The pH Calculation of Anilinium Hydrochloride Solution

How to calculate the pH of 0.626 M anilinium hydrochloride solution in water?

Given that Kb for aniline is 3.83 × 10^-4, what is the pH of the solution?

Answer:

The pH of a 0.626 M solution of anilinium hydrochloride in water can be calculated using an ICE table and the relationship between Ka, Kb, and Kw. The calculation will first reveal the hydronium ion concentration, and the pH is then derived from this value using the pH formula. The pH of the solution is 5.39.

To calculate the pH of the solution, we first need to understand that anilinium chloride is an acid and will transfer a proton to water upon dissociation: C6H5NH3+ + H2O → C6H5NH2 + H3O+

From here, you can calculate the Hydronium Ion Concentration [H3O+] using the known value of the base dissociation constant (Kb) for aniline and the concentration of the anilinium ion. We then apply the relationship between Ka, Kb and Kw (the ion product of water 1.0 *10^-14 at 25 degree Celsius).

The acid dissociation constant (Ka) can be calculated as Ka=Kw/Kb. Solving this will give Ka approximately equal to 2.61 × 10^-10. Applying the expression for Ka in an ICE table (Initial, Change, Equilibrium concentrations), you will find that the hydronium ion concentration is equal to sqrt(Ka*M), where M is the initial concentration of the weak acid. Solve the expression gives [H3O+] = sqrt((2.61 × 10^-10)*(0.626 M)) = 1.29 * 10^-5.

Finally, the pH is then calculated from this value using the pH formula, pH=-log[H3O+]. Substituting the value gives pH = 5.39.