Ka of benzoic acid = 6.3*10^-5
Ka = 6.3*10^-5
pKa = - log (Ka)
= - log(6.3*10^-5)
= 4.201
use:
pH = pKa + log {[conjugate base]/[acid]}
pH = pKa + log {[C6H5COO-]/[C6H5COOH]}
pH = pKa + log {mol(C6H5COO-)/mol(C6H5COOH)}
= 4.201+ log {0.075/0.05}
= 4.377
Answer: 4.38
10. What is the pH of a solution which contains 0.05 moles benzoic acid and 0.075 moles sodium benzoate dissolved t...
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