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I have this exercise from my H.W: enter image description here

and I am asked: if we split the solution into 3 different containers equally -

  • how many moles will the solution have in each container?
  • will the Cl- concentration change?

for the first question, I answered 0.08/3 moles. my common sense say if I split something to 3 it will split equally. for the second I wrote that the Cl- concentration will not change from the original solution, but im not sure how to explain this.

Am i right?

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    $\begingroup$ Using photos/screenshots of text ( especially if handwritten ) instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. Additionally, it can be challenge to decipher. Consider copy/pasting or rewriting of the essential parts and using of MathJax for eventual formatting of mathematical/chemical formulas or equations. $\endgroup$
    – Poutnik
    May 22, 2021 at 6:35
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    $\begingroup$ Concentration is intensive and amount of substance is extensive. $\endgroup$
    – Nisarg Bhavsar
    May 22, 2021 at 6:59
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    $\begingroup$ I wonder if you really need that sketch to answer something obvious. $\endgroup$
    – Alchimista
    May 22, 2021 at 8:49

2 Answers 2

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Ignoring the handwritten notes, which are difficult to read (you can use "Markdown" language, instead), your common-sense answers certainly seem correct!

To explain the thought-process behind stating the concentration stays the same when splitting a solution, consider that if the ratio of solvent:solute is X, then the ratio of solvent/3:solute/3 is still X. You've simply divided the top and bottom of the equation by the same amount. Without the math, you've split off the solute and solvent in the same proportions that they originally had.

One might quibble that when dealing with picoscopic quantities, i.e. just a few atoms of each, one sample might have only one ion while the other had two, because chemistry doesn't split ions, but that's splitting hairs.

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If the solution is Standarized, distributing equally the ions; then the concentration [C] is a ration of volumes for your complex formation.

[C]i*(Vf/Vi)=[C]f

When Vf1=Vf2=Vf3, then all 3 [C]f will have an approximate Concentration.

Ignoring solutions fugacity, Cl- ion concentration must be constant as well; it is likely to reach equilibrium way before complex formation.

The formation constant Kf for CuCl2- is 3e5, an amount will be in relation to the solubility product of CuCl Ksp 1.72e-7

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  • $\begingroup$ Don't make things more complicated than they are. The answer doesn't change in whatever solution or if activity is considered. The only requisite is that of homogeneity. Nor the solute nature plays a role. $\endgroup$
    – Alchimista
    May 22, 2021 at 8:56

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