Calculate the Percent Ionization of Benzoic Acid Solution

How to calculate the percent ionization of a 0.15 mol/L benzoic acid solution?

Given data: Ka(C6H5COOH) = 6.3×10^-5, molar concentration of benzoic acid = 0.15 mol/L, molar concentration of sodium benzoate = 0.15 mol/L

What equation should be used to find the percent ionization?

Answer:

The percent ionization of a 0.15 mol/L benzoic acid solution in a solution containing 0.15 mol/L sodium benzoate is calculated using the Henderson-Hasselbalch Equation and amounts to approximately 0.648%.

To calculate the percent ionization of the benzoic acid (C6H5COOH) in a solution containing sodium benzoate, we need to use the Ka expression and the Henderson-Hasselbalch Equation. The Henderson-Hasselbalch Equation helps re-writing the ionization-constant expression for a solution of a weak acid: [H3O+] [A-]/[HA].

Given Ka(C6H5COOH) = 6.3×10^-5 and the molar concentrations for both compounds, we can calculate the ionization using the formula: [H3O+] = √Ka × [HA], and then calculate the percent ionization: ([H3O+]/[HA]) * 100. From the given information, [H3O+] is approximately 0.0081 M.

Plugging these values into our formula gives a percent ionization of 8.1 × 10^-3/0.125 = 0.00648, or 0.648%.