Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
use:
pH = pKa + log {[conjugate base]/[acid]}
4.4 = 4.745+log {[CH3COO-]/[CH3COOH]}
log {[CH3COO-]/[CH3COOH]} = -0.3447
[CH3COO-]/[CH3COOH] = 0.4521
Answer: 0.452
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