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Chem 330 notes Week 1

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by: Leslie Pike

Chem 330 notes Week 1 Chem 330

Leslie Pike
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Chem 222 review, competing equilibria, effects of electrolytes on equilibria
Quantitative Analysis Chemistry
Dr. Darwin Dahl
Class Notes
General Chemistry, quantitative_analysis_chemistry, Equilibrium




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This 6 page Class Notes was uploaded by Leslie Pike on Tuesday June 7, 2016. The Class Notes belongs to Chem 330 at Western Kentucky University taught by Dr. Darwin Dahl in Fall 2015. Since its upload, it has received 58 views. For similar materials see Quantitative Analysis Chemistry in Chemistry at Western Kentucky University.


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Date Created: 06/07/16
General review from 222: (If you need a more thorough review, I have oodles of Chem 222 materials available) Molarity is in units of moles solute per liter solution, or millimoles of solute per milliliters of solution. Molarity times volume equals moles. You will be making your own solutions in lab now, the TA will not make it for you, so it is important that you know how to prepare a dilute solution from a concentrated stock solution. % by mass: mass solute/mass total * 100%. This is unitless. Parts per million: mass solute/mass total * 1,000,000. Units are mg/kg or mg/L for dilute aqueous solution. (One liter of water weighs one kilogram, so liter and kilogram can be used interchangeably.) To prepare a 20 ppm solution NaCl, you would weigh out 20 mg NaCl and add it to 1 liter of water. Your containers in lab will be marked in % by mass, and specific gravity. In order to perform a dilution, you must convert these figures to moles. Specific gravity is density, and it is in units of grams solution/mL solution. If you were to calculate the density of a solution, you would measure out a certain volume and then weigh it, and divide mass by volume. The mass of the solute is not used in finding the density of a solution; therefore, the units of density are grams solution and mL solution. For example, a solution of HNO that is 70% HNO by mass and has a specific 3 3 gravity of 1.4: 70 g HNO /130 g solution * 1.4 g solution/mL solution * 1000 mL/L * mol HNO /63 g 3 HNO =315.6 mol HNO . 3 Use the ideal gas law, PV=nRT, to find the number of moles of a gas from the pressure, volume, and temperature. Pressure must be in atm, volume in L, and temperature in K. Dahl is tricky, he will give you the wrong units on purpose to see if you are paying enough attention. 2+ 3+ Sample problem: You have a solution that is 0.1 M Mg and 0.1 M Fe . Both of these metals form insoluble hydroxides. You want to separate the two ions by precipitating one out of solution as an insoluble hydroxide. Is this feasible, and if so, by how much do you need to raise the pH in order to accomplish separation? (Dahl made these numbers up) Ksp for magnesium hydroxide is 10^-8, and Ksp for iron(III) hydroxide is 10^-14. Solve for the solubilities of these hydroxides using ICE. (You learned how to do this in 222.) Iron has the lower solubility; therefore, it is the one we will precipitate out. A precipitation is considered complete when only 1% of the original concentration is remaining. The iron is 0.1 M, so 1% of this is 0.001 M. Solve the Ksp expression for the hydroxide concentration needed to achieve this. 10^-14 = (0.001)[OH ] , [OH ]=0.000215 M We don’t want any of the magnesium to precipitate out. Therefore, our hydroxide concentration must not get high enough to cause this. The highest acceptable concentration of hydroxide is: (once again, solve the Ksp expression) 10^-8 = (0.1)[OH ] , [OH] = 0.000316 M Our minimum hydroxide needed to precipitate the iron is smaller than the maximum hydroxide concentration feasible without precipitating the magnesium. Thus, this is doable. Converting the [OH] to pH gives a pH range of 10.3-10.5. The pH must be kept within this range to keep the iron precipitated without allowing the magnesium to precipitate. Chapter 11: Competing equilibria In practice, the solubility of barium carbonate in water far exceeds the value that would have been calculated from the Ksp of barium carbonate. Why is this? Carbonate reacts with water to form carbonic acid. This removes carbonate ion from the water, and more barium carbonate dissolves to compensate. To calculate the observed solubility of barium carbonate, we solve what is known as a competing equilibria problem. This has 8 steps: 1. Find a variable to express solubility. Since the amount number of moles of aqueous barium is equal the number of dissolved moles of barium carbonate (one mole of barium per one mole of barium carbonate), the concentration of barium is equal to the solubility. If we were calculating the solubility of silver carbonate, the solubility would be equal to half of the silver ion concentration (silver carbonate contains two silver atoms, so the silver concentration is double the amount of compound dissolved, so the solubility is half the amount of silver in solution). 