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 0.080 mol L−1 test solution of X-281. The pH of the solution is determined to be 2.40. What is the pKa of X-281?

Express your answer numerically.

Answer 1

[X-281] = 0.080 M ...... Initial concentration of unknown monoprotic acid.

pH = 2.40

[H⁺] = 10^-2.40 = 0.004 M = 4.0*10^-3 M

This means at equilibrium, out of initial 0.080 M X-281, 0.004 M Ionized.

Let the chemical formula of monoprotic X-281 is HA which ionizes as

HA (aq) H⁺ (aq) + A^- (aq)

Ka = [H⁺][A^-] / [HA] ......... (1)

Expressed are the equilibrium concentrations and at equilibrium,

[H⁺] = [A^-] = 0.004 M . ( By stoichiometry of ionization)

[HA] = 0.080 - 0.004 = 0.076 M

Placing the values in the expression of ka w get,

Ka = (0.004)(0.004) / (0.076)

Ka = 2.11*10^-4.

Then,

pKa = -log Ka = -log (2.11*10^-4) = 3.68

For X-281, Ka = 2.11*10^-4 & pKa = 3.68.

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