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A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka of the acid. g

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Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

  • HA ⇄ H⁺ + A⁻
  • Ka = [H⁺][A⁻]/[HA]

We calculate [H⁺] from the pH:

  • pH = -log[H⁺]
  • [H⁺] = [tex]10^{-pH}[/tex]
  • [H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

  • [HA] * 0.66/100 = [H⁺]

We calculate [HA]:

  • [HA] = 0.138 M

Finally we calculate the Ka:

  • Ka = [tex]\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}[/tex] = 6.02x10⁻⁶

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