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Dilutions, Solution Stoichiometry and Titrations Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 1/1
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/2 Diluting Concentrated Solutions dilute solutionconcentrated solution + solvent M i V i = M f V f finalinitial Since the number of moles of solute remains constant, all that changes is the volume of solution by adding more solvent.
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/6 Diluting Concentrated Solutions Add 6.94 mL 18.0 M sulfuric acid to enough water to make mL of M solution. M i = 18.0 MM f = M V i = ? mLV f = mL = 6.94 mL 18.0 M mL V i = MiMi Mf VfMf Vf M = Sulfuric acid is normally purchased at a concentration of 18.0 M. How would you prepare mL of M aqueous H 2 SO 4 ? x Why?
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/11 Solution Stoichiometry aA + bBcC + dD Moles of A Volume of Solution of A Moles of B Volume of Solution of B Mole Ratio Between A and B (Coefficients) Molar Mass of B Molarity of A
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/12 Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq)Na 2 SO 4 (aq) + 2H 2 O(l) What volume of M H 2 SO 4 is needed to react with 50.0 mL of M NaOH? Moles of H 2 SO 4 Volume of Solution of H 2 SO 4 Moles of NaOH Volume of Solution of NaOH Mole Ratio Between H 2 SO 4 and NaOH Molarity of NaOH Molarity of H 2 SO 4
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/13 Solution Stoichiometry H 2 SO 4 (aq) + 2NaOH(aq)Na 2 SO 4 (aq) + 2H 2 O(l) 2 mol NaOH 1 mol H 2 SO mol H 2 SO 4 1 L solution 1 L mol 1 L 1000 mL = mol NaOH Volume of H 2 SO 4 needed: 1000 mL 1 L 10.0 mL solution (0.250 M H 2 SO 4 ) mol NaOH 50.0 mL NaOH Moles of NaOH available: x x x xx
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/25 Titration How can you tell when the reaction is complete? HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. Once the reaction is complete you can calculate the concentration of the unknown solution.
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/26 Titration unknown concentration solution Erlenmeyer flask buret standard solution (known concentration) An indicator is added which changes color once the reaction is complete
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/31 Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 48.6 mL of a M NaOH solution is needed to react with 20.0 mL of an unknown HCl concentration. What is the concentration of the HCl solution? Moles of NaOH Volume of Solution of NaOH Moles of HCl Volume of Solution of HCl Mole Ratio Between NaOH and HCl Molarity of HCl Molarity of NaOH
Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/32 Titration HCl(aq) + NaOH(aq)NaCl(aq) + 2H 2 O(l) 20.0 mL solution mol HCl = M HCl Concentration of HCl solution: Moles of NaOH available: 1 L mol = mol NaOH 48.6 mL NaOH 1000 mL 1 L Moles of HCl reacted: 1 mol NaOH 1 mol HCl = mol HCl mol NaOH 1 L 1000 mL x x x x
H12 – C3 3.19, 3.20, 3.22, 3.87*, 3.89, 3.91*, 3.93*, 3.120*
NEUTRALIZATION REACTIONS Section Neutralization Reactions Acid + Base Water + salt Hydronium ions + hydroxide ions water + water H 3 O + + OH -
Chapter 14 Section 14.2 Solution Concentration. The concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent.
MOLARITY – Ch 13, p. 412 Quantifies the concentration of a solution. Molarity (M) = mol solute = n = mol volume solution V L Read as “moles solute per.
The Care and Feeding of Burets. Burets A buret is a long cylindrical tube with a valve at the bottom end, which is used for making very precise volumetric.
Acid-Base Equilibria and Solubility Equilibria Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Stoichiometry. Stoichiometry is the part of chemistry that studies amounts of substances that are involved in reactions. It could be amounts of substances.
Titrations......help you see neutralization reactions. Acid and base react to form salt and water....determine concentration of a solution by reacting.
Solution Concentration. Calculations of Solution Concentration Mole fraction Mole fraction – the ratio of moles of solute to total moles of solution.
1 Chapter 4 Aqueous solutions Types of reactions.
Acid-Base Equilibria and Solubility Equilibria Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
What is the concentration of the solution?. What is in the flask?
© Copyright R.J. Rusay Aqueous Solutions Concentration / Calculations Dr. Ron Rusay Spring 2008.
1 Chemical Reactions Chapter 4 Stoichiometry. 2 Chemical Equations æ A chemical reaction shows the formulas and relative amounts of reactants and products.
Chapter 16 Properties of solutions. Making solutions l What the solute and the solvent are – Whether a substance will dissolve. – How much will dissolve.
Volumetric Analysis: Acid-Base Chpt. 13. Quantitative Analysis: is analysis which involves investigating the quantities or amounts of materials present.
Physical Properties of Solutions Chapter 12 Solution Stoichiometry end of Chapter 4 Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36,
II III I II. Solution Concentration (p. 480 – 486) Ch. 16 – Solutions.
COMMON ION EFFECT. COMMON ION an ion common with one in a system at equilibrium which places a stress on the equilibrium Common Ion Common Ion.
Calculations in chemistry: stoichiometry Chapter 15.
Making Molar Solutions From Liquids (More accurately, from stock solutions)
1 Chapter 12 Solutions 12.6 Solutions in Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.
MIXTURES & SOLUTIONS Today you need: Materials to take notes.
II III I II. Solution Concentration (p. 480 – 488) Ch. 14 – Mixtures & Solutions.
III. Titration (p. 493 – 503) Ch. 15 & 16 – Acids & Bases.
Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry Properties of Solutions.
1. Neutralization 2. NaOH + HCl H 2 O + NaCl 2NaOH + H 2 SO 4 2H 2 O + Na 2 SO 4 3. Twice as much HCl was required. Because it takes twice as much HCl.
Chapter 4: Aqueous Reactions and Solution Stoichiometry.
Chapter 9 Chemical Quantities. Chapter 9 Table of Contents Copyright © Cengage Learning. All rights reserved Information Given by Chemical Equations.
is the study of the relative quantities stoichiometry A. Using Mole Ratios Stoichiometry and Quantitative Analysis of reactants and products in a chemical.
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