a solution is 0.10M in free NH3, 0.10M in NH4Cl, and 0.015M in [cu(NH3)4]. Will Cu(OH)2 (s) precipitate from this solution? Ksp of Cu(OH)2 is 2.2810^-20
I would write the reaction for the formation of Cu(NH3)4^+2. Now write the expression for Kformation for this reaction. You have Cu(NH3)4^+2 and NH3; the only unknown is (Cu^+2). Solve for that.
Now write the expression for NH3 + HOH ==> NH4^+ + OH^-
Write Kb expression.
Kb = (NH4^+)(OH^-)/(NH3).
You know (NH4^+) from the NH4Cl, you know NH3 from the problem, calculate (OH^-).
Now see if (Cu^+2)(OH^-)^2 exceeds Ksp.
[note:since the problem says that the FREE ammonia is 0.10 M, I assume we don't need to make a correction for the amount of NH3 used up forming the copper ammine complex.]
i did all the steps you said, however the final answer i get is 1.36*10^-11
the answer is percipitate does not form
To determine if Cu(OH)2 (s) will precipitate from this solution, we need to calculate the concentration of Cu2+ and OH- ions and compare it to the Ksp value.
1. Write the balanced equation for the dissociation of Cu(OH)2:
Cu(OH)2 (s) ⇌ Cu2+ (aq) + 2 OH- (aq)
2. Find the concentration of Cu2+ ion (Cu2+ (aq)):
The concentration of [cu(NH3)4] is given as 0.015M, so the concentration of Cu2+ ion is also 0.015M.
3. Find the concentration of OH- ion (OH- (aq)):
The NH4Cl will dissociate to NH4+ and Cl- ions, but it doesn't contribute OH- ions.
The NH3 will react with NH4+ to form NH4OH, which will then ionize to release OH- ions.
Since NH3 and NH4+ are present in equal concentrations (0.10M), the concentration of OH- ions from NH3 is also 0.10M.
4. Calculate the [Cu2+][OH-]^2:
[Cu2+] = 0.015M
[OH-] = 0.10M
[Cu2+][OH-]^2 = 0.015M * (0.10M)^2
= 0.0015M * 0.01M
= 1.5 × 10^-5
5. Compare the calculated [Cu2+][OH-]^2 to Ksp:
Ksp = 2.28 × 10^-20
Since [Cu2+][OH-]^2 = 1.5 × 10^-5 is greater than the Ksp value, Cu(OH)2 will precipitate from the solution.
Therefore, Cu(OH)2 (s) will precipitate from the given solution.
To determine if Cu(OH)2 (s) will precipitate from the given solution, we need to calculate the ion product (Qsp) and compare it to the solubility product constant (Ksp) for the compound.
The balanced chemical equation for the dissolution of Cu(OH)2 is:
Cu(OH)2 (s) ⇌ Cu2+(aq) + 2OH-(aq)
First, let's determine the concentrations of Cu2+ and OH- ions in the solution using the given information:
From [Cu(NH3)4], we have 0.015M copper ions.
From the free NH3, the concentration of OH- ions can be determined using the fact that NH3 acts as a weak base and reacts with water to produce OH- ions. However, the concentration of free NH3 is not provided, so we cannot calculate the OH- concentration directly.
However, we are provided with NH4Cl, which is a salt formed from a strong acid (HCl) and a weak base (NH3). When NH4Cl dissolves in water, it undergoes hydrolysis. The NH4+ ions react with water, resulting in the formation of NH3 and H3O+ ions. Since NH3 acts as a base, it will react with water to produce OH- ions. Therefore, we can assume that the concentration of NH3 is equal to the concentration of OH- ions produced from the hydrolysis of NH4Cl.
Since the concentration of NH4Cl is 0.10M, the concentration of NH3 and OH- ions will also be 0.10M.
Now that we have the concentrations of Cu2+ and OH- ions, we can calculate the ion product (Qsp):
Qsp = [Cu2+][OH-]^2
Substituting the concentrations:
Qsp = (0.015M)(0.10M)^2 = 0.00015
Finally, we compare Qsp to Ksp to determine if Cu(OH)2 will precipitate:
If Qsp > Ksp, then Cu(OH)2 will precipitate.
If Qsp < Ksp, then Cu(OH)2 will not precipitate.
Comparing Qsp (0.00015) to Ksp (2.28 x 10^-20), we can see that Qsp is significantly smaller than Ksp. Therefore, Cu(OH)2 will not precipitate from this solution.