A 0.01 M solution of an acid has pH = 5.0. What is the Ka value?

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The question is –

A 0.01 M solution of an acid has pH = 5.0. What is the Ka value?

⇒ Ka = 1 x 10-8.

Explanation:

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

⇒ Considering pH 5.0, 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 = 5.0. So we can use equation 4 to determine the concentration of hydronium ions formed in the 0.01 M acidic solutions.

[H3O+] = 10-5.0 = 1 x 10-5 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] = 1 x 10-5 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+] = 1 x 10-5 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., 0.01 M.

[HA] equilibrium = (0.01) – (1 x 10-5) = 9.99 x 10-3 M

⇒ Plugging in all the required values in equation 5 successfully gives us the value of Ka for a 0.01 M acid solution.

$K_{a}= \frac{(1\times 10^{-5})(1\times 10^{-5})}{(9.99\times 10^{-3})}$

$K_{a}= \frac{(1\times 10^{-10})}{(9.99\times 10^{-3})}$

Ka = 1.00 x 10-8.

Answer: Ka = 1 x 10-8 (rounded off to one significant figure).

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

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