Calculate the pH for an aqueous solution of acetic acid that contains 2.15 x10^-3 mol L^-1 hydronium ion. A) 4.65 × 10^-12 B) 2.15 × 10^-3 C) 2. 67 D) 11. 33

Question

Quizplus · Accepted Answer

Acetic acid is a weak acid, and its dissociation in water is described by the following equation:

CH3COOH + H2O ⇌ CH3COO- + H3O+

The equilibrium constant for this reaction is known as the acid dissociation constant (Ka) of acetic acid and is equal to 1.8 x 10^-5 at 25°C.

Therefore, we can write the following expression for the equilibrium constant:

Ka = [CH3COO-][H3O+] / [CH3COOH]

Since we know the concentration of H3O+ in the solution, we can rearrange the equation to solve for the concentration of CH3COO-:

[CH3COO-] = (Ka x [CH3COOH]) / [H3O+]

Substituting the given values:

[CH3COO-] = (1.8 x 10^-5 x 2.15 x 10^-3) / 10^-3 = 3.87 x 10^-8 M

Now we can calculate the pH of the solution:

pH = -log[H3O+]

pH = -log(2.15 x 10^-3) = 2.67

Therefore, the correct choice is C, 2.67.

Calculate the PH for an Aqueous Solution of Acetic Acid