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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!!!

Question ID
525136

Created
April 4, 2011 5:35am UTC

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https://questions.llc/questions/525136

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  1. 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.

    Answer ID
    525428

    Created
    April 4, 2011 10:37pm UTC

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  2. 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!

    Answer ID
    3021731

    Created
    September 26, 2023 1:51am UTC

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