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![]() Problem: Panting and out of breath, you run into your darkroom. In the palm of your sweaty hands you have a roll of film on which you have both proof of the existence of extraterrestrial life and some voyeuristic shots of your neighbor's Swedish Au-Paire. Your roll of unexposed black-and-white film contains about 0.27 g of AgBr and in order to develop you must dissolve the AgBr as Ag(S2O3)2-3 . What mass of (hypo) Na2S2O3*5H2O must you add to a 1.5 L developing tray of solution to salvage this priceless roll of film? (adding too much will ruin everything!) Kf of Ag(S2O3)2-3 = 4.7 E13, Ksp of AgBr = 3.3 E-13 |
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![]() Solution: Even though you're a pervert, I'll help you on this one. So you want to dissolve something through the formation of a soluble complex ion? Sounds like a Type 3 solubility problem. Much like a Type 2 problem, this sucker isn't so bad once you distill it down to the basics and figure out what's really going on. Let's start with the two equations at work: |
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![]() Here's what's goin' on: You start out with equal molar amounts of Ag+ and Br¯. When you start adding S2O3-2 you reach a point when Ag+ begins getting used up as part of the complex ion formed in rxn. 2. This causes the concentration of Ag+ to start dropping and it continues dropping until a point when the product of the concentrations of Ag+ and Br¯ hit the threshhold point known as Ksp. At this point, you have the maximum number of Ag+ and Br¯ ions that can be in solution before a precipitate is formed. This is also the point at which the AgBr starts to dissolve, the point you are looking for. To start the math, you gotta figure out how much Ag+ and Br¯ you start out with, and the concentration of Ag+ at the point when the AgBr starts to dissolve. The key here is that the same Br¯ concentration that you start out with is the same Br¯ concentration you have when AgBr starts to dissolve. |
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![]() Gmm AgBr = 107.868 + 79.904 = 187.772 g/mol 0.27 g Ag Br * 1/ (187.772 g/mol) = 0.0014 mol (initial molar amounts of Ag+ and Br¯) [Br¯] = 0.0014 mol / 1.5 L = 9.6 E-4 (because we know the amount of Br¯ doesn't change) Ksp(AgBr) = 3.3 E-13 = [Ag+][Br¯] Ksp(AgBr) = [Ag+](9.6 E-4) [Ag+] = 3.3 E-13 / 9.6 E-4 [Ag+] = 3.4 E-10 (concentration of Ag+ when AgBr starts to dissolve) |
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![]() Moving right along (I know you can't wait to see those pictures and notify the Daily Sun), we now know a piece of the puzzle. Actually, the puzzle in this problem is the following equilibrium equation using the formation reaction of equation 2: |
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![]() Kf = 4.7 E13 = [Ag(S2O3)2-3] / ( [ S2O3-2][Ag+] ) |
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![]() In this puzzle, there are three unknowns, and assuming this is all we got to work worth, we gotta know two of them to solve for the other. We know [Ag+] and we have no clue about [S2O3-2], but we can figga out da udda one. If we started out with 0.0014 mol of silver, and when all of the AgBr has dissolved we have a concentration of only 3.4 E -10 M, the rest must be a part of the complex ion that was formed, [Ag(S2O3)2-3]. After all, the point of adding the salt is to freeze up the Ag+ as a part of the ion, allowing AgBr to dissolve. Confused? Well, um, read over that again. This math might also clear some stuff up. |
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![]() [Ag(S2O3)2-3] = (0.0014 mol / 1.5L ) - 3.4 E -10 M = 9.3 E -4 (M complex ion = starting mol Ag / volume - final M Ag) |
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![]() So now we know two of the three unknowns and we solve for the other one like so: |
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![]() Kf = 4.7 E13 = 1.4 E-4 / ( [ S2O3-2] * 3.3 E-10) [S2O3-2] = 1.4 E-4/ (3.3 E-10 * 4.7 E13) = 9.0 E-9 M mol S2O3-2 = 9.0 E-9 M *1.5 L = 1.4 E-8 mol |
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![]() Before we calculate the grams of salt used, and get a really, really, wrong answer, we can't forget that much of the S2O3-2 we added is now part of the complex ion, in which there are two moles of S2O3-2 for every one mole of the ion. |
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![]() Total moles of S2O3-2 = 1.4 E-8 mol + 2(0.0014 mol) = 0.0028 mol |
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![]() So we used .0.0028 mol of salt. Now don't screw up now and use the wrong molar mass. |
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![]() Gmm of Na2S2O3*5H2O = 2(22.9) + 2(32.1) + 3(16.0) + 10(1.00) + 5 (16.0) = 248 g/mol 0.0028 mol * 248 g/mol = 0.70 g of Na2S2O3*5H2O |
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![]() Good work, young grasshopper! You've finished the material! We have given you all the knowledge we possess. Now, you must go forth and face your destiny. But first, why don't you read our last words of wisdom in our exciting Conclusion? |
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