Question

in the laboratory, a general chemistry student measured the ph of a 0.413 m aqueous solution of hypochlorous acid to be 3.935. use the information she obtained to determine the $k_{a}$ for this acid. $k_{a}(experiment)=$

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 \(pH = 3.935\), we have \([H^+]=10^{- 3.935}\)

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Step2: Set up the equilibrium expression for \(HClO\) dissociation

The dissociation of \(HClO\) is \(HClO
ightleftharpoons H^++ClO^-\). Let \(c = 0.413\space M\) be the initial concentration of \(HClO\). At equilibrium, \([H^+]=[ClO^-]=x = 1.16\times 10^{-4}\space M\) and \([HClO]=c - x\). Since \(c\gg x\) (because \(0.413\gg1.16\times 10^{-4}\)), \([HClO]\approx c\)

The acid - dissociation constant \(K_a=\frac{[H^+][ClO^-]}{[HClO]}\)

Substituting the values: \(K_a=\frac{(1.16\times 10^{-4})(1.16\times 10^{-4})}{0.413}\)

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Answer:

\(3.26\times 10^{-8}\)