Question

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

Answer 1

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⁺] = $10^{-pH}$

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