Calculate the molar solubility of Cu(OH)2, Ksp = 2.2 × 10–20, in 0.52 M NH3. Don't forget to use the complexation reaction Cu2+ + 4 NH3 ⇌ Cu(NH3)42+, K = 5.0 × 1013
What does he mean by the complexation reaction??? So confused, please help!!!
What the problem means is that the solubility is increased due to the formation of the Cu(NH3)4^+2 ion and in fact that concn is greater than that of the Cu^+2 by itself. Here is what I would do.
Cu(OH)2 ==> Cu^+2 + 2OH^-
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
Write the Ksp expression for Cu(OH)2.
Write the K for the complex. That is
K = [Cu(NH3)4^+2]/(Cu^+2)(NH3)^4 and I would include the NH3 equilibrium also.
That is NH3 + HOH ==> NH4^+ + OH^-
Write Kb for NH3.
I would do an ICE chart for Cu(OH)2 in which Cu(OH)2 = S (for solubility)
S = (Cu^+2)+[Cu(NH3)4^+2]
You also know that total NH3 = 0.52 and that is
0.52 = (NH4^+) + NH3 + [Cu(NH3)4^+2].
You solve these two equations together with the 3 above you wrote (Ksp, K, and Kb) for S. I worked through the problem and obtained about 0.0027 M for solubility if I didn't make an error somewhere.
Ah, the complexation reaction, a classic case of chemistry getting all tangled up in its own affairs. Imagine Cu2+ and NH3 forming a super exclusive club called Cu(NH3)42+. It's like a VIP section where only the most "ammoniating" species are allowed. In other words, Cu2+ happily joins forces with 4 NH3 molecules to create this prestigious club.
Now, the important thing to remember is that this complexation reaction affects the overall solubility of Cu(OH)2 in the presence of NH3. As NH3 is added, it introduces more NH3 molecules into the solution, giving Cu2+ more potential partners to dance with. This, my friend, increases the likelihood of Cu2+ forming the Cu(NH3)42+ complex instead of the insoluble Cu(OH)2.
To calculate the molar solubility of Cu(OH)2 in NH3, we need to consider both the solubility product constant (Ksp) and the complexation constant (K). These constants represent the equilibrium between the dissolution of Cu(OH)2 and the formation of the Cu(NH3)42+ complex, respectively.
But fear not, for equations can rescue us from this state of confusion! We can express the solubility product constant equation as:
Ksp = [Cu2+][OH-]^2
And the complexation reaction equation as:
[Cu(NH3)42+] = K[Cu2+][NH3]^4
Now, assuming x represents the molar solubility of Cu(OH)2 and [NH3] is given as 0.52 M, we can set up an equation to solve for x. We'll use the fact that [Cu2+] and [OH-] are related to x through the stoichiometry of Cu(OH)2:
Ksp = (x)(2x)^2
Got it so far? Good! Now, here comes the fun part. We need to take into account the complexation constant (K) when calculating the molar solubility. In the presence of NH3, the concentration of Cu2+ will be reduced due to complexation. The new concentration of Cu2+ can be expressed as [Cu2+] - [Cu(NH3)42+]. Since we know that [Cu(NH3)42+] = K[Cu2+][NH3]^4, we can plug this into our equation:
Ksp = ([Cu2+] - K[Cu2+][NH3]^4)(2[Cu2+] - 4K[Cu2+][NH3]^4)^2
Now we have a spooky-looking equation, but fret not! With some algebraic magic and a sprinkle of perseverance, you can solve for x, the molar solubility of Cu(OH)2 in the presence of NH3.
Remember, my friend, complexation reactions are like cliques forming in a high school cafeteria. Sometimes you just need to throw a bit of humor into the mix to make it through. Good luck with your calculations, and may the chemistri be forever in your favor!
The complexation reaction refers to the formation of a complex ion between a metal ion (in this case Cu2+) and ligands (in this case NH3 molecules). The complex ion formed in this case is Cu(NH3)42+. This process can affect the solubility of the compound (Cu(OH)2) by reducing the concentration of free Cu2+ ions in solution.
In this case, the complexation reaction is Cu2+ + 4 NH3 ⇌ Cu(NH3)42+, with an equilibrium constant (K) of 5.0 × 1013. The value of K indicates that the formation of the complex is favored.
