Question

The scenario is this. I have 50mL of 0.1M {NH₄+} at a certain temperature which gives it a Ka = 5.2 * 10⁻⁸. To this solution, I add 0.02 moles of {Cd(NO₃)₂}. It is known that cadmium ions undergo the following reaction:

{Cd²+ + 4NH₃ <=> [Cd(NH₃)₄]²+}
which has an equilibrium constant Kf = 7.3 * 10¹⁷ at this temperature. Assuming no other reactions, Calculate the pH of the resulting solution.

Answer 1

The pH of the solution after adding cadmium nitrate can be calculated by considering the dissociation of ammonium ions and the formation of the cadmium-ammonia complex, using ICE tables and applying the equilibrium constants for both reactions.

Step-by-step explanation:

To calculate the pH of the resulting solution after adding 0.02 moles of {Cd(NO₃)₂} to 50mL of 0.1M {NH₄+} with a Ka of 5.2 * 10⁻¸ and a Kf of 7.3 * 10¹⁷ for the complex {Cd(NH₃)₄²+}, we must first consider the ammonia liberated from the ammonium ion by the addition of cadmium ions and then its reaction with cadmium to form the complex. The equilibrium calculations for this system would involve the set up of an ICE table (Initial, Change, Equilibrium) for both equilibria (dissociation of NH₄+ and formation of [Cd(NH₃)₄]²+), and application of the respective equilibrium constants.

However, the high Kf for the complex formation suggests that nearly all available ammonia will form the complex with cadmium, shifting the ammonium equilibrium to produce more ammonia. After determining the changes in concentrations, the concentration of H+ ions can be found from the dissociation of NH₄+ and used to calculate the pH of the solution.