What is the concentration of H3O in a .30 M urea solution, a weak base. Kb = 1.5 X10^-14
Let α be the dissociation of the weak base, urea(BOH)
BOH
<---> B + + OH-
initial conc. c 0 0
change -cα +cα +cα
Equb. conc. c(1-α) cα cα
Dissociation constant, Kb = (cα x cα) / ( c(1-α)
= c α2 / (1-α)
In the case of weak bases α is very small so 1-α is taken as 1
So Kb = cα2
==> α = √ ( Kb / c )
Given Kb = 1.5x10-14
c = concentration = 0.30 M
Plug the values we get α = 2.24x10-7
So the concentration of [OH-] = cα
= 0.3 x1.5x10-14
= 6.71 x 10-8 M
pOH = - log [OH-]
= - log (6.71 x 10-8 )
= 8 - log 6.71
= 7.17
So pH = 14 - 7.17 = 6.83
pH = - log[H3O+] = 6.83
---> [H3O+] = 10 -6.83
= 1.48x10-7 M
Therefore the concentration of [H3O+]= 1.48x10-7 M
What is the concentration of H3O in a .30 M urea solution, a weak base. Kb...
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