Answer:
The pKa of X-281 is 3.73.
Explanation:
X-281 is a monoprotic weak acid so that it will not dissociate completely in solution (weak acid) and will only produce 1 mol of protons per mol compound (monoprotic acid).
The dissociation equation could be written as follows:
X-281-H ⇆ X-281 + H⁺
Note the equilibrium arrows indicating that not all X-281-H dissociates at equilibrium.
Initially, the concentration of X-281-H is 0.089 M. At equilibrium, the concentration of the dissociation products, X-281 and H⁺, is unknown but both must be the same, since the drug is a monoprotic acid. We can call "X" to the concentration of the products. The concentration of X-281-H is the initial concentration minus the concentration of the products: 0.089 M - X. Then, at equilibrium, these are the concentrations of each species present:
X-281-H ⇆ X-281 + H⁺
(0.089 M-X) X X
We also know that the pH is 2.40. Then:
pH = -log[H⁺] = 2.40
where [H⁺] is the molar concentration of the protons or "X". Then:
-log X = 2.40
log X = -2.40
X = 10^(-2.40) = 3.98 x 10⁻³ M
The concentration of each species present in the equilibrium is then:
[H⁺] = 3.98 x 10⁻³ M
[X-281] = 3.98 x 10⁻³ M
[X-281-H] = 0.089 M - 3.98 x 10⁻³M = 0.085 M
At equlibrium, the acidity constant Ka is:
Ka = [X-281] * [H⁺] / [X-281-H]
Ka = (3.98 x 10⁻³ M * 3.98 x 10⁻³ M ) / 0.085 M = 1.86 x 10⁻⁴
Then the pKa is:
pKa = -log Ka = -log (1.86 x 10⁻⁴) = <u>3.73</u>
