Question

in the laboratory, a general chemistry student measured the ph of a 0.413 m aqueous solution of phenol (a weak acid), c₆h₅oh to be 5.177. use the information she obtained to determine the kₐ for this acid. kₐ(experiment) =

Explanation:

Step1: Calculate the concentration of \(H^+\) ions

The formula for \(pH\) is \(pH =-\log[H^+]\). Rearranging for \([H^+]\), we get \([H^+]=10^{-pH}\).
Substituting \(pH = 5.177\), we have \([H^+]=10^{- 5.177}\)

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For a weak acid \(HA\) (here \(HA = C_6H_5OH\)), the dissociation reaction is \(HA
ightleftharpoons H^++A^-\). At equilibrium, \([H^+]=[A^-]\) (from the stoichiometry of the dissociation reaction), and \([HA]=C - [H^+]\), where \(C\) is the initial concentration of the acid. Given \(C = 0.413\space M\) and \([H^+]=6.65\times 10^{-6}\space M\), since \(C\gg[H^+]\), \([HA]\approx C = 0.413\space M\)

Step2: Calculate the acid - dissociation constant \(K_a\)

The formula for \(K_a\) is \(K_a=\frac{[H^+][A^-]}{[HA]}\). Since \([H^+]=[A^-]=6.65\times 10^{-6}\space M\) and \([HA]\approx0.413\space M\)

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Answer:

\(1.07\times 10^{-10}\)