A 0.0100 M solution of a weak base has pH = 9.78 at T = 25 C. What is the numerical value for Kb for the weak base?
we have below equation to be used:
pH = -log [H+]
9.78 = -log [H+]
log [H+] = -9.78
[H+] = 10^(-9.78)
[H+] = 1.66*10^-10 M
we have below equation to be used:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(1.66*10^-10)
[OH-] = 6.026*10^-5 M
Lets write the dissociation equation of BOH
BOH -----> B+ + OH-
1*10^-2 0 0
1*10^-2-x x x
Kb = [B+][OH-]/[BOH]
Kb = x*x/(c-x)
Kb = 6.026*10^-5*6.026*10^-5/(0.01-6.026*10^-5)
Kb = 3.65*10^-7
Answer: 3.65*10^-7
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