Benzoic acid has a Ka of 6.3x10^-5. Calculate the pH of a 35mL sample of 0.130M

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Explanation / Answer

Dissociation of an acid: HA (equilibrium arrow) H(+) + A(-) --> HA is the acid and A(-) is the conjugate base. Equilibrium constant Ka for an acid: Ka = [H+] * [A-] / [HA] Ka = 6.3 x 10^-5 From the dissociation equation, there is a 1:1 molar ratio between [H+] and [A-]. Therefore [H+] = [A-] [HA] = 0.130 M (given in question) Solving for [H+] and [A-]: Ka = [H+] * [A-] / [HA] 6.3 x 10^-5 = [H+] / 0.130 --> Disregard [A-] since it will be the same as [H+] [H+] = (6.3 x 10^-5)(0.130) = 8.19 x 10^-6 M pH = -log[H+] = -log(8.19 x 10^-6 M) pH = 5.1 Hope this helps! :-)

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