Answer :
The principal species at pH 6.00 is [tex]HA^-[/tex], the pH at which [tex][H_2A] = [HA^-][/tex] is 6.00, and the pH at which [[tex]HA^-[/tex]] = [[tex]A2^-[/tex]] is 8.00.
1. Compare the pH value with the [tex]pK_a[/tex] values of the acid to find the principal species at 6.00 pH.
[tex]pK_1 = 4.00\\pK_2 = 8.00[/tex]
6.00 pH is between [tex]pK_1[/tex] and [tex]pK_2[/tex] and at that the acid is partially ionized.
The principal species at 6.00 pH contains the highest concentration.
Here, the concentration of [tex]HA^-[/tex] is higher than that of [tex]H_2A[/tex], and the concentration of [tex]A2^-[/tex] is negligible.
Therefore, the principal species at pH 6.00 is B. [tex]HA^-[/tex].
2. Compare the concentrations of the two species for finding the pH at which [[tex]H_2A[/tex]] = [[tex]HA^-[/tex]].
The acid is exactly halfway between being fully ionized and being in its molecular form when [[tex]H_2A[/tex]]=[[tex]HA^-[/tex]]. This occurs at the average of the two [tex]pK_a[/tex] values.
[tex]pH = \frac{pK_1 + pK_2}{2}[/tex]
[tex]= \frac{4.00 + 8.00}{2}\\ = 6.00[/tex]
Therefore, the pH at which [[tex]H_2A[/tex]] = [[tex]HA^-[/tex]] is D. 6.00.
3. To find the pH at which [[tex]HA^-[/tex]] = [[tex]A2^-[/tex]], compare the concentrations of the two species.
[[tex]HA^-[/tex]] = [[tex]A2^-[/tex]] when the acid is fully ionized and there is no molecular form left. This occurs at a pH higher than [tex]pK_2[/tex], which means the acid is completely deprotonated.
Therefore, the pH at which [[tex]HA^-[/tex]] = [[tex]A2^-[/tex]] is higher than the [tex]pK_2[/tex] value of 8.00. Therefore, the pH at which [[tex]HA^-[/tex]] = [[tex]A2^-[/tex]] is C. 8.00.
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