Question

A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries.
To find the pKa of X-281, you prepare a 0.089 M test solution of X-281 at 25.0 ∘C. The pH of the solution is determined to be 2.40.
a. What is the pKa of X-281? Express your answer numerically.
At 25∘C, for any conjugate acid-base pair
pKa + pKb = 14.00
b. What is pKb of the conjugate base of X-281? (Assume 25 ∘C.) Express your answer numerically.

Answer 1

Answer:

a. pka = 3,73.

b. pkb = 10,27.

Explanation:

a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:

HX + H₂O ⇄ H₃O⁺ + X⁻

Where ka is defined as:

$ka = \frac{[H_3O^+][X^-]}{[HX]}$

In equilibrium, molar concentrations are:

[HX] = 0,089M - x

[H₃O⁺] = x

[X⁻] = x

pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:

$[H_3O^+]} = 10^{-2,40}$

[H₃O⁺] = 0,004M

Thus:

[X⁻] = 0,004M

And:

[HX] = 0,089M - 0,004M = 0,085M

$ka = \frac{[0,004][0,004]}{[0,085]}$

ka = 1,88x10⁻⁴

And pka = 3,73

b. As pka + pkb = 14,00

pkb = 14,00 - 3,73

pkb = 10,27

I hope it helps!