3.6 Solubility Solution: homogeneous mixture or mixture in which components are uniformly intermingled Solution: homogeneous mixture or mixture in which.

Slides:



Advertisements
Similar presentations
Chapter 15 Solutions. Chapter 15 Table of Contents Copyright © Cengage Learning. All rights reserved Solubility 15.2 Solution Composition: An Introduction.
Advertisements

SOLUTIONS Chapter 15.
UNIT 5 Aqueous Reactions and Solution Stoichiometry Molarity.
Ch 13 Solutions. Solutions Solution: homogeneous mixture of two substances. Each substance retains its identity. Solution: homogeneous mixture of two.
Chapter 15 Solutions Solution- homogeneous mixture w/ components uniformly intermingled Solute- substance in the smallest amount Solvent- substance in.
Chemical Calculations for Solutions (Ch 12) Dr. Harris Lecture 12 Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81.
Chapter 4 Types of Chemical Reactions and Solution Stoichiometry.
Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.
Molarity 2. Molarity (M) this is the most common expression of concentration M = molarity = moles of solute = mol liters of solution L Units are.
Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.
Metal + Acid Displacement. Activity Series of Metals.
Topic E conservation of atoms and mass
Chapter 13 Solutions I have no doubt that in reality the future will be vastly more surprising than anything I can imagine. Now my own suspicion is that.
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 / V = mol / L 2.
Solutions are homogeneous mixtures consisting of two or more components. The major component of a solution is known as the solvent and the minor component.
Concentration of Solutions
Chapter 4B Solutions.
Molarity, Molality, Dilutions, Percent Solutions, & Mole Fractions
Examples-Molarity and Dilutions. Example What is the molarity of a solution that contains 4.5 moles of sucrose in L of water?
Solutions Solubility -the amount of solute that can be dissolved to form a solution. Solvent – the substance in a solution present in the greatest amount.
Chapter 15: Solutions 15.1 Solubility
What quantities are conserved in chemical reactions? grams and atoms.
Solutions Chapter 14. Common Solutions Chemical solutions encountered in everyday life: aircoffee tap watergasoline shampoocough syrup orange sodaGatorade.
Chapter 15 Solutions.
Unit 14 – Solutions 14.1 Solubility 14.2 Solution Composition 14.3 Mass Percent 14.4 Molarity 14.5 Dilution 14.6 Stoichiometry of Solution Reactions 14.7.
Chapter 4 Chemical Quantities and Aqueous Reactions.
Chemical Calculations for Solutions (Ch 12) Dr. Harris Lecture 12 Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81.
1 SOLUTIONS A N I NTRODUCTION. 2 Objectives 1. What are different ways of expressing the concentration of a solution? When is each used?
Aqueous Reactions Aqueous Reactions and Solution Stoichiometry CDO High School.
Solutions: Molarity. A. Concentration – measure of the amount of solute that is dissolved in a given amount of solvent I. Concentration of Solutions Solutions:
Notes 15.2 Describing Solution Composition. Mass Percent Mass percent= mass of solute X 100 mass of solution = grams of solute X 100 grams of solute +
Helpful Definitions Solutions: homogeneous mixture of two or more substances physically mixed together in a uniform way. Solute: substance being dissolved.
Concentration Units: Terms like “dilute” and “concentrated” are not specific. Percent by Mass: Mass % = mass of solute x 100 Total mass of solution Recall:
Section 15.2 Describing Solution Composition 1. To understand mass percent and how to calculate it Objective.
