4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solutions 4.4 Types of Chemical Reactions 4.5 Precipitation Reactions 4.6 Describing Reactions in Solution 4.7 Stoichiometry of Precipitation Reactions 4.8 Acid–Base Reactions 4.9 Oxidation–Reduction Reactions 4.10 Balancing Oxidation–Reduction Equations Copyright © Cengage Learning. All rights reserved

Moles of solute before dilution = moles of solute after dilution The process of adding water to a concentrated or stock solution to achieve the molarity desired for a particular solution. Dilution with water does not alter the numbers of moles of solute present. Moles of solute before dilution = moles of solute after dilution M1V1 = M2V2 Copyright © Cengage Learning. All rights reserved

Concept Check A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution? Add water to the solution. Pour some of the solution down the sink drain. c) Add more sodium chloride to the solution. d) Let the solution sit out in the open air for a couple of days. e) At least two of the above would decrease the concentration of the salt solution. Copyright © Cengage Learning. All rights reserved

Exercise What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution? 60.0 mL Copyright © Cengage Learning. All rights reserved

Precipitation Reaction A double displacement reaction in which a solid forms and separates from the solution. When ionic compounds dissolve in water, the resulting solution contains the separated ions. Precipitate – the solid that forms. Copyright © Cengage Learning. All rights reserved

The Reaction of K2CrO4(aq) and Ba(NO3)2(aq) Ba2+(aq) + CrO42–(aq) → BaCrO4(s) Copyright © Cengage Learning. All rights reserved

Precipitation of Silver Chloride Copyright © Cengage Learning. All rights reserved

Soluble – solid dissolves in solution; (aq) is used in reaction. Precipitates Soluble – solid dissolves in solution; (aq) is used in reaction. Insoluble – solid does not dissolve in solution; (s) is used in reaction. Insoluble and slightly soluble are often used interchangeably. Copyright © Cengage Learning. All rights reserved

Simple Rules for Solubility Most nitrate (NO3) salts are soluble. Most alkali metal (group 1A) salts and NH4+ are soluble. Most Cl, Br, and I salts are soluble (except Ag+, Pb2+, Hg22+). Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4). Most OH are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble). Most S2, CO32, CrO42, PO43 salts are only slightly soluble, except for those containing the cations in Rule 2. Copyright © Cengage Learning. All rights reserved

Concept Check Which of the following ions form compounds with Pb2+ that are generally soluble in water? a) S2– b) Cl– c) NO3– d) SO42– e) Na+ Copyright © Cengage Learning. All rights reserved

Formula Equation (Molecular Equation) Gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solution. Reactants and products generally shown as compounds. Use solubility rules to determine which compounds are aqueous and which compounds are solids. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Copyright © Cengage Learning. All rights reserved

Complete Ionic Equation Represents as ions all reactants and products that are strong electrolytes. Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)  AgCl(s) + Na+(aq) + NO3(aq) Copyright © Cengage Learning. All rights reserved

Ag+(aq) + Cl(aq)  AgCl(s) Net Ionic Equation Includes only those solution components undergoing a change. Show only components that actually react. Ag+(aq) + Cl(aq)  AgCl(s) Spectator ions are not included (ions that do not participate directly in the reaction). Na+ and NO3 are spectator ions. Copyright © Cengage Learning. All rights reserved

Concept Check Write the correct formula equation, complete ionic equation, and net ionic equation for the reaction between cobalt(II) chloride and sodium hydroxide. Formula Equation: CoCl2(aq) + 2NaOH(aq)  Co(OH)2(s) + 2NaCl(aq) Complete Ionic Equation: Co2+(aq) + 2Cl(aq) + 2Na+(aq) + 2OH(aq)  Co(OH)2(s) + 2Na+(aq) + 2Cl(aq) Net Ionic Equation: Co2+(aq) + 2Cl(aq)  Co(OH)2(s) Copyright © Cengage Learning. All rights reserved

Solving Stoichiometry Problems for Reactions in Solution Identify the species present in the combined solution, and determine what reaction if any occurs. Write the balanced net ionic equation for the reaction. Calculate the moles of reactants. Determine which reactant is limiting. Calculate the moles of product(s), as required. Convert to grams or other units, as required. Copyright © Cengage Learning. All rights reserved

What precipitate will form? lead(II) phosphate, Pb3(PO4)2 Concept Check (Part I) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What precipitate will form? lead(II) phosphate, Pb3(PO4)2 What mass of precipitate will form? 1.1 g Pb3(PO4)2 Copyright © Cengage Learning. All rights reserved

Where are we going? How do we get there? Let’s Think About It To find the mass of solid Pb3(PO4)2 formed. How do we get there? What are the ions present in the combined solution? What is the balanced net ionic equation for the reaction? What are the moles of reactants present in the solution? Which reactant is limiting? What moles of Pb3(PO4)2 will be formed? What mass of Pb3(PO4)2 will be formed? Copyright © Cengage Learning. All rights reserved

Concept Check (Part II) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of nitrate ions left in solution after the reaction is complete? 0.27 M Copyright © Cengage Learning. All rights reserved

Where are we going? How do we get there? Let’s Think About It To find the concentration of nitrate ions left in solution after the reaction is complete. How do we get there? What are the moles of nitrate ions present in the combined solution? What is the total volume of the combined solution? Copyright © Cengage Learning. All rights reserved

Concept Check (Part III) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of phosphate ions left in solution after the reaction is complete? 0.011 M Copyright © Cengage Learning. All rights reserved

Where are we going? How do we get there? Let’s Think About It To find the concentration of phosphate ions left in solution after the reaction is complete. How do we get there? What are the moles of phosphate ions present in the solution at the start of the reaction? How many moles of phosphate ions were used up in the reaction to make the solid Pb3(PO4)2? How many moles of phosphate ions are left over after the reaction is complete? What is the total volume of the combined solution? Copyright © Cengage Learning. All rights reserved

Continue next time with Acids and Bases. Copyright © Cengage Learning. All rights reserved