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

Ka = 6.02x10⁻⁶

Step-by-step 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 =
  `([9.12x10^(-4)]*[9.12x10^(-4)])/([0.138])` = 6.02x10⁻⁶