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### Topic: Common ion, complex ion, molar solubility problem.  (Read 5756 times)

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#### zainsiddiqui14

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##### Common ion, complex ion, molar solubility problem.
« on: February 18, 2012, 05:27:48 PM »
The copper(I) ion forms a chloride salt that has Ksp = 1.2 * 10^-6. Copper(I) also forms a complex ion with Cl-.
Cu+(aq) + 2 Cl-(aq) yields CuCl2-(aq) K = 8.7 * 10^4
(a) Calculate the solubility of copper(I) chloride in pure water. (Ignore CuCl2- formation for part a.)
I got this one, it's .0011 molar.
(b) Calculate the solubility of copper(I) chloride in 0.36 M NaCl.
Where do I start?

#### Borek

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##### Re: Common ion, complex ion, molar solubility problem.
« Reply #1 on: February 18, 2012, 06:48:01 PM »
Concentration of Cl- is given, just put it into Ksp formula. What is known, what is unknown?
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#### zainsiddiqui14

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##### Re: Common ion, complex ion, molar solubility problem.
« Reply #2 on: February 18, 2012, 10:58:23 PM »
That's what I thought initially, too.

Ksp = [Cu+][.37], right?
No, actually.

Because Cu+ and Cl- then make a complex and ion, that further affects the solubility, which is where I'm confused.

#### Borek

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##### Re: Common ion, complex ion, molar solubility problem.
« Reply #3 on: February 19, 2012, 05:23:45 AM »
OK. Assume concentration of Cl- doesn't change. You will get two equations in two unknowns - free copper(I) and copper(I) complex. Try to solve.
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#### zainsiddiqui14

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##### Re: Common ion, complex ion, molar solubility problem.
« Reply #4 on: February 19, 2012, 10:07:53 AM »
Oh, haha, I overlooked something really quite simple. Only aqueous and gaseous species appear in the K expression, and I erroneously put CuCl(s) in the expression and kept thinking, "there are too many unknowns!". In reality, if you add the two equations together and multiply K values (Hess's law), you get quite simply:

.1044=[CuCl2-]/[Cl-]

.1044 =
• /[.36-x]

x=.034 M

Thanks for the pointers!