pKa = 4.96
If X-281 is a weak acid HA, it will hydrolyze in water to produce some H3O+ and an equal amount of A-.
HA + H2O ==> H3O+ + A-
Initial molarity ; 0.095 . . . . . . . . . .0 . . . .0
Change; -x . . . . . . . . . . . .x . . . .x
At Equilibrium; 0.095 - x . . . . . . . . .x . . . .x
Ka = [H3O+][A-] / [HA]
= x^2 / (0.095 - x)
But since the pH = 3.00, then [H3O+]
= 10^-pH = 1 x 10^-3 M
= x = 0.001.
Ka = (0.001)^2 / (0.095 - 0.001)
= 1.1 x 10^-5
But; pKa = -log (Ka)
pKa = - log (1.1 x 10^-5)
= 4.96
pka value tells about the strength of an acid. The pka of x-281 at 25.0 ∘C is 4.96.
pka is the method to show the strength of an acid. A lower pka value shows that the acid is strong.
Given,
The solution is a weak acid, so it will fully dissociate into water.
[tex]\bold{pka= \dfrac{[H_3O^+][A^-]}{ [HA] } }\\[/tex]
[tex]\bold{pka= \dfrac{ X^2}{ 0.095-x} }\\[/tex]
But, since the pH is 3.00
Thus, [tex]pH = 10^-^p^H \\\\pH= 1 \times 10^-^3 M\\\\pH= x= 0.001[/tex]
The pka is
[tex]\bold{pka= \dfrac{ 0.001^2}{ 0.095-0.001}= 1.1\times 10^-^5 }\\[/tex]
Now, the pka = -log (Ka)
[tex]\bold{ - log (1.1 x 10^-^5) = 4.96}[/tex]
Thus, the pka of x-281 at 25.0 ∘C is 4.96.
Learn more about, pka value, here:
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