Question

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

Answer 1

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=$\frac{[H^+][A^-]}{[HA]}$

Keep in mind that [H⁺] = [A⁻].

We can calculate [H⁺] from the pH:

• pH = 1.80 = -log[H⁺]

• [H⁺]=$10^{-pH}$=0.0158

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

Ka = $\frac{(0.0158)^2}{0.079}$ = 3.18x10⁻³