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Question on Silver Chloride (gravimetric analysis)

  1. Mar 8, 2006 #1
    Forgive me for asking such a broad question, but does anybody know by how much (on average) oxygen discoloration will affect the percent chloride obtained from the precipitation of chloride from a sample using an aqueous solution of Silver Nitrate? More specifically, if the experimenter was to do this precipitation twice, the first time getting a white grey precipitate (no observeable purple/oxygen discoloration at all), and the second time the experimenter exposed his sample to light and recieved a purplish precipitate, by how much would the calculated percent Chloride result differ in the 2 samples?

    As an example, say that I precipitated chloride from an unknown sample and obtained a pure white precipitate and calculated out that the original sample contained 56.4% Chloride in it. If the precipitate had been purple, by how much would I expect my result to deviate from 56.4% (or would it not deviate noticeably at all?)? I wish I could try this in the lab for myself, but my school is cheap and I only get to do this once (in case you are wondering, I recieved a white precipitate by obsessesively shielding my sample from the light, while everyone around me had purple precipitates because they didn't shield their samples from the light and I was wondering how much of a difference this would have made on the final result).

    Thanks in advance for any help :)
  2. jcsd
  3. Mar 9, 2006 #2


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    You'll need to consider the equation for the decomposition and then apply the stoichiometry, but as for the percent error due to this reaction, I'm not quite sure how one would figure that out without knowing anything about the original unknown. What you can probably do though, is to find someway to determine the mass of the unknown sample minus the precipitate upon isolation, before you start drying it. That is some sensitive experimental procedure relating to weighing by difference; if you know what the mass of the precipitate had to be through such a gravimetric analysis, then you can determine the loss of the precipitate mass, which also includes the loss of water. Then perhaps solve a set of simultaneous equations.

    Other ways, may involve time dependent regression analysis and so on, but we can circumvent all of this simply by minimizing the samples exposure to light. I don't quite know the literature values or any standard values that may have bee proposed through research for the magnitude of error due to such a decomposition.
  4. Mar 9, 2006 #3


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    That is, minimize your determinate errors.
  5. Mar 9, 2006 #4


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    There's no measurable difference. If you've played with equilibrium constants, you can look at the reaction

    2AgCl + 1/2 O2 = Ag2O + Cl2
    (change in standard Gibbs free energy is around + 200 kJ/mol(Ag2), and make a few assumptions about chlorine and oxygen activities in the solution from which the precipitate formed (O2 activity is around 0.2, atmosphere; chlorine activity is going to be of the same order of magnitude as the amount of silver oxide formed (activity = 1)).

    You also have the photolytic decomposition of AgCl to consider. This is the reaction producing the "purple" color of the precipitate, finely divided elemental silver. Same game as for the oxide formation, 100 kJ/mol, same chlorine activity estimate, same unit activities for the solid phases. Still a very small number.
  6. Mar 10, 2006 #5
    Thanks GCT and Bystander. I was wondering why nobody seemed to care if they left their samples exposed to the light (including the instructor). I guess if the error was small, that would explain why :wink:
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