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A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.250 M , [H+]=2.00×10−4 M , and [A−]=2.00×10−4 M . Calculate the value of pKa for the acid HA .

Respuesta :

Answer:

[tex]pK_{a}[/tex] of HA is 6.80

Explanation:

[tex]pK_{a}=-logK_{a}[/tex]

Acid dissociation constant ([tex]K_{a}[/tex]) of HA is represented as-

                [tex]K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}[/tex]

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

[tex]K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}[/tex]

So, [tex]K_{a}=1.6\times 10^{-7}[/tex]

Hence [tex]pK_{a}=-log(1.6\times 10^{-7})=6.80[/tex]

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