2. Generate all K expressions for reactions taking place. (Shown in photo.) 3. Generate a mass balance expression for the reaction. (Shown in photo.) Barium carbonate contains equal amounts of barium and carbonate. However, some of the carbonate is present as bicarbonate ion or carbonic acid. 4. Generate a charge balance expression for the reaction. (Shown in photo.) Cations on the left, anions on the right. Anything with a charge of +2 or +3 is multiplied by the corresponding coefficient. (An ion with a +2 charge contains twice the charge of an ion with a +1 charge; therefore, its concentration is multiplied by 2 to compensate for this.) 5. Make sure you have an equation for every unknown. 6. Simplify everything that can be simplified. In this case, since barium carbonate produces a basic solution, we know automatically that this solution will contain very few (if any) acidic molecules. Thus, we remove carbonic acid from the mass balance equation, and hydronium from the charge balance equation. 7. Solve. If you get a nasty expression, it is not necessary to solve it numerically for the exam, but you must at least come up with an expression for solubility. Since the barium concentration is equal to the solubility, we devise an equation in which the only unknown is barium concentration. (All K will be given to you for the exam.) 8. Check your work and make sure that any simplifications you made did not significantly affect your result. (We ran out of time to do this one today.) Mass balancing an equation for ammonium sulfate, (NH ) S: 4 2 [NH ]3+ [NH ] =42([S ] + [HS ] + [H S]) 2 There is a 2 in front of all the sulfur things because ammonium sulfate contains two ammonium ions and one sulfur ion, meaning that the ammonium concentration would be double the sulfur concentration. To make the ammonium concentration and the sulfur concentration equal, sulfur must be multiplied by 2. Charge balancing barium phosphate: 2+ + 3- 2- - - 2[Ba ] + [H ] = 3[PO ] + 2[4PO ] + [H PO 4 + [OH ] 2 4 Competing Equilibria with a buffer involved Solution of calcium oxalate is buffered at a pH of 4. What is the solubility of calcium oxalate? 1. Oxalate will react with water to form bioxalate and oxalic acid. Set solubility equal to the calcium ion concentration. 2. Write all K expressions. Ksp=[Ca ][C O ], Kb1=[2C 4 ][OH ]/[C O ],2 4- - 2 42- Kb2=[H C O2]2OH4]/[HC O ], Kw2[4 ][OH ] + - 2+ - 2- 3. Mass balance. [Ca ]=[H C O ]+[H2 O2]+4C O ], 2H+4=10^-42 [4H-]=10^- 10 (we can get these because we are given the pH at the beginning of the problem) 4. Cannot charge balance because we do not know what the buffer is. 5. Number of equations equals number of unknowns, despite that we do not have a charge balance equation. 6. No simplifications can be made, but that is ok because pH problems are simpler anyway. 7. Using the mass balance equation: [Ca ]= Ksp/[Ca ] + Kb1*Ksp/([Ca ]*10^-10) + Kb2*Kb1*Ksp/([Ca ]*10^- 2+ 20) Other Competing Equilibria problems Silver can form complex ions with thiocyanate. What is the solubility of AgSCN in a 4 M solution of SCN-? Solubility is equal to the total amount of dissolved silver (silver ion plus the different silver complex ions). Write all K of formation expressions. (Not shown because the previous problems contain examples.) Mass balance (shown in picture below). The mass balance equation can then be solved by substituting in K expressions for the variables. We know the concentration of SCN-, 4 M, because it was given to us in the problem. Another competing equilibria problem: Chapter 10: Effects of electrolytes on equilibria An electrolyte’s interaction with water molecules affects its solubility. Ksp for an electrolyte is actually the product of activities, not the product of concentrations. Activity is equal to the activity constant times the concentration of the ion, Activity constant is calculated with this equation: Mu symbolizes ionic strength. This is calculated as .5 of the summation of the concentration of each ion times its charge squared. Ionic strength for 0.1 M magnesium chloride is calculated this way: Alpha, the effective hydrated radius, is a constant that would be looked up in a table. A hydrated ion is surrounded by water molecules that travel with it (they are attracted to its charge), making its effective radius larger. Z is the charge on the ion. Sample problem: what is the solubility of barium iodate in a solution of 0.02 M magnesium iodate? Activity constants were calculated with the activity constant equation (calculations not shown).


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