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Question

The Ka of propionic add is 1.34×105. What is the pH of a solution containing 0.5 M propionic acid and 0.5 M sodium propionate?

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Solution

As we know that,
pKa=logKa
Ka=1.34×105(Given)
pKa=log(1.34×105)
pKa=50.127=4.873
Now,
pH=pKa+log([salt][acid])
Given:-
[salt]=[acid]=0.5M
pH=4.87+log0.50.5=4.873
Hence the pH of the solution is 4.873.

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