If a buffer solution is 0.170 M in a weak base ( 𝐾b=2.9×10−5) and 0.590 M in its conjugate acid?

pOH = pKb + log ([conj.acid]/[Base])
pOH = -logkb + log ([conj.acid]/[Base])
pOH = 4.59 + log(0.590/0.17) = 5.13
pH=14-pOH=14-5.13=8.87...Show more

For the equilibrium: B + H2O <--> BH+ + OH-

Kb = 2.9X10^-5 = [BH+][OH-]/[B]

2.9X10^-5 = (0.170) [OH-]/(0.590)
[OH-] = 1.0X10^-4 M
pOH = 4.00
pH = 14.00 - pOH = 10.00...Show more