Question

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

Answer 1

ka = 2x10⁻⁹

Step-by-step 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 =
`([H^+][A^-])/([HA])`

ka =
`((2.24*10^(-5))(2.24*10^(-5)))/(0.25-2.24*10^(-5))` = 2x10⁻⁹