Use the information below to answer the following question. What is the pH of a 0.100 M solution of benzoic acid?
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Benzoic acid solutions are sometimes used in experiments to determine the molarity of a basic solution of unknown concentration.
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PhCOOH(aq) --> PhCOO-(aq) + H+(aq)
Ka =[PhCOO-][H+]/[PhCOOH]
x = amount of acid dissociate, then [PhCOO-] = [H+] = x
Ka = x^2/( 0.1M - x ) = 0.000065
6.5 * 10 ^-6 - 6.5 * 10^-5x - x^2 = 0
solve for x = 0.00252 M
so pH = - log(0.00252) = 2.5 answer
Hope u understood well
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