Chapter 15: Solutions 15.1 Solubility 15.2 Solution Composition 15.3 Mass Percent 15.4 Molarity 15.7 Neutralization Reactions.
4.4 Solution Concentration and Stoichiometry. Solution Key Terms What type of mixture is also considered a solution? Give an example. – A homogeneous.
1 Chapter 12 Solutions 12.1 Solutions. 2 Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one.
INTRODUCTION TO SOLUTIONS/ACIDS AND BASES HONORS CHEMISTRY MAY 27 TH, 2014.
Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81. Aqueous Solutions Much of the chemistry that affects us occurs among substances dissolved in water (proteins,
SOLUTIONS Chapter 15. Solution = homogeneous mixture Solute = gets dissolved (minor component) Solvent = dissolving agent (major component)
Solutions Concentration of Solutions. Objectives 1.Define molarity and calculate its value. 2.Define molality and calculate its value. 3.Apply the principles.
Daily Inquiry: 1.Many solutions are described as concentrated or dilute. Distinguish between these two terms. 2.A solution is made by dissolving 4.75 g.
Chapter 11 Solutions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.
Daily Inquiry: 1.Many solutions are described as concentrated or dilute. Distinguish between these two terms. 2.A solution is made by dissolving 4.75 g.
John E. McMurry Robert C. Fay C H E M I S T R Y Sixth Edition Chapter 3 Mass Relationships in Chemical Reactions These Sections Will NOT be on EXAM 1.
Chapter 6: Mass Relationships in Chemical Reactions
John E. McMurry Robert C. Fay C H E M I S T R Y Sixth Edition Chapter 3 Mass Relationships in Chemical Reactions These Sections Will NOT be on EXAM 1.
Chapter 4 Chemical Quantities and Aqueous Reactions.
Molarity, pH, and Stoichiometry of Solutions Chapter 5 part 4.
John E. McMurry Robert C. Fay C H E M I S T R Y Sixth Edition Chapter 3 Mass Relationships in Chemical Reactions These Sections Will NOT be on EXAM 1.
Solubility Solubility: – Amount of a solute that can dissolve in a given amount of solvent Unsaturated Solution: – Contain less than the maximum amount.
Chapter 14 Solutions.
Copyright © Houghton Mifflin Company. All rights reserved. 15 | 1 Solutions Solutions are uniform mixtures. Solvent: the substance present in the highest.
Acid-Base Reactions and Titrations Chemistry. Examples of Acid-Base Rxns HNO 3 + KOH  H 2 O + KNO 3 H 2 SO NH 4 OH  (NH 4 ) 2 SO H 2 O LiOH.
Chapter 41 Aqueous Reactions and Solution Stoichiometry Chapter 4.
Topic 6. 5 Solutions. Solutions are homogeneous mixtures. The most common way to express solution concentration is by molarity.
Stoichiometry: Quantitative Information About Chemical Reactions Chapter 4.
Chapter Four Reactions in Aqueous Solutions. Chapter Four / Reactions in Aqueous Solutions Solutions and concentrations Solution is a homogenous mixture.
1.5 Solutions. Assessment Objectives Distinguish between the terms solute, solvent, solution and concentration (g dm–3 and mol dm–3) Solve.
Solutions, Solubility Rules, and Molarity. Solutions Solutions are defined as homogeneous mixtures of two or more pure substances. Aqueous solution –
Solutions - Quantitatively. Solutions Mixture of at least two components Mixture of at least two components Solute Solute Solvent Solvent Components can.
Solutions, Acids, & Bases
Solutions, Acids, & Bases
A homogeneous mixture in which the components are uniformly mixed
Solution Concentration (Molarity) Topic 1.5
Which one of these is more concentrated?
Unit 12. Solutions (including Molarity)
II. Molarity.
Solutions.
Warm-up Define solute and solvent What is a super saturated solution?
Presentation transcript:

