Respuesta :

sebassandin

Answer:

The acid with about [tex]Ka=6.31x10^{-10}[/tex].

Explanation:

Hello,

By assuming that the dissociation of a weak acid is:

[tex]H^+X^-<-->H^++X^-[/tex]

From the following equation one computes pKa as long as the concentration of the conjugate base equals the concentration of the acid for this buffered solution:

[tex]pH=pKa+Log(\frac{[Conj.Base]}{[Acid]})\\pKa=pH-Log(1)\\pKa=pH=9.2\\Ka=10^{-9.2}\\Ka=6.31x10^{-10}[/tex]

So the acid will be that with a [tex]Ka=6.31x10^{-10}[/tex] or close since the possible answers are not shown.

Best regards.

fichoh

An aqueous solution containing the mixture of a weak acid and its conjugate base is called a buffer solution. Hence, the weak acid to use will be [tex] Ka = 6.31 \times 10^{-10} [/tex]

  • pH of buffer solution = 9.20

Using the relation:

[tex] pH = pKa + log(\frac{[base conjugate]}{[acid]}) [/tex]

[tex] 9.20 = - log Ka + log (1) [/tex]

[tex] 9.20 = - log Ka [/tex]

[tex] Ka = 10^{-9.20} [/tex]

[tex] Ka = 6.31 \times 10^{-10} [/tex]

Therefore, the acid dissociation constant should be [tex] Ka = 6.31 \times 10^{-10} [/tex]

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