Question

A new potential heart medicine, code-named X-281, is being testedby a pharmaceutical company, Pharma-pill. As a research technicianat Pharma-pill, you are told that X-281 is a monoprotic weak acid,but because of security concerns, the actual chemical formula mustremain top secret. The company is interested in the drug'sK_a value because only the dissociated form of the chemicalis active in preventing cholesterol buildup in arteries.To find the pKa of X-281, you prepare a 0.084 \it M test solution of X-281. The pH of the solution isdetermined to be 2.40. What is the pKa of X-281?

Answer 1

Answer:

pKa = 3.72

Explanation:

Let's consider the dissociation of a generic monoprotic weak acid.

HA(aq) ⇄ H⁺(aq) + A⁻(aq)

For a weak acid, we can find the value of the acid dissociation constant (Ka) using the following expression:

$Ka=\frac{[H^{+}]^{2} }{Ca}$

where,

[H⁺] is the molar concentration of H⁺

Ca is the initial concentration of the acid

First, we need to find [H⁺] from pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.40 = 3.98 × 10⁻³ M

Then,

$Ka=\frac{[H^{+}]^{2} }{Ca}=\frac{(3.98 \times 10^{-3})^{2} }{0.084} =1.89 \times 10^{-4}$

Finally,

pKa = -log Ka = -log 1.89 × 10⁻⁴ = 3.72