For the weak acid HX, what is the pH of a 0.10 M solution if the acid is 4.0% ionized?

HX ==> H^+ + X^-

(H^+) = 0.1*0.04 = ??
pH = -log(H^+)

2.398

To find the pH of a weak acid solution, we need to consider the ionization of the acid. In this case, we are given that the acid HX is 4.0% ionized. This means that only 4.0% of the HX molecules dissociate into their ions.

To solve this problem, we need to set up an equilibrium expression for the ionization of the weak acid HX. Let's assume that the concentration of ionized acid (HX-) at equilibrium is represented by [HX-] and the concentration of the unionized acid (HX) is represented by [HX].

The equilibrium expression for the ionization of the weak acid HX can be written as:

HX ⇌ H+ + X-

The expression for the ionization constant (Ka) of the weak acid is:

Ka = [H+][X-] / [HX]

Since only 4.0% of the HX is ionized, we can say that [HX-] = 0.04 * [HX].

Substituting this information into the equilibrium expression, we get:

Ka = [H+][0.04 * [HX]] / [HX]

Simplifying the equation, we have:

Ka = 0.04 * [H+]

Now, we can use the fact that pH = -log[H+] to find the pH of the solution. So, we need to determine the concentration of H+ ions.

Rearranging the equation, we get:

[H+] = Ka / 0.04

Substitute the given value of Ka into the equation:

[H+] = Ka / 0.04

Now, we can calculate the concentration of H+ ions. Once we have that, we can find the pH by taking the negative logarithm of the concentration.

Thus, to find the pH of the 0.10 M solution of the weak acid HX when it is 4.0% ionized, we need to calculate [H+], and then take the negative logarithm to find the pH.