Question

Learning Goal:

To understand the relation between the strength of an acid or a base and its pKa and pKb values.

The degree to which a weak acid dissociates in solution is given by its acid-ionization constant, Ka. For the generic weak acid, HA,

HA(aq)⇌A−(aq)+H+(aq)

and the acid-ionization constant is given by

Ka=[A−][H+][HA]

Similarly, the degree to which a weak base reacts with H2O in solution is given by its base-ionization constant, Kb. For the generic weak base, B,

B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)

and the base-ionization constant is given by

Kb=[BH+][OH−][B]

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 0.079 M test solution of X-281. The pH of the solution is determined to be 2.70.

What is the pKa of X-281?

Answer 1

Let X-281 as HA, because its monoprotonic weak acid, the equation is

HA = H⁺ + A^-

Initially at t=0 0.079 0 0

at end of reacction at t=t 0.079-x x x

Thus, acid dissociation constant Ka is,

Ka = [H⁺] [A^-] / [HA]

as,[H⁺] = [A^-], and pH = 2.70 = -log [H⁺]

Thus , [H⁺] = 1.99 * 10^-3 mol = [A^-]

now,

Ka = ( 1.99 * 10^-3 * 1.99 * 10^-3 ) /( 0.079 -   1.99 * 10^-3)

= 3.96 * 10^-6 / 0.07701

= 5.142 * 10^-5

pKa = -log Ka = -log 5.142 * 10^-5 = 4.288

Thus, pKa of X-281 is 4.288