Chapter 4 Solution Stoichiometry Dr. Sapna Gupta
Concentrations of Solutions A solution is solute dissolved in a solvent. To quantify and know exactly how much of a solute is present in a certain amount of solvent, one will need to calculate concentrations. Concentrations are given in percent solutions mass/mass % - (g of solute/g of solution) x 100% mass/volume % - (g of solute/ml of solution) x 100% volume/volume % - (ml of solute/ml of solution) x 100% Molarity (mol/l of solution)- used more in Chemistry Molality (mol/kg of solution) this is used more in Biology Dr. Sapna Gupta/Solution Stoichiometry 2
Molar Concentration, Molarity (M) In this chapter we will study Molarity which is moles in a L of solution. Molarity is represented by M and the formula is given below moles of solute Molarity = L ofsolution Moles are converted to grams in order to make the solution in the lab. To prepare a solution, add the measured amount of solute to a volumetric flask, then add water to bring the solution to the mark on the flask. A 3M solution of NaCl means there are 3 moles of NaCl in the solution. If you have a 200 ml of 2 M HCl that means that there are 2 mols of HCl in 1 L solution. If you want to know how many grams of HCl you have in 200 ml then you will have to calculate the amount of moles in 200 ml of that solution using the Molarity equation; then you can calculate the grams from those moles. Dr. Sapna Gupta/Solution Stoichiometry 3
Example: You place a 1.52 g of potassium dichromate, K 2 Cr 2 O 7, into a 50.0 ml volumetric flask. You then add water to bring the solution up to the mark on the neck of the flask. What is the molarity of K 2 Cr 2 O 7 in the solution? Molar mass of K 2 Cr 2 O 7 is 294 g/mol 1 mol 1.52 g 294 g 3 50.0 10 L 0.103 M Example: A solution of sodium chloride used for intravenous transfusion (physiological saline solution) has a concentration of 0.154 M NaCl. How many moles of NaCl are contained in 500. ml of physiological saline? How many grams of NaCl are in the 500. ml of solution? mol M = mol L M L 0.154 0.0770 M 0.500 L mol NaCl Molar mass NaCl 58.4 g 0.0770 mol 1mol 4.50 g NaCl 58.4 g Dr. Sapna Gupta/Solution Stoichiometry 4
Example: Calculate the molarity of a solution prepared by dissolving 45.00 grams of KI into a total volume of 500.0 ml. 45.00 g KI 500.0 ml 1mol KI 166.0 g KI 1000 ml 1L 0.5422 M Example: How many milliliters of 3.50 M NaOH can be prepared from 75.00 grams of the solid? 1mol NaOH 1L 1000 ml 75.00 g NaOH 536 ml 40.00 g NaOH 3.50 mol NaOH 1L Dr. Sapna Gupta/Solution Stoichiometry 5
Dilution When a higher concentration solution is used to make a less concentration solution, the moles of solute are determined by the amount of the higher concentration solution. The number of moles of solute remains constant when more water is added. M i V i = M f V f Note: The units on V i and V f must match. Dr. Sapna Gupta/Solution Stoichiometry 6
Diluting a solution quantitatively requires specific glassware. The photo at the right shows a volumetric flask used in dilution. Dr. Sapna Gupta/Solution Stoichiometry 7
Example: A saturated stock solution of NaCl is 6.00 M. How much of this stock solution is needed to prepare 1.00 L of physiological saline solution (0.154 M)? M V V i i i M V f M M V i f f f V V i i (0.154 M)(1.00 L) 6.00 M 0.0257 L or 25.7 ml Example: For the next experiment the class will need 250. ml of 0.10 M CuCl 2. There is a bottle of 2.0 M CuCl 2. Describe how to prepare this solution. How much of the 2.0 M solution do we need? Concentrated: 2.0 M use? ml (V c ) Diluted: 250. ml of 0.10 M M c V c = M d V d (2.0 M) (V c ) = (0.10 M) (250.mL) V c = 12.5 ml 12.5 ml of the concentrated solution are needed; add enough distilled water to prepare 250. ml of the solution. Dr. Sapna Gupta/Solution Stoichiometry 8
Solution Stoichiometry In solution stoichiometry you have to presume that soluble ionic compounds dissociate completely in solution. Then using mole ratios we can calculate the concentration of all species in solution. There are three common types of stoichiometric calculations Quantitative Analysis: The determination of the amount of a substance or species present in a material. Volumetric Analysis: A type of quantitative analysis based on titration. Gravimetric Analysis: A type of quantitative analysis in which the amount of a species in a material is determined by converting the species to a product that can be isolated completely and weighed. Dr. Sapna Gupta/Solution Stoichiometry 9
