Tutorial 4 SOLUTION STOICHIOMETRY. Solution stoichiometry calculations involve chemical reactions taking place in solution.

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1 T-27 Tutorial 4 SOLUTION STOICHIOMETRY Solution stoichiometry calculations involve chemical reactions taking place in solution. Of the various methods of expressing solution concentration the most convenient for general laboratory use is molarity, which is defined: Moles of solute Molarity = or M = Liters of solution n solute L soln Chemical reactions are written in terms of moles of reactants and products; this molarity concentration unit relates moles of solute to volume of solution. Thus, easily measured solution volumes provide a simple method of measuring moles of reactants. EXAMPLE: What is the molarity of a solution made by dissolving 5.67 g of potassium chloride in enough water to make ml of solution? This data gives a relationship between amount of solute and volume of solution: 5.67 g KCl / ml. To find molarity we must convert grams KCl to moles KCl and ml solution to L: 5.67 g KCl 1 mol KCl 1000 ml mol KCl x x = or M KCl ml 74.6 g KCl L L Whenever the solution concentration is given in molarity, M, you must change to the equivalent units, mol/l or mol/1000 ml, to use as a conversion factor. EXAMPLE: What mass of solute is contained in 15.0 ml of a M KCl solution? The conversion sequence is: ml solution L solution mol KCl g KCl 1 L mol KCl 74.6 g KCl 15.0 ml x x x = g KCl 1000 ml L mol KCl

2 T-28 OR, use the conversion sequence: ml solution mol KCl g KCl mol KCl 74.6 g KCl 15.0 ml x x = g KCl 1000 ml mol KCl EXAMPLE: What mass of potassium chloride would be needed to prepare ml of a M solution? OR 1 L mol KCl 74.6 g KCl ml x x x = 9.32 g KCl (needed) 1000 ml L mol KCl mol KCl 74.6 g KCl ml x x = 9.32 g KCl (needed) 1000 ml mol KCl Please note: the preceding two EXAMPLES are the same kind of problem worded differently: conversion from volume of solution to mass of solute. EXAMPLE: Silver nitrate solution is added to ml of a M potassium chloride solution until no more precipitate forms. What mass of silver chloride will be formed? The chemical equation for the reaction is: KCl (aq) + AgNO 3 (aq) AgCl (s) + KNO 3 (aq) Since the concentration and volume of silver nitrate solution are not specified, we can assume it is in excess. First find the number of moles of KCl in the ml of M solution: mol KCl (A) ml x = mol KCl 1000 ml

3 T-29 From the mole ratio in the balanced chemical equation: 1 mol AgCl (B) mol KCl x = mol AgCl formed 1 mol KCl Then convert to grams of AgCl: g AgCl (C) mol AgCl x = 1.79 g AgCl formed mol AgCl Steps (A), (B), and (C) may be combined: mol KCl 1 mol AgCl g AgCl ml x x x = 1.79 g AgCl 1000 ml 1 mol KCl mol AgCL Please note: the steps in solving this problem are essentially the same as used in Tutorial 3; the only difference is the method of first finding moles: thereafter they are identical problems. If volume and concentration of both reactant solutions are given then you are dealing with a limiting reagent problem. The first step is to find the moles of each reactant from the volume and concentration of the solution. Once moles are found the process is the same as shown in Tutorial 3, p T-23. The following equations are needed for some of the problems in this Tutorial. Na 2 SO 4 (aq) + BaCl 2 (aq) BaSO 4 (s) + 2NaCl (aq) BaCl 2 (aq) + 2AgNO 3 (aq) 2AgCl (s) + Ba(NO 3 ) 2 (aq) Ca(OH) 2 (aq) + 2HCl (aq) CaCl 2 (aq) + 2H 2 O (l) 2Al(OH) 3 (s) + 3H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (aq) + 6H 2 O (l) AlCl 3 (aq) + 3NaOH (aq) Al(OH) 3 (s) + 3NaCl (aq) 3KOH (aq) + H 3 PO 4 (aq) K 3 PO 4 (aq) + 3H 2 O (l)

