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Section 4.5 Solution Stoichiometry

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Presentation on theme: "Section 4.5 Solution Stoichiometry"— Presentation transcript:

1 Section 4.5 Solution Stoichiometry

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

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

4 Molarity Conversion 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 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 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 g of an unknown diprotic acid. What is the molar mass of the acid?

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