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MOLARITY A measurement of the concentration of a solution

MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = n solute / V solution = mol solute / L solution.

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MOLARITY A measurement of the concentration of a solution

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  1. MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = nsolute / Vsolution = molsolute / Lsolution Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in enough water to make 500.0 mL of solution. First calculate the moles of solute: 1.5 g NaCl (1 mole NaCl) = 0.0257 moles of NaCl 58.45 g NaCl Next convert mL to L: 0.500 L of solution Last, plug the appropriate values into the correct variables in the equation: M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L = 0.051M

  2. MOLARITY M = n / V = mol/ L How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution? First calculate the moles of solute needed: M = n / V, now rearrange to solve for n:n = MV n = (1.25 mol / L) (0.2500 L) = 0.3125 moles of solute needed Next calculate the molar mass of LiOH: 23.95 g/mol Last, use diminsional analysis to solve for mass: 0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

  3. MOLARITY M = n / V = mol/ L What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL? First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute therefore: 47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = 0.705 g/mL density of solute Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass. 0.705 g/mL (1 mole/ 128 g) = 0.00551 mol/mL Next convert mL to L: 0.00551 mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M

  4. MOLARITY & DILUTION M1V1 = M2V2 The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged. Since the amount or moles of solute before dilution (nb) and the moles of solute after the dilution (na) are the same: nb = na And the moles for any solution can be calculated by n=MV A relationship can be established such that MbVb = nb = na= MaVa Or simply : MbVb = MaVa

  5. MOLARITY & DILUTION Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M1 = 0.05 mol/L M2 = ? V1 = 25.0 mL V2 = 50.0 + 25.0 = 75.0 mL M1V1 = M2V2 M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V2 75.0 mL

  6. MOLARITY & DILUTION Given a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl? M1 = 6.00 mol/L M2 = 0.150 V1 = ? mL V2 = 250.0 mL M1V1 = M2V2 M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl M1 6.00 mol/L You would need 6.25 mL of the 6.00 M HCl reagent which would be added to about 100 mL of DI water in a 250.0 mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at 250.0 mL. Mix well.

  7. MOLARITY & Stoichiometry In the past chapters, you measured out a solid reactant and needed to predict the mass of the product. You learned to use the molar mass and mole ratio to achieve that goal. The same reasoning applies with solution chemistry but instead to using the molar mass you will use the molarity to convert to moles. Stoichiometry Flow chart

  8. MOLARITY & Stoichiometry How much calcium carbonate will be precipitated by adding 25.0 mL calcium chloride to 25.0 mL of 0.56 M potassium carbonate? (assume the CaCl2 is in excess) CaCl2 + K2CO3 CaCO3 + 2 KCl V = 25.0 mL M = 0.56 mol/L m=? V = 25.0 mL First convert volume of A to moles of A: 0.025 L K2CO3 (0.56 mol/L) = 0.014 moles of K2CO3 Now convert moles of A to moles of B: 0.014 mol K2CO3(1 mol CaCO3/1 mol K2CO3) =0.014 mol CaCO3 Next convert moles of B to grams of B: 0.014 mol CaCO3 (100 g/mol) =1.40 g of CaCO3

  9. MOLARITY & Stoichiometry What would be the molarity of the potassium chloride solution from the last problem? CaCl2 + K2CO3 CaCO3 + 2 KCl V = 25.0 mL M = 0.56 mol/L M = ? V = 25.0 mL First convert volume of A to moles of A: 0.025 L K2CO3 (0.56 mol/L) = 0.014 moles of K2CO3 Now convert moles of A to moles of B: 0.014 mol K2CO3(2 mol KCl /1 mol K2CO3) =0.028 mol KCl Next convert moles of B to molarity of B: 0.028 mol KCl / 0.050 L =0.56 M of KCl

  10. PRACTICE PROBLEMS #24a 2.4 M _________1. What is the concentration of 250.0 mL of 0.60 moles of HCl? _________ 2. What volume of 0.7690 M LiOH will contain 55.3 g of LiOH? _________ 3. How many grams of barium sulfate that will precipitate when 500.0 mL of 0.340 M BaCl2 and 300.0 mL of 1.70 M Na2SO4 are mixed? 4. How would you prepare 850.0 mL of a 0.020 M ferric chloride solution if you start with crystals of FeCl3. 6H2O? 3.00 L 39.6 g Weigh out 4.59 g of the hydrated salt to a graduated cylinder with 800.0 mL then add enough DI water to make 850 mL exactly.

  11. PRACTICE PROBLEM #24b 1. What is the concentration of 35.0 mL of 0. 0556 moles of KCl? 2. How many grams of KCl is needed to prepare 50.0 mL of a 0.10 M solution? 3. How many milliliters of water must be added to 30.0 mL of 9.0 M KCl to make a solution that is 0.50 M KCl? 4. How many grams of calcium carbonate will precipitate when 500.0 mL of 0.340 M CaCl2 and 300.0 mL of 1.70 M Na2CO3 are mixed? 5. What is the concentration of the product solution (assuming the volumes are additive) when 500.0 mL of 0.340 M CaCl2 and 300.0 mL of 1.70 M Na2CO3 are mixed? 1.59 M 0.38 g 510 mL 17.0 g 0.425 M

  12. Class Problems ______1. How many grams of K3PO4 is needed to prepare 500.0 mL of a 0.00567 M solution? ______2. What volume of 0.7690 M LiOH will contain 55.3 g of LiOH? ______3. How many milliliters of water must be added to 600.0 mL of 3.40 M HCl to make a solution that is 0.500 M HCl? ______ 4. What is the concentration of the product solution (assuming the volumes are additive) when 24.9 mL of 0.70 M BaCl2 and 45.0 mL of 0.15 M Na3PO4 are mixed?

  13. Class problems ______1. What is the concentration of 250.0 mL of 0.60 moles of NaOH? ______2. What mass of MgCl2 will be required to prepare 500 mL of 0.150 M solution? ______3. How many milliliters of water must be added to 50.0 mL of 1.97 M NaOH to make a solution that is 0.025 M NaOH? ______ 4. How many grams of barium sulfate that will precipitate when 500.0 mL of 0.340 M BaCl2 and 300.0 mL of 1.70 M Na2SO4 are mixed? ______ 5. What is the concentration of the product solution (assuming the volumes are additive) when 500.0 mL of 0.340 M BaCl2 and 300.0 mL of 1.70 M Na2SO4 are mixed?

  14. Oxalic acid, H2C2O4 ACID-BASE REACTIONSTitrations H2C2O4(aq) + 2 NaOH(aq) ---> acidbase Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION.

  15. Setup for titrating an acid with a base

  16. Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. • Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION.

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