Calculate the pH of a solution that is 0.195M benzoic acid and 0.200M sodium benzoate, a salt whose anion is the conjuga...
Calculate the pH of a solution that is 0.195M benzoic acid and 0.200M sodium benzoate, a salt whose anion is the conjugate base of benzoic acid.
Homework Answers
Ka of benzoic acid = 6.30e-5
Ka = 6.3*10^-5
pKa = - log (Ka)
= - log(6.3*10^-5)
= 4.201
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.201+ log {0.2/0.195}
= 4.212
Answer: 4.21
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