solution contains 0.50M Nh3 and 1.50 M Nh4cl. What is the maximum concentration of mg2+ that can be present in such a solution without precipitating Mg(oh)2?
To find the maximum concentration of Mg2+ that can be present in the solution without precipitating Mg(OH)2, we need to consider the solubility product (Ksp) of Mg(OH)2.
The balanced equation for the precipitation reaction is:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
The Ksp expression for Mg(OH)2 is:
Ksp = [Mg2+][OH-]^2
First, we need to calculate the concentration of OH- ions in the solution. Since NH3 is a weak base, it will react with water to form OH- ions according to the following equation:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
Given that the solution contains 0.50 M NH3, we can assume that [OH-] is equal to [NH3] because NH3 is a weak base and does not completely dissociate.
So, [OH-] = 0.50 M
Next, we need to find the concentration of NH4+ ions in the solution. NH4Cl is a strong electrolyte, meaning it completely dissociates into its ions in water.
Given that the solution contains 1.50 M NH4Cl, the concentration of NH4+ ions is also 1.50 M.
Since the concentration of Mg2+ ions is the limiting factor, we can calculate the maximum concentration of Mg2+ using the Ksp expression for Mg(OH)2.
Ksp = [Mg2+][OH-]^2
Substituting the known values:
Ksp = [Mg2+](0.50 M)^2
We need to rearrange the equation to solve for [Mg2+]:
[Mg2+] = Ksp / (0.50 M)^2
Now, we need to find the value of Ksp for Mg(OH)2. The Ksp value for Mg(OH)2 is approximately 1.8 x 10^-11.
Substituting the Ksp value and calculating:
[Mg2+] = (1.8 x 10^-11) / (0.50 M)^2
[Mg2+] ≈ 7.2 x 10^-10 M
Therefore, the maximum concentration of Mg2+ that can be present in the solution without precipitating Mg(OH)2 is approximately 7.2 x 10^-10 M.