It’s time to learn about . . . There she blows!!! Stoichiometry: Molar Solutions At the conclusion of our time together, you should be able to: 1. Define molarity 2. Determine the molarity of a given solution 3. Make a solution with a given molarity Some Definitions A solution is a homogeneous mixture of 2 or more substances in a single phase. The larger constituent is usually regarded as the SOLVENT and the others as SOLUTES. IONIC COMPOUNDS Compounds in Aqueous Solution Many reactions involve ionic compounds, in water these are — aqueous solutions. KMnO4 in water K+(aq) + MnO4-(aq) What Do You Get From a Pampered Cow? Spoiled Milk. Concentration of Solute The amount of solute in a solution is given by its concentration. Molarity(M ) = moles solute liters of solution PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl2•6H2O 1 mol 5.00 g • = 0.0210 mol 237.7 g Step 2: Calculate Molarity 0.0210 mol = 0.0841 M 0.250 L [NiCl2•6 H2O ] = 0.0841 M Concentration of Solute Molarity(M ) = moles solute liters of solution Therefore: x both sides by Liters = MV = moles = grams/molar mass SAME PROBLEM: Using moles = MV. 5.00 g NiCl2 x 1 mol NiCl2 237.71 g NiCl2 = 0.0841 M NiCl2*6H2O = M x 0.250 L USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? moles = M•V X g H2C2O4 x 1 mol H2C2O4 90.04 g H2C2O4 = 1.13 g H2C2O4 = 0.0500 x 0.250 L Page 25:2 moles = M•V X g NaNO3 x 1 mol NaNO3 85.00 g NaNO3 = 42.5 g NaNO3 = 0.50 M x 1.00 L Over-Worked Mouse Learning Check the Old Way?!! How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? moles = MV mol = 3.0 M x 0.400 L mol = 1.2 1.2 mol NaOH x 40.00 g NaOH 48 g NaOH 1 mol NaOH The Better Way moles = M•V X g NaOH x 1 mol NaOH 40.00 g NaOH = 48 g NaOH = 3.0 M x 0.400 L Hopefully, you’re not this lost!!! Preparing Solutions Determine the mass of solute. Place in the appropriate volumetric flask. Add deionized water and swirl until solute is dissolved. Add water to the mark on the neck of the flask. Stopper and mix thoroughly. Interesting Vanity Plate… Stoichiometry: Molar Solutions Let’s see if you can: 1. Define molarity 2. Determine the molarity of a given solution 3. Make a solution with a given molarity Concentration of Solute Molarity(M ) = moles solute liters of solution Therefore: MV = moles = grams/molar mass Support Bacteria!! They're the only culture some people have. PROBLEM: Mr. T accidentally dropped 15.14 g of silver (I) nitrate into 100.0 mL of deionized water. Rather than throw the solution away, help him by determining its molar concentration so that he can label it and still use it.. 15.14 g AgNO3 x 1 mol AgNO3 = 169.88 g AgNO3 = 0.8912 M AgNO3 M x 0.100 L PROBLEM: Mr. T needs to make 0.500 L of a 0.100 M solution of lead (II) nitrate for a future lab. Please help him by calculating the amount of solute needed and by outlining in detail the steps he would need to take to make this molar solution. X g Pb(NO3)2 x 1 mol Pb(NO3)2 331.22 g Pb(NO3)2 = 16.6 g Pb(NO3)2 = 0.100 M x 0.500 L Preparing Solutions Measure 16.6 g Pb(NO3)2. Place in a 0.500 L volumetric flask. Add deionized water and swirl until solute is dissolved. Add water to the mark on the neck of the flask. Stopper and mix thoroughly. Most Caring Person!! An elderly gentleman had recently lost his wife. Upon seeing the man cry, a little 4 year old neighbor boy went into the old gentleman's yard, climbed onto his lap, and just sat there. When his mother asked him what he had said to the neighbor, the little boy just said, "Nothing, I just helped him cry." PROBLEM Page 26-4: Grams to make 100. mL of 0.100 M calcium hydroxide. X g Ca(OH)2 x 1 mol Ca(OH)2 74.10 g Ca(OH)2 = 0.741 g Ca(OH)2 = 0.100 M x 0.100 L PROBLEM Page 26-5: 4.00 moles of nitric acid in 1.50 L of solution = what M. 4.00 mol HNO3 = 2.67 M HNO3 = X M x 1.50 L Entrance Quiz #1 1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water. 2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution. 1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water. 0.75 mol NaCl = X M x 0.3000 L = 2.5 M NaCl 2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution. X mol HCl = 2.45 M x 0.0122 L = 0.0299 mol HCl 0.0299 mol HCl x 36.46 g HCl 1 mol HCl = 1.09 g HCl Entrance Quiz #2 1. Calculate the molarity if 0.50 mol of KBr in placed in 750 mL of water. 2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl in solution. 1. Calculate the molarity if 0.50 mol of KBr in placed in 750 mL of water. 0.50 mol KBr = X M x 0.75 L = 0.67 M KBr 2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl. 2.45 g HCl x 1 mol HCl = 0.0672 mol HCl 36.46 g HCl 0.0672 mol HCl = X M x 0.0122 L = 5.51 M HCl Stoichiometry: Molar Solutions At the conclusion of our time together, you should be able to: 1. 2. 3. 4. Define molarity Determine the molarity of a given solution Make a solution with a given molarity Dilute a given solution to a new molarity Kid’s Letters to God: For Dilution: the Amount (moles) of Solute #1 = #2 And Using the Formula: Molarity(M ) = moles solute liters of solution M1V1 = moles and M2V2 = moles Therefore if moles of solute are constant: M1V1 = M2V2 The Scientific Method begins with Questions about the World Around You. Ever Wonder Why?... there are flotation devices under plane seats instead of parachutes? PROBLEM Page 30-1: Using M1V1 = M2V2 0.150 M NaOH x = 0.125 M NaOH 0.125 L = M x 0.150 L PROBLEM Page 30-5: Using M1V1 = M2V2 2.40 M KCl x 0.500 L = 1.00 M KCl x = 1.20 L Therefore: 0.700 L needs to be added XL How We Doing?? Need Help???? Stoichiometry: Molar Solutions Let’s see if you can: 1. 2. 3. 4. Define molarity Determine the molarity of a given solution Make a solution with a given molarity Dilute a given solution to a new molarity I knew I shouldn’t have done that!! How much water do I need to add to 250 mL of 3.0 M HCl to dilute it to 1.0 M HCl? 3.0 M HCl x 0.250 L = = 0.75 L Total, therefore 0.50 L 1.0 M HCl x L Name ________Class Period _____ Clicker Number Name ________Class Period _____ Clicker Number "Making Molar Solutions A1" (10 points) Make 50.00 mL of a 0.100M BaCl22H2O solution. __________ grams mass of solute needed __________ Instructor initials (one point) What are the products of the reaction? Balance the equation. Write the balanced molecular, complete ionic and net ionic equations below. Place the precipitate on the 3rd line of the first row of lines. (5 points) ________ + ________ ________ + ________ ___ ____ + ___ ____ ______ + ___ ____ ____ + ___ _____ What is this the best molar ratio based on the stoichiometry? Should you have put all 50.00 mL of each reactant together to form the most product? Circle yes or no. (2 points) ____________ : ____________ Yes No Preparing Solutions Determine the mass of solute. Place in the appropriate volumetric flask. Add deionized water and swirl until solute is dissolved. Add water to the mark on the neck of the flask. Stopper and mix thoroughly.