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2 years ago

5

A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni

cian 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 0.086 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?

1 answer:

iogann1982 [59]2 years ago

8 0

Answer:

pKa X-281 = 3.714

Explanation:

pKa = - log Ka

∴ HA ↔ H3O+ + A-

⇒ Ka = [H3O+].[A-] / [HA]

∴ <em>C</em> X-281 = 0.086 M

∴ pH = 2.40 = - Log [H3O+]

⇒ [H3O+] = 3.981 E-3 M

mass blance:

⇒ <em>C</em> X-281 = 0.086 M = [HA] + [A-]

charge balance:

⇒ [H3O+] = [A-] + [OH-].......[OH-], its comes from water.

⇒ [H3O+] = [A-]

⇒ [HA] = 0.086 - [H3O+]

⇒ [HA] = 0.086 M - 3.981 E-3 M = 0.082 M

⇒ Ka = [H3O+]²/[HA]

⇒ Ka = (3.981 E-3)² / 0.082 = 1.9323 E-4

⇒ pKa = - Log Ka = - Log ( 1.9323 E-4)

⇒ pKa = 3.714

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The given question is incomplete. The complete question is as follows.

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Therefore, putting the given values into the above formula as follows.

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Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0

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