If a solution is made 0.080 M in Mg(NO3)2, 0.075 M in aqueous ammonia, and 3.5 M in NH4NO3, will Mg(OH)2 precipitate? What is the pH of this solution?

Question

If a solution is made 0.080 M in Mg(NO3)2, 0.075 M in aqueous ammonia, and 3.5 M in NH4NO3, will Mg(OH)2 precipitate? What is the pH of this solution?

Answer 1

To determine whether Mg(OH)2 will precipitate, we can calculate the ion product (IP) of Mg(OH)2 using the concentrations of Mg2+ and OH- ions in the solution. If the IP exceeds the solubility product constant (Ksp) of Mg(OH)2, it will precipitate.

The balanced equation for the dissociation of Mg(OH)2 is:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

The Ksp expression for Mg(OH)2 is:
Ksp = [Mg2+][OH-]^2

To find the concentrations of Mg2+ and OH- ions in the solution, we need to account for the contribution from Mg(NO3)2 and NH4NO3.

Since Mg(NO3)2 dissociates into Mg2+ and 2 NO3- ions, the concentration of Mg2+ is equal to the initial concentration of Mg(NO3)2:
[Mg2+] = 0.080 M

Ammonia (NH3) reacts with water to produce OH- ions:
NH3 + H2O ⇌ NH4+ + OH-

Since NH4NO3 dissociates into NH4+ and NO3- ions, the concentration of NH4+ is equal to the initial concentration of NH4NO3:
[NH4+] = 3.5 M

The concentration of OH- ions can be determined from the concentration of NH4+ using the equilibrium reaction above. However, we need to account for the fact that NH3 is a weak base, and not all of it will dissociate. To calculate the concentration of OH- accurately, we need to consider the pKa of NH4+/NH3.

The pKa of NH4+/NH3 is approximately 9.25. We can use this information to calculate the concentration of NH3 and OH-:

pH = pKa + log([NH4+]/[NH3])

Solving for [OH-]:
[OH-] = 10^(-pH) * [NH3] = 10^(-pH) * [NH4+]

Now that we have the concentrations of Mg2+ and OH- ions, we can calculate the IP of Mg(OH)2:

IP = [Mg2+][OH-]^2

If the IP exceeds the Ksp value of Mg(OH)2 (which is approximately 1.2 x 10^(-11) at 25°C), then Mg(OH)2 will precipitate.

To calculate the pH of the solution, we'll use the equation above:
pH = pKa + log([NH4+]/[NH3])

Plug in the values for [NH4+] and [NH3] to solve for the pH.

By following these steps, you will be able to determine if Mg(OH)2 will precipitate and find the pH of the solution.

Answer 2

Use the Henderson-Hasselbalch equation to solve for pH of the solution. Use NH4Cl as the acid and NH3 as the base. Convert the pH to (OH^-) and set up Ksp with Mg(OH)2 and see if Qsp is greater than Ksp for Mg(OH)2. In the regular qual scheme of things, Mg ions are in group V and the solution in group IV is treated with NH3 and NH4Cl to keep Mg(OH)2 from pptng in group IV instead of group V.