A .12 M solution of an unknown, monoequivalent base has a pH of 9.99. Calculate the Kb for this base.
2007-03-13 16:42:49 · 2 answers · asked by me 2 in Science & Mathematics ➔ Chemistry
B + H2O <--> BH+ + OH-
or
B(aq) <--> BH+(aq) + OH-(aq)
Kb = [BH+][OH-]/[B]
say the equilibriums of [BH+]=[OH-]=x
the equilibrium concentration of [B] = 0.12 -x
x2/(0.12-x) = Kb
but pH = -log[H+] = 9.99 or pOH = 4.01
so x = [OH-] = 10^-4.01 = 9.77*10-5
therefore Kb = (9.77*10-5)^2/.12 = 7.95*10-8
2007-03-13 16:57:55 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
A base dissociation is
BOH + -> OH- + B+ and
[OH-][B+]/[BOH]= Kb
Since pH=9.99, pOH=4.01 and [OH-]= 1x10^-4
Since [B+]=[OH-] and both << 0.12,
(1x10^-4)^2/.12 = Kb
So Kb = 8.33x10^-8
2007-03-13 23:58:35 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