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In this same situation, what is the percent Ca 2+ ion left when Cd 2+ begins to ppt?
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Note that the concentration for Ca 2+ was given at the beginning of the problem. In other words, at what does each metal fall out?įirst look at each compound and build the K sp equation. So this question is extremely easy! When asked from a list of compounds which will precipitate out first, or the order of precipitation, look at the K sp values and go from smallest to largest.Ī harder question would be one that asks when each of the ions fall out of solution. The second largest is CdC 2O 4, so this will ppt out second, and MgC 2O 4 last. In this case, CaC 2O 4 has the smallest K sp, so it will be reached first, and CaC 2O 4 will form ppt first. To see which ppt will form first, look at which K sp will be reached first, in other words, compare to see which K sp is smaller. So an easy way to see when a ppt will form is to set Q c = K sp. In other words, just past the equilibrium. Three compounds with low solubility will form: CdC 2O 4, CaC 2O 4, MgC 2O 4.Ī precipitate will form when Q c is greater than K sp. If Na 2C 2O 4 is added slowly, what is the order of ppt? Recognize that each of the three ions will react with the oxalate anion in Na 2C 2O 4, which is C 2O 4 2. Each will dissociate to give 0.200M each of Cd 2+, Ca 2+, and Mg 2+. Start with a solution of 0.200M each of Cd(NO 3) 2, Ca(NO 3) 2, Mg(NO 3) 2. Q c is greater than K sp, the reaction will shift to the left and a ppt will form.Ī technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth. When these four ions are floating around in the solution, the only possible ppt is AgCl.īecause we are mixing, the first thing we need to do is convert to mols then divide by the new volume to find the new concentrations of AgNO 3 and NaCl: Write out what happens when these two compounds break up: The first step is to think about what the possible ppt might be from these compounds. Will there be a ppt if 50.0mL of 0.450 M AgNO 3 is mixed with 50.0mL of 0.350 M NaCl? When Q c is equal to K sp, the reaction is at equilibrium. When Q c is greater than K sp, there is too much product, so the reaction will shift to the left, and a ppt will form.
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When Q c is less than K sp, there is not enough product yet to be at equilibrium, so the reaction will shift to the right, and no ppt (precipitate) will form. The Q c, or ion product as it is called when considering solubility, is found the same way as the K sp, except the K sp is found at equilibrium and the Q c is not. (We went over Q c and comparing it to K c in Chapter 14). To try to find out whether or not precipitation will occur in a reaction, we need to compare the Q c to the K c, or in this case, the K sp. So we have to convert mols to grams using the periodic table: Remember the question asked for solubility, not molar solubility. (always check, but the approximation should hold for K sp problems)
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Notice that there is an initial value for C 2O 4 2-, contributed by the Na 2C 2O 4. What is the solubility of MgC 2O 4 in this solution? MgC 2O 4is added to a 0.437 M solution of Na 2C 2O 4. This change in solubility is explained by Le Chatelier's principle:ĬaC 2O 4 (s) ⇌ Ca 2+ (aq) + C 2O 4 2- (aq)Ĭalcium chloride will also contribute Ca 2+ ion, which will act as a stress on the right hand side of the equation and the equilibrium will shift to the left, and calcium oxalate will precipitate from the solution. For example, when adding calcium oxalate, CaC 2O 4, to a solution of calcium chloride, CaCl 2, both salts will contribute a Ca 2+ ion. If you have a salt in a solution of another salt having the same cation or anion, the salt will be less soluble in this solution than it would be in pure water.