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]=0.160 M , [H+]=3.00×10^−4 M , and [A−]=3.00 ×10^−4 M. Calculate the ????a value for the acid HA .

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Alleei Alleei · Answer 1 · 2019-08-12T12:00:35+02:00

Answer : The value of [tex]K_a[/tex] for the acid HA is, [tex]5.625\times 10^{-7}[/tex]

Solution :

The equilibrium reaction for dissociation of weak acid will be,

[tex]HA\rightleftharpoons H^++A^-[/tex]

The expression for dissociation constant will be,

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

Now put all the given values in this expression, we get:

[tex]K_a=\frac{(3\times 10^{-4})\times (3\times 10^{-4})}{(0.160)}[/tex]

[tex]K_a=5.625\times 10^{-7}[/tex]

Therefore, the value of [tex]K_a[/tex] for the acid HA is, [tex]5.625\times 10^{-7}[/tex]