
can someone help with this problem. thank you
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
use:
pH = pKa + log {[conjugate base]/[acid]}
pH = pKa + log {[CH3COO-]/[CH3COOH]}
= 4.745+ log {0.5/0.5}
= 4.745
Answer: 4.75
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