Answer:
3.18x10⁻³
Explanation:
The equilibrium of a weak acid can be written as:
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:
- [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
