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Section 4.5 Solution Stoichiometry

Section 4.5 Solution Stoichiometry. In This Section : Using Molarity Determining concentration Making Solutions Direct Addition Dilution Solution Stoichiometry and Titrations. 2 HCl + CaCO 3  H 2 O + CO 2 + CaCl 2. mol mol mol mol mol.

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Section 4.5 Solution Stoichiometry

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  1. Section 4.5 Solution Stoichiometry

  2. In This Section: Using Molarity Determining concentration Making Solutions Direct Addition Dilution Solution Stoichiometry and Titrations

  3. 2 HCl + CaCO3 H2O + CO2 + CaCl2 mol mol mol mol mol g g g g g Stoichiometry so Far

  4. MolarityConversion Factor: mol <-> liters A solution is 0.35 molar (3.5 mol/l) in NiCl2 Converting from Liters to Moles: How many moles are contained in 450 mL of this solution? Converting from Moles to Liters: What volume of this solution contains 2.24 mol NiCl2?

  5. Solution Stoichiometry

  6. Determining Solution Concentration: Molarity = mol solute/liter solution What is the concentration of a solution made by dissolving 23.5 g NiCl2 into a volume of 250 mL?

  7. Preparing Solutions: Direct Addition How many grams of NiCl2 would you use to prepare 100 mL of a 0.300 M solution?

  8. Preparing Solutions: Dilution from a concentrated solution How many mL of a 2.60 M NiCl2 solution would you use to prepare 100 mL of a 0.300 M solution?

  9. Titrations: The idea

  10. Titrations: This week’s lab Part 1. “Standardizing” a solution of base: 23.8 mL of an NaOH solution is used to neutralize 1.020 g H2C2O4-2H2O. What is the concentration of the NaOH solution?

  11. Titrations: This week’s lab Part 2. Determining the molar mass of an acid: 35.2 mL of the same NaOH solution is used to neutralize 1.265 g of an unknown diprotic acid. What is the molar mass of the acid?

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