Ka = 6.8*10^-4
pKa = - log (Ka)
= - log(6.8*10^-4)
= 3.167
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.167+ log {0.35/0.1}
= 3.71
use:
pH = -log [H3O+]
3.71 = -log [H3O+]
[H3O+] = 1.95*10^-4 M
Answer:
[H3O+] = 2.0*10^-4 M
pH = 3.71
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