4 T-30 1) Calculate the molarity of the following solutions: a) 15.5 g of potassium chloride in ml of solution. b) 1.25 x 10 2 g of silver nitrate in ml of solution. c) g of barium chloride in ml of solution. d) 15.0 mg of calcium hydroxide in 50.0 ml of solution. e) g of aluminum chloride in ml of solution. f) 1.00 g of potassium hydroxide in L of solution. g) g of sodium carbonate in ml of solution. h) 2.50 x 10 6 kg of potassium phosphate in ml of solution. 2) What mass of solute is contained in ml of a M sodium hydroxide solution? 3) What mass of solute is needed to prepare ml of a 1.50 M potassium nitrate solution? 4) What mass of solute is contained in ml of a 1.00 M barium nitrate solution? 5) What volume of M silver nitrate contains gram of silver nitrate? 6) What volume of M aluminum chloride solution contains 5.00 g of solute? 7) ml of a M sodium chloride solution was evaporated to dryness. What mass of solid remained? 8) In what total volume must 5.00 g of sodium nitrate be dissolved to make a M solution? 9) Barium chloride solution is added to ml of a M sodium sulfate solution until no more precipitate forms. What mass of barium sulfate forms? 10) Excess barium chloride solution is added to ml of M silver nitrate solution. What mass of silver chloride is formed? 11) What mass of aluminum hydroxide is required to completely react with 50.0 ml of M sulfuric acid solution? 12) Excess sodium hydroxide solution is added to ml of M aluminum chloride solution. What mass of aluminum hydroxide is formed?

5 T-31 13) What volume of 1.25 M sulfuric acid is needed to dissolve g of aluminum hydroxide? What is the molarity of the resulting solution of aluminum sulfate? (Assume final solution volume = volume of sulfuric acid used.) 14) Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution is needed to form g of silver chloride? 15) What volume of M potassium hydroxide solution is needed to react completely with a solution containing 1.00 g of phosphoric acid? 16) What volume of 1.25 M hydrochloric acid is needed to react completely with a solution containing 2.50 g of calcium hydroxide? 17) ml of M barium chloride solution is mixed with ml of silver nitrate solution. What mass of silver chloride will be formed? 18) 15.0 ml of M barium chloride solution is mixed with 25.0 ml of M silver nitrate solution. What mass of silver chloride will be formed? 19) ml of M aluminum chloride solution is mixed with ml of M sodium hydroxide solution. What mass of aluminum hydroxide will be formed? 20) ml of a solution containing 2.55 g of potassium hydroxide is mixed with ml of a 1.50 M phosphoric acid solution. Will the resulting mixture be acidic or basic? What will be the molarity of the potassium phosphate formed? (Assume the volumes are additive.) 21) ml of 1.50 M phosphoric acid solution is titrated with M potassium hydroxide solution. What volume of base solution is needed for complete neutralization? 22) ml of a barium chloride solution is titrated with M silver nitrate solution ml of the silver nitrate solution is required to completely precipitate the chloride ion as silver chloride. What is the molarity of the barium chloride solution? 23) ml of a phosphoric acid solution required ml of a M potassium hydroxide solution for neutralization. What is the molarity of the acid? 24) ml of a M calcium hydroxide solution is titrated with M hydrochloric acid solution. What volume of acid is needed for complete neutralization? 25) ml of M hydrochloric acid is titrated with a calcium hydroxide solution, and ml of the base is required for complete neutralization. What is the molarity of the calcium hydroxide solution?

6 T-32 Answers to Problems 1) a) M KCl 10) g AgCl b) 7.36 x 10 4 M AgNO 3 11) g Al(OH) 3 c) 5.33 x 10 4 M BaCl 2 12) g Al(OH) 3 d) 4.05 x 10 3 M Ca(OH) 2 13) 11.5 ml H 2 SO 4 ; M Al 2 (SO 4 ) 3 e) M AlCl 3 14) 1.22 ml BaCl 2 f) M KOH 15) 122 ml KOH g) M Na 2 CO 3 16) 54.0 ml HCl h) 2.36 x 10 5 M K 3 PO 4 17) AgNO 3 limiting; 1.79 g AgCl 2) g NaOH 18) BaCl 2 limiting; g AgCl 3) 37.9 g KNO 3 19) NaCl limiting; g Al(OH) 3 4) 26.1 g Ba(NO 3 ) 2 20) acidic; M K 3 PO 4 5) 20.8 ml AgNO 3 21) 247 ml KOH 6) 166 ml AlCl 3 22) M BaCl 2 7) 2.92 g NaCl 23) M H 3 PO 4 8) 235 ml 24) 23.6 ml HCl 9) 2.92 g BaSO 4 25) M Ca(OH) 2

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