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

3.18x10⁻³

Explanation:

The equilibrium of a weak acid can be written as:

HA ⇔H⁺ + A⁻

And Ka can be expressed as:

Ka=

Keep in mind that [H⁺] = [A⁻].

We can calculate [H⁺] from the pH:

pH = 1.80 = -log[H⁺][H⁺]==0.0158

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

Ka = = 3.18x10⁻³

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Ais stronge