What is the pH at 25.0 degrees Celsius of a 0.100 M aqueous solution of acetic acid (HC2H3O2) if the Ka is 1.8 x 10^-5?

To find the pH of a 0.100 M solution of acetic acid (HC₂H₃O₂), we start by using its dissociation constant (K_a) to determine the concentration of hydrogen ions ([H⁺]). The dissociation of acetic acid in water can be represented as:

HC₂H₃O₂ ⇌ H⁺ + C₂H₃O₂^–

The expression for the acid dissociation constant K_a is given by:

K_a = rac{[H⁺][C₂H₃O₂^–]}{[HC₂H₃O₂]}

For a weak acid like acetic acid, when it dissociates, if we let ‘x’ be the amount that dissociates, we can express the concentrations at equilibrium as:

• [HC₂H₃O₂] = 0.100 – x
• [H⁺] = x
• [C₂H₃O₂^–] = x

Substituting these expressions into the K_a equation gives us:

1.8 x 10^-5 = rac{x imes x}{0.100 – x}

Assuming that x is small compared to the initial concentration (which is valid for weak acids), we can simplify it to:

1.8 x 10^-5 ≈ rac{x^2}{0.100}

By rearranging this, we find:

x² ≈ 1.8 x 10^-6

Taking the square root:

x ≈ 1.34 x 10^-3 M

[H⁺] = x ≈ 1.34 x 10^-3 M

Now, we can calculate the pH using the formula:

pH = -log[H⁺]

pH = -log(1.34 x 10^-3) ≈ 2.87

Thus, the pH of the 0.100 M aqueous solution of acetic acid at 25.0 degrees Celsius is approximately 2.87.