Question

A \(0.040 M\) solution of a monoprotic acid is 14 percent ionized. Calculate the ionization constant of the acid.

Short answer

The Ionization constant (Ka) of the acid is approximately \(8.8 * 10^{-3}\).

Step-by-step solution

Understand what is given and what is asked

We know that the molar concentration of acid is \(0.040 M\) and the percentage of ionization is \(14\%\). We're asked to determine the Ionization constant (Ka) of the acid.

Correlate the percentage ionization with the amount of hydronium ions

The 14 percent ionized means 14 percent of the acid molecules have donated their protons and therefore became hydronium ions. So, we have \(0.040 M * 0.14 = 0.0056 M\) hydronium ions. Note that for every one acid molecule that ionizes, it donates one proton (H+) and becomes one hydronium ion (H3O+). So, the concentration of protons is equal to the concentration of hydronium ions.

Determine the Ionization Constant

The ionization constant for a weak acid (Ka) is given by: \[Ka = [H3O+][A–] / [HA]\] Where \([HA]\) is the concentration of the un-ionized acid, \([H3O+]\) is the concentration of the hydronium ion, \([A–]\) is the concentration of the conjugate base. For a monoprotic weak acid, \([H3O+]\) = \([A–]\). Therefore, \[Ka = [H3O+]^2 / (0.040 - [H3O+])\] Substituting the value we have for \([H3O+]\) (which is 0.0056), we find \[Ka= [(0.0056)]^2 / (0.040-0.0056)\] After calculation, we find that the Ionization Constant, Ka, is approximately \(8.8 * 10^{-3}\).