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

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

$K_{a}= \frac{[A^{-}][H_{3}O^{+}]}{HA}$  ………. 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 Ka for a 0.50 M acid solution.

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

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

Ka = 6.69 x 10-5

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

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

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