Question

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] Another way to express acid strength is by using pKa: pKa=−logKa Another way to express base strength is by using pKb: pKb=−logKb Part A 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 at 25.0 ∘C. The pH of the solution is determined to be 2.40. What is the pKa of X-281?

Answer 1

Answer:

pKa = 3.675

Explanation:

• pKa = - Log Ka

∴ C X-281 = 0.079 M

∴ pH = 2.40

let X-281 a weak acid ( HA ):

∴ HA ↔ H+ + A-

⇒ Ka = [H+] * [A-] / [HA]

mass balance:

⇒ C HA = 0.079 M = [HA] + [A-]

⇒ [HA] = 0.079 - [A-]

charge balance:

⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water

⇒ [H+] = [A-]

∴ pH = - log [H+] = 2.40

⇒ [H+] = 3.981 E-3 M

replacing in Ka:

⇒ Ka = [H+]² / ( 0.079 - [H+] )

⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )

⇒ Ka = 2.113 E-4

⇒ pKa = - Log ( 2.113 E-4 )

⇒ pKa = 3.675