Question #1809d

1 Answer
Stefan V.
Oct 2, 2015

Kb=3.7⋅10−6

Explanation:

Start by using the known pH of the solution to find the molarity of the hydrogen ions, H+. This will help you find the acid dissociation constant, Ka, for quinine's conjugae acid, QH+.

So, you know that quinine's conjugate acid, QH+, will ionize in aqueous solution to give

OH+(aq]→Q(aq]+H+(aq]

The concentration of the conjugate acid is simply the ratio between the number of moles you have and the volume of the solution

[QH+]=0.23 moles1.0 L=0.23 M

The concentration of H+ will be

[H+]=10−pH=10−4.58=2.63⋅10−5M

Since the dissociation reactrion produces equal numbers of moles of Q and H+, you get that

[Q]=[H+]=2.63⋅10−5M

The acid dissociation constant for this equilibrium will thus be

Ka=[Q]⋅[H+][QH+]

Ka=2.63⋅10−5⋅2.63⋅10−50.23=2.7⋅10−9

Finally, to get the base dissociation constant, Kb, use the fact that

Ka⋅Kb=KW , where

KW - the water self-ionization constant, equal to 10−14.

This means that Kb will be equal to

Kb=KWKa=10−142.7⋅10−9=3.7⋅10−6

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