Plug the values we get Î± = 0.0127

So the fraction of dissociation is Î± = 0.0127

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