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.

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error