Calculate the pH of a Weak Monoprotic Acid Solution

To find the pH of a 0.010 mol/L solution of a weak monoprotic acid that is 4.0% ionized, we need to determine the concentration of the acid in its ionized and unionized forms.

Let's assume the acid is represented by the formula HA.

Since the acid is 4.0% ionized, it means that 4.0% of the initial concentration of HA has dissociated into H+ and A- ions. Thus, the concentration of H+ ions is 4.0% of 0.010 mol/L, which is 0.040 * 0.010 = 0.0004 mol/L.

Since the acid is monoprotic, the concentration of A- ions is also 0.0004 mol/L.

The concentration of unionized acid, HA, can be calculated by subtracting the concentration of the ionized species from the initial concentration:

Concentration of HA = Initial concentration - Concentration of H+ ions = 0.010 mol/L - 0.0004 mol/L = 0.0096 mol/L.

Now, we can calculate the pKa of the acid, which is the negative logarithm of the acid dissociation constant (Ka).

Ka = [H+][A-] / [HA]

Since the concentration of H+ ions and A- ions is the same, we can substitute their concentration as [H+]^2 into the equation:

Ka = [H+]^2 / [HA]

0.0004^2 / 0.0096 = 0.0000167 mol/L.

Taking the negative logarithm of Ka gives us the pKa:

pKa = -log(0.0000167) = 4.78.

Since the acid is weak, we can assume that [H+] = [A-] = 0.0004 mol/L.

Using the equation for pH:

pH = -log[H+] = -log(0.0004) = 3.40.

Therefore, the pH of the 0.010 mol/L solution of the weak monoprotic acid that is 4.0% ionized is approximately 3.40.