A 0.50 m solution of an acid ha has ph = 2.24. what is the value of ka for the acid?

The question is –

A 0.50 m solution of an acid ha has ph = 2.24. what is the value of ka for the acid?

Answer:

⇒ K_a = 6.7 x 10^-5 = 0.000067

Explanation:

Acids are proton donors that get ionized to release H⁺ ions in water.

⇒ To solve this question, we assume that the acid given above is a monoprotic weak acid, i.e., it partially dissociates in an aqueous solution to produce a single H⁺ ion per HA molecule, as shown below.

HA _(aq) ⇌ A^– _(aq) + H⁺ _(aq)   ………… Equation 1

⇒ An H⁺ ion combines with an H₂O molecule to form a hydronium (H₃O⁺) ion in an acidic solution. 

HA _(aq) + H₂O _(l) ⇌ A^– _(aq) + H₃O⁺ _(aq) ………… Equation 2      (Ionization Equilibrium)

⇒ The concentration of hydronium [H₃O⁺] ions present in the solution determines its acidic strength. The higher the [H₃O⁺], the lower the pH and the greater the strength of the acidic solution.

pH of the acidic solution = -log[H⁺]

or

pH = -log [H₃O⁺] …………Equation 3

⇒ Rearranging equation 3 to make [H₃O⁺] the subject of the formula.

[H₃O⁺] = 10^-pH ……. Equation 4

⇒ As the pH of the solution is given, i.e., pH = 2.24. So we can use equation 4 to determine the concentration of hydronium ions formed in the 0.50 M acidic solutions.

[H₃O⁺] = 10^-2.24 = 5.75 x 10^-3 M.

⇒ As per equation 2, at equilibrium,1 mole of an acid (HA) dissociates to produce an equal concentration of A^– (the conjugate base) and H₃O⁺ (hydronium ions).

∴ So in our case, at equilibrium [H₃O⁺] = [A^–] = 5.75 x 10^-3 M.

The initial HA concentration decreases to produce A^– and H₃O⁺ ions in the aqueous solution.

An equilibrium is achieved when there is no further change in concentrations at either the reactant and/or the product side.

⇒ The equilibrium constant for the dissociation of a weak acid in an aqueous solution is known as the acid dissociation constant, i.e., K_a.

K_{a}= \frac{[A^{-}][H_{3}O^{+}]}{HA}  ………. Equation 5

In our example, [A^–] = [H₃O⁺] = 5.75 x 10^-3 M (already determined).

⇒ The concentration of the weak acid at equilibrium [HA] = Initial acid concentration- concentration of H₃O⁺ ions released in the aqueous solution.

[HA]_equilibrium = [HA]_initial – [H₃O]⁺ ………. Equation 6

⇒ You may have noticed that [HA]_initial is given in the question statement, i.e., 0.50 M.

[HA] _equilibrium = (0.50) – (5.75 x 10^-3) = 0.494 M

∴ Plugging in all the required values in equation 5 successfully gives us the value of K_a for a 0.50 M acid solution.

K_{a}= \frac{(5.75\times 10^{-8})(5.75\times 10^{-8})}{(0.494)}

K_{a}= \frac{(3.31\times 10^{-5})}{(0.494)}

K_a = 6.69 x 10^-5

Answer: K_a = 6.7 x 10^-5 (rounded off to one decimal place).

Note:  As it is a ratio, so there are no units for K_a.