Sarahkristine2118 Sarahkristine2118
  • 22-10-2020
  • Chemistry
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An unknown weak acid with a concentration of 0.079 M has a pH of 1.80. What is the Ka of the weak acid

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maacastrobr
maacastrobr maacastrobr
  • 22-10-2020

Answer:

3.18x10⁻³

Explanation:

The equilibrium of a weak acid can be written as:

  • HA ⇔H⁺ + A⁻

And Ka can be expressed as:

  • Ka=[tex]\frac{[H^+][A^-]}{[HA]}[/tex]

Keep in mind that [H⁺] = [A⁻].

We can calculate [H⁺] from the pH:

  • pH = 1.80 = -log[H⁺]
  • [H⁺]=[tex]10^{-pH}[/tex]=0.0158

Now we compute the calculated [H⁺] (and [A⁻]) with the [HA] given by the problem:

Ka = [tex]\frac{(0.0158)^2}{0.079}[/tex] = 3.18x10⁻³

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