3.6 Solubility Solution: homogeneous mixture or mixture in which components are uniformly intermingled Solution: homogeneous mixture or mixture in which components are uniformly intermingled Solute: substance that is being dissolved in solvent Solute: substance that is being dissolved in solvent Solvent: substance that dissolves solvent and present in a large amount Solvent: substance that dissolves solvent and present in a large amount Aqueous solutions: solutions with water as the solvent Aqueous solutions: solutions with water as the solvent

Solubility Saturated: A solution in which the maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the container Saturated: A solution in which the maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the container Unsaturated: A solution in which more of solute can be dissolved Unsaturated: A solution in which more of solute can be dissolved Concentrated: a relative large amount of solute is being dissolved in solvent Concentrated: a relative large amount of solute is being dissolved in solvent Diluted: a relative small amount of solute is being dissolved in solvent Diluted: a relative small amount of solute is being dissolved in solvent

Solution_Molarity Molarity: the number moles of solute per volume of solution in liters Molarity: the number moles of solute per volume of solution in liters moles of solute Molarity = Molarity = Liters of solution  unit = mol/L or M (molar)  Standard solution: is a solution whose concentration is accurately known.

Example What is the molarity of solution made by dissolving 2.355g of H 2 SO 4 in water and dilutin to a final volume of 50.0mL What is the molarity of solution made by dissolving 2.355g of H 2 SO 4 in water and dilutin to a final volume of 50.0mL Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution

Calculating volume and mass Determine how much volume (in ml) will be needed to dissolved 2.50 g of solid NaCl to make 0.050M solution. Determine how much volume (in ml) will be needed to dissolved 2.50 g of solid NaCl to make 0.050M solution. How many grams of solute would you use to prepare 1.50L of M glucose (C 6 H 12 O 6 ) How many grams of solute would you use to prepare 1.50L of M glucose (C 6 H 12 O 6 )

3.8 Dilution Reducing the original concentration of a chemical solution Reducing the original concentration of a chemical solution A process of transferring solution to achieve a the desired molarity by diluting with solvent A process of transferring solution to achieve a the desired molarity by diluting with solvent Moles of solute after dilution = moles of solute before dilution Moles of solute after dilution = moles of solute before dilution Formula  M 1 V 1 = M 2 V 2 Formula  M 1 V 1 = M 2 V 2

Example What volume of 16 M sulfuric acid must be used to prepare 1.5L of 0.10 M of H 2 SO 4 solution? What volume of 16 M sulfuric acid must be used to prepare 1.5L of 0.10 M of H 2 SO 4 solution? What is the final concentration if 75.0 mL of a 3.50M glucose solution is dilute to a volume of 400.0mL? What is the final concentration if 75.0 mL of a 3.50M glucose solution is dilute to a volume of 400.0mL? Calculate the new molarity if a dilution is made for: Calculate the new molarity if a dilution is made for: 25.0 ml of water is added to 10.0 mL of M CaCl ml of water is added to 10.0 mL of M CaCl 2

Examples Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO 3 NaHCO 3, according to the equation Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO 3 NaHCO 3, according to the equation HCl(aq) + NaHCO 3 (aq)  NaCl(aq) + H 2 O(l) + CO 2 (g) How many milliliters of 0.125M NaHCO 3 solution are needed to neutralized with 18.0mL of 0.100M HCl?

Examples A 25.0mL sample of oxalic acid is titrated and found to react with 94.7 mL of 0.200M NaOH. What is the molarity of the oxalic acid solution? The reaction is A 25.0mL sample of oxalic acid is titrated and found to react with 94.7 mL of 0.200M NaOH. What is the molarity of the oxalic acid solution? The reaction is 2NaOH(aq) + H 2 C 2 O 4 (aq)  C 2 O 4 Na 2 (aq) + 2H 2 O(l)

Calculating the mass When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. Calculate the mass of PbSO 4 formed when 1.25 L of M Pb(NO 3 ) 2 and 2.00 L of M Na 2 SO 4 are mixed When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. Calculate the mass of PbSO 4 formed when 1.25 L of M Pb(NO 3 ) 2 and 2.00 L of M Na 2 SO 4 are mixed Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq)  PbSO 4 (s) + 2NaNO 3 (aq)

Example What volume of M HNO 3 is required to neutralize with 24.9 ml of M Ba(OH) 2 (aq)? What volume of M HNO 3 is required to neutralize with 24.9 ml of M Ba(OH) 2 (aq)?