Question

What is the ka for a 0.25 m solution of a monoprotic weak acid with a ph of 4.65

Answer 1

Answer:

ka = 2x10⁻⁹

Explanation:

Using the pH we can calculate the molar concentration of H⁺, [H⁺]

• pH = -log[H⁺] = 4.65

• $10^{-4.65}$ =  [H⁺] = 2.24 x 10⁻⁵ M

Then we use the expression of Ka for the equilibrium of a weak monoprotic acid:

HA ↔ H⁺ + A⁻

Where:

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

ka =  $\frac{(2.24*10^{-5})(2.24*10^{-5})}{0.25-2.24*10^{-5}}$ =  2x10⁻⁹