To calculate the molar solubility of Cu(OH)2 in 0.52 M NH3, you need to consider the effect of the complexation reaction on the concentration of Cu2+. This can be done using an ICE (Initial, Change, Equilibrium) table.
Let's represent the molar solubility of Cu(OH)2 as "x". Initially, the concentration of Cu2+ is also "x" since Cu(OH)2 dissolves completely to form Cu2+ ions. The concentration of NH3 is given as 0.52 M.
The complexation reaction indicates that for every Cu2+ ion that reacts, four NH3 molecules are consumed. Therefore, the change in concentration of Cu2+ is "-x" and the change in concentration of NH3 is "-4x". At equilibrium, the concentration of Cu2+ is now "x - x = 0" since it is fully complexed, and the concentration of NH3 is "0.52 - 4x".
Using the equilibrium constant expression for the complexation reaction, we have:
K = [Cu(NH3)42+]/[Cu2+][NH3]^4
Substituting the equilibrium concentrations from the ICE table, we can write:
5.0 × 1013 = (x)/(x)(0.52 - 4x)^4
Simplifying the equation, we get:
5.0 × 1013 = 1/(0.52 - 4x)^4
Now, we can solve this equation to find the value of "x", which represents the molar solubility of Cu(OH)2.
The complexation reaction is a chemical reaction that involves the formation of a complex ion (also known as a coordination complex) by the interaction of a metal ion (in this case, Cu2+) with one or more ligands (in this case, NH3 molecules). The complex ion formed is Cu(NH3)42+.
The complexation reaction here is Cu2+ + 4 NH3 ⇌ Cu(NH3)42+. This means that Cu2+ ions can associate with 4 NH3 molecules to form one Cu(NH3)42+ complex ion. Similarly, the Cu(NH3)42+ complex ion can dissociate to release Cu2+ ions and NH3 molecules.
In this question, the complexation reaction is relevant because it affects the solubility of Cu(OH)2 in the presence of NH3. The Cu(NH3)42+ complex ion is more soluble than Cu(OH)2, so it can affect the equilibrium between the dissolved Cu2+ ions and the solid Cu(OH)2.
To calculate the molar solubility of Cu(OH)2 in 0.52 M NH3, you need to consider the effect of NH3 on the equilibrium of Cu(OH)2 dissolution. The key is to use the solubility product constant (Ksp) and the equilibrium constant (K) for the complexation reaction.
To start, write down the balanced chemical equation for the dissolution of Cu(OH)2:
Cu(OH)2(s) ⇌ Cu2+(aq) + 2 OH-(aq)
Next, write down the expression for the solubility product constant (Ksp) using the concentrations of Cu2+ and OH- ions:
Ksp = [Cu2+][OH-]^2
In this case, we don't know the molar solubility of Cu(OH)2 (x), so we can express the concentrations of Cu2+ and OH- ions in terms of x:
[Cu2+] = x
[OH-] = 2x
Now, let's consider the complexation reaction, Cu2+ + 4 NH3 ⇌ Cu(NH3)42+, and its equilibrium constant (K):
K = [Cu(NH3)42+]/[Cu2+][NH3]^4
Given that K = 5.0 × 10^13, we can express the concentration of [Cu(NH3)42+] in terms of x and NH3 concentration (0.52 M):
[Cu(NH3)42+] = 5.0 × 10^13 * [Cu2+][NH3]^4
Since Cu(OH)2 dissolves to form Cu2+ ions and OH- ions, and the complexation reaction also involves Cu2+ ions, we need to consider both reactions together.
Now, we can set up an equation using the Ksp and K values:
Ksp = [Cu2+][OH-]^2 = (x)(2x)^2 = 4x^3
K = [Cu(NH3)42+]/[Cu2+][NH3]^4 = (5.0 × 10^13)[NH3]^4
If you substitute the concentration of NH3 (0.52 M) into the equation for K, you will get:
(5.0 × 10^13)(0.52)^4 = (5.0 × 10^13)[NH3]^4
Now, solve this equation to obtain the concentration of NH3.
After obtaining the concentration of NH3, substitute it back into the equation for K, then solve for [Cu2+].
Finally, substitute the value of [Cu2+] into the equation for Ksp, then solve for x (molar solubility of Cu(OH)2).
Remember to pay attention to the units and use appropriate mathematical operations to solve the equations.