I'm lazy so let's call sodium propanoate NaPr. The Pr anion is hydrolyzed.
Pr^- + HOH ==> HPr + OH^-
Kb= (Kw/Ka) = (HPr)(OH^-)/(Pr^-)
Substitute for Kw and Ka (for propionic acid) and (Pr^-). Let X = HPr and OH. Solve for x, convert to pOH, then to pH.
What is the pH of 0.2 M solution of sodium propanoate?
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