Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {0.3/0.2}
= 4.921
use:
PH = 14 - pOH
= 14 - 4.9208
= 9.0792
Answer: 9.08
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