What is the ph of a 0.500 m solution of benzoic acid, pka = 4.19?

Solution:

 Benzoic acid is C6H5COOH  

In finding pH

C6H5COOH(aq) <=> C6H5COO^- + H^+ pKa = 4.19, pKa = -logKa so Ka = 10^(-4.19)  

Ka = 6.45 x 10^-6  

[C6H5COO^-] = x = [H^+]; [C6H5COOH] = 0.5 - x (we are able to make an estimate of [C6H5COOH] = 0.5.

Ka = [H^+][C6H5COO^-]/[C6H5COOH] = x^2/(0.5 - x) = 6.45 x 10^-6  

Now,

According to the quadratic equation. x^2 = 3.23 x 10^-5 - 6.45 x 10^-6x  

x^2 + (6.45 x 10^-6)x - 3.23 x 10^-5 = 0  

enter a = 1, b = 0.00000645, c = 0.0000323  

x = 5.68 x 10^-3 = 0.00568 M  expression is [C6H5COOH] = 0.5 M is the correct answer.  

[H^+] = 0.00568 M, so pH = -log(0.00568 M ) = 2.25  

This is the required solution.

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