A new potential heart medicine, code-named X-281, is being tested by a pharmaceu

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 0.095 M test solution of X-281. The pH of the solution is determined to be 3.00. What is the pKa of X-281?

Explanation / Answer

If X-281 is a weak acid HA, it will hydrolyze in water to produce some H3O+ and an equal amount of A-.

Molarity . . . .HA + H2O ==> H3O+ + A-
Initial . . . . .0.095 . . . . . . . . . .0 . . . .0
Change . . . .-x . . . . . . . . . . . .x . . . .x
At Equil. .0.095 - x . . . . . . . . .x . . . .x

Ka = [H3O+][A-] / [HA] = x^2 / (0.095 - x)

But since the pH = 3.00, then [H3O+] = 10^-pH = 1 x 10^-3 M = x = 0.001.

Ka = (0.001)^2 / (0.095 - 0.001) = 1.1 x 10^-5

To expand on HPV, Ka = 1.1 x 10^-5 however we still need to find pKa.
pKa = -log (Ka)
pKa = - log (1.1 x 10^-5) = 4.96

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