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A 0.22 M solution of a weak acid HA dissociates such that 99.5% of the weak acid remains intact (i.e., remains as HA). To the nearest hundredths, what is the pH of the solution?

Respuesta :

Answer:

pH = 0.65975

Explanation:

  • HA ↔ A- + H+
  • pH = - Log [H+]
  • %α = ([A-] / [A-] + [HA]) × 100

∴ C HA = 0.22 M

mass balance:

⇒ C HA = [A-] + [HA] = 0.22 M

charge balance:

⇒ [H+] = [A-].........[OH-] is neglected, is come frome the water.

replacing in %α:

∴ %α = 99.5%

⇒ 99.5% = ( [A-] / [A-] + [HA])×100

⇒ 0.995 = [A-] / 0.22 M

⇒ [A-] = (0.22 M)*(0.995) = 0.2189 M

∴ [A-] = [H+] = 0.2189 M

⇒ pH = - Log(0.2189)

⇒ pH = 0.65975

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