If the pH of 0.043 M BH+ is 6.44, find PKb for the weak base, B
2006-10-23 16:11:48 · 2 answers · asked by justjbk 3 in Science & Mathematics ➔ Chemistry
.. .. .. .. .. BH(+) <=> B + H(+)
Initial.. .. .C
Dissoc.. x
Produce.. .. .. .. .. .. ..x .. .x
At equilb C-x.. .. .. .. x.. ..x
Ka= [B][H+]/[BH+] = x^2/(C-x)
and x=[H+]= 10^-pH
But also Ka=Kw/Kb => pKb=pKw-pKa=>
pKb= pKw - (-log( x^2/(C-x)) )=
= 14+log(((10^-6.44)^2) / (0.043-10^-6.44)) = 2.49
2006-10-23 23:40:57 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋
pH + pOH = pKw = 14
pOH = 14-6.44 = 7.56 = - log[OH-], find OH- concentration
log[OH-] = -7.56 ----------> [OH-] = 10^-7.56
Kb = [BH+][OH-]/[B] = (10^-7.56)^2/[0.043] = 1.76 * 10^-14, that is one weak base.
please check for errors.
2006-10-23 16:22:25 · answer #2 · answered by Anonymous · 0⤊ 2⤋