# chemistry

A weak base that encompasses a concentration of 1.4 mol per litre has a percent ionization of 0.63%. Calculate the Kb of this weak base

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asked by Laura
1. Let's call a weak base BOH.
BOH ==> B^+ + OH^-

Kb = (B^+)(OH^-)/(BOH)

Look at the ICE chart.
I = initial concn (before ionization):
BOH = 1.4 mols/L.
(B^+) = 0
(OH^-) = 0

C = change in concn:
(B+) = +1.4*0.0063
(OH^-) = +1.4*0.0063
(BOH^-) = 1.4*0.9937 (0.9937 is the fraction from 100%-0.63% = 99.37%)

E = equilibrium concn:
(B^+) = 1.4*0.0063 = ??
(OH^-) = 1.4^0.0063 = ??
(BOH) = 1.4-(1.4*0.0063) OR (1.4*0.9937= ??)

Substitute the equilibrium numbers into Ka expression and calculate Ka.

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posted by DrBob222

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