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### AuthorTopic: Solubility - Complex ion equilibrium problem  (Read 3350 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### nefsicle

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##### Solubility - Complex ion equilibrium problem
« on: March 09, 2016, 05:03:49 PM »

Hello, this is a problem from the 2015 USNCO exam on solubility equilibrium that I'm having a bit of trouble on...

Silver chloride, AgCl (Ksp = 1.8 x 10-10), can be dissolved
in solutions containing ammonia due to the formation of
the soluble complex ion Ag(NH3)2+ (Kf = 1x108)
What is the minimum amount of NH3 that would need to
be added to dissolve 0.010 mol AgCl in 1.00 L of
solution?
(A) 0.010 mol (B) 0.020 mol
(C) 0.095 mol (D) 0.13 mol

I am able to combine the equations to solve for the correct answer, but my trouble is more with the concept. So, when we add the two equations, we obtain the equation of AgCl + ammonia -> complex ion + Cl-. However, how does this allow us to solve for the molar solubility of AgCl at varying concentrations of ammonia, for example? [Ag+] is absent from the equation, and from how I see people who solve these types of questions do it, they use the final concentration of the formed complex ion as the molar solubility. Does this not neglect the portion of silver ions that did not form the complex?

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#### AWK

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##### Re: Solubility - Complex ion equilibrium problem
« Reply #1 on: March 09, 2016, 07:54:15 PM »

You have two equilibria in solution. Then you have 2 equation. After some approximations the system of equation is easily solvable.
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#### nefsicle

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##### Re: Solubility - Complex ion equilibrium problem
« Reply #2 on: March 09, 2016, 09:05:50 PM »

Thanks for the response, but that still doesn't really clear up my question. Could you expand a bit on what "after some approximations" implies?

Do you mean that when we solve for x with x being the amount of complex ion formed, we are assuming x >> than the remaining [Ag+]?

For reference, this is what I meant by "people who solve it" in case my explanation was vague: https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/copy-of-solubility-equilibria-mcat/v/solubility-and-complex-ion-formation
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#### AWK

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##### Re: Solubility - Complex ion equilibrium problem
« Reply #3 on: March 09, 2016, 10:00:12 PM »

Quote
Do you mean that when we solve for x with x being the amount of complex ion formed, we are assuming x >> than the remaining [Ag+]?
Just the case.
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