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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.

  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?

  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?

  4. MOLARITY M = n / V = mol/ L • Find the molar concentration of each of the ions in a solution that contains 0.165 moles of barium chloride in 520 mL solution.

  5. MOLARITY M = n / V = mol/ L • Find the molarity of each of the ions present in a solution prepared by mixing 35 mL of 0.42 M K2SO4 with 27 mL of 0.17 M K3PO4. (Assume volumes are additive.)

  6. 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

  7. 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.

  8. MOLARITY & DILUTION Given a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl?

  9. 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

  10. 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

  11. 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

  12. 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.

  13. 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

  14. Titration • The controlled addition of the measured amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration.

  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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