Volumetric Analysis - Titrations A procedure for determining the amount of substance A by adding a carefully measured volume with a known concentration of B until the reaction of A and B is just complete. This can be for precipitation, neutralization or redox reactions. Standardization is the determination of the exact concentration of a solution. Equivalence point represents completion of the reaction. Endpoint is where the titration is stopped. An indicator is used to signal the endpoint. Dr. Sapna Gupta/Solution Stoichiometry 10
Gravimetric Analysis In gravimetric analysis precipitation reactions are carried out. After the reaction the product is precipitated and collected in a crucible or filter paper. The precipitate is weighed and then using mole ratios we can calculate the concentration of all species in original solution. The figure below shows the reaction of Ba(NO 3 ) 2 with K 2 CrO 4 forming the yellow BaCrO 4 precipitate. The BaCrO 4 precipitate is being filtered. It can then be dried and weighed. Then concentration of Ba 2+ ions can be calculated Dr. Sapna Gupta/Solution Stoichiometry 11
Example: Find the concentration of all species in a 0.25 M solution of MgCl 2 MgCl 2 Mg 2+ + 2Cl Given: MgCl 2 = 0.25 M [Mg 2+ ] = 0.25 M (1:1 ratio) [Cl ] = 0.50 M (1:2 ratio) Example: A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compound gave 1.788 g of silver chloride, what is the mass percent of silver in the compound? Strategy: g AgCl -> mol AgCl -> mol Ag -> g Ag -> % Ag Molar mass of silver chloride (AgCl) = 143.32 g 1mol AgCl 1mol Ag 107.9 g Ag 1.788 g AgCl = 1.346 g Ag in the compound 143.32 g AgCl 1mol AgCl 1mol Ag 1.583 1.346 g Ag g silver compound 100% = 85.03% Ag Dr. Sapna Gupta/Solution Stoichiometry 12
Example: Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas: ZnS(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 S(g) How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS? Strategy: g ZnS -> mol ZnS -> mol HCl (mol ratio from eq) -> vol HCl (using Molarity) Molar mass of ZnS = 97.45 g 1mol ZnS 2 mol HCl 1L solution 0.392 g ZnS 97.45 g ZnS 1mol ZnS 0.0512 mol HCl = 0.157 L = 157 ml HCl solution Dr. Sapna Gupta/Solution Stoichiometry 13
Example: A dilute solution of hydrogen peroxide is sold in drugstores as a mild antiseptic. A typical solution was analyzed for the percentage of hydrogen peroxide by titrating it with potassium permanganate: 5H 2 O 2 (aq) + 2KMnO 4 (aq) + 6H + (aq) 8H 2 O(l) + 5O 2 (g) + 2K + (aq) + 2Mn 2+ (aq) What is the mass percent of H 2 O 2 in a solution if 57.5 g of solution required 38.9 ml of 0.534 M KMnO 4 for its titration? Strategy: mols KMnO 4 -> mols H 2 O 2 -> mass H 2 O 2 -> % H 2 O 2 Molar mass of H 2 O 2 = 34.01 g 0.534 mol KMnO 5 mol H O 34.01g H O 1L 2 mol KMnO 1mol H O 3 38.9 10 L 4 2 2 2 2 4 2 2 = 1.77 g H 2 O 2 1.77 g H2O2 100% 57.5 g solution = 3.07% H 2 O 2 Dr. Sapna Gupta/Solution Stoichiometry 14
Example: A student measured exactly 15.0 ml of an unknown monoprotic acidic solution and placed in an Erlenmeyer flask. An indicator was added to the flask. At the end of the titration the student had used 35.0 ml of 0.12 M NaOH to neutralize the acid. Calculate the molarity of the acid. Strategy: mols NaOH -> mols acid (from eq) -> Molarity of acid 0.035 M NaOH x 0.12 mol NaOH 1 L x 1 mol acid 1 mol NaOH = 0.0042 mol acid 0.0042 mol 0.015 L = 0.28 M acid Example: Calculate the molarity of 25.0 ml of a monoprotic acid if it took 45.50 ml of 0.25 M KOH to neutralize the acid. 0.25 mol KOH 0.04550 L L 1mol acid 1mol KOH 0.01338 mol acid 0.01338 mol acid 0.0250 L 0.455 M Dr. Sapna Gupta/Solution Stoichiometry 15
Key Words/Concepts Solutions o Solvent o Solute Molarity (mol/l) Dilutions (M i V i = M f V f ) Solution stoichiometry o Volumetric analysis (titration) o End point o Equivalence point o Gravimetric analysis Dr. Sapna Gupta/Solution Stoichiometry 16