Chapter 4 & 11—Reactions & Solutions

Chapter 4 & 11—Reactions & Solutions

Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent.

Various Types of Solutions
Example State of Solution State of Solute State of Solvent Air, natural gas Gas Vodka, antifreeze Liquid Brass Solid Carbonated water (soda) Seawater, sugar solution Hydrogen in platinum Copyright © Cengage Learning. All rights reserved

Electrolytes Strong Electrolytes – conduct current very efficiently (bulb shines brightly). Weak Electrolytes – conduct only a small current (bulb glows dimly). Nonelectrolytes – no current flows (bulb remains unlit). Copyright © Cengage Learning. All rights reserved

Water, the common solvent
One of the most important substances on Earth. Can dissolve many different substances. A polar molecule because of its unequal charge distribution. Copyright © Cengage Learning. All rights reserved

Formation of a Liquid Solution
Separating the solute into its individual components (expanding the solute). Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). Allowing the solute and solvent to interact to form the solution. Copyright © Cengage Learning. All rights reserved

Steps in the Dissolving Process
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Steps in the Dissolving Process
Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic. Copyright © Cengage Learning. All rights reserved

Enthalpy (Heat) of Solution
Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔHsoln = ΔH1 + ΔH2 + ΔH3 ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released). Copyright © Cengage Learning. All rights reserved

Enthalpy (Heat) of Solution

If nature prefers exothermic processes, why do endothermic processes occur?

Example 1 Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role. Oil is a mixture of nonpolar molecules that interact through London dispersion forces, which depend on molecule size. ΔH1 will be relatively large for the large oil molecules. The term ΔH3 will be small, since interactions between the nonpolar solute molecules and the polar water molecules will be negligible. However, ΔH2 will be large and positive because it takes considerable energy to overcome the hydrogen bonding forces among the water molecules to expand the solvent. Thus ΔHsoln will be large and positive because of the ΔH1 and ΔH2 terms. Since a large amount of energy would have to be expended to form an oil-water solution, this process does not occur to any appreciable extent. Copyright © Cengage Learning. All rights reserved

The Energy Terms for Various Types of Solutes and Solvents
Hsoln Outcome Polar solute, polar solvent Large Large, negative Small Solution forms Nonpolar solute, polar solvent Large, positive No solution forms Nonpolar solute, nonpolar solvent Polar solute, nonpolar solvent Copyright © Cengage Learning. All rights reserved

In General One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed. Processes that require large amounts of energy tend not to occur. Overall, remember that “like dissolves like”. Copyright © Cengage Learning. All rights reserved

Factors Affecting Solubility
Structural Effects: Polarity Pressure Effects: Henry’s law Temperature Effects: Affecting aqueous solutions Copyright © Cengage Learning. All rights reserved

Pressure Effects C = concentration of dissolved gas k = constant
Henry’s law: C = kP C = concentration of dissolved gas k = constant P = partial pressure of gas solute above the solution Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. Copyright © Cengage Learning. All rights reserved

Temperature Effects (for Aqueous Solutions)
Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature. Copyright © Cengage Learning. All rights reserved

The Solubilities of Several Solids as a Function of Temperature

The Solubilities of Several Gases in Water

Example 3 A g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? 1.57 M 500.0 g is equivalent to mol K3PO4 (500.0 g / g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L). Copyright © Cengage Learning. All rights reserved

Example 4 What is the percent by mass concentration of glucose in a solution made by dissolving 5.5 g of glucose in 78.2 g of water? 6.6% [5.5 g / (5.5 g g)] × 100 = 6.6% Copyright © Cengage Learning. All rights reserved

Example 5 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 0.0145 8.00 g H3PO4 × (1 mol / g H3PO4) = mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 mol / g H2O) = 5.55 mol H2O Mole Fraction (H3PO4) = mol H3PO4 / [ mol H3PO mol H2O] = Copyright © Cengage Learning. All rights reserved

Example 6 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) 0.816 m 8.00 g H3PO4 × (1 mol / g H3PO4) = mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = kg H2O Molality = mol H3PO4 / kg H2O] = m Copyright © Cengage Learning. All rights reserved

Concentration of Ions For a 0.25 M CaCl2 solution: CaCl2 → Ca2+ + 2Cl–
Ca2+: 1 × 0.25 M = 0.25 M Ca2+ Cl–: 2 × 0.25 M = 0.50 M Cl–. Copyright © Cengage Learning. All rights reserved

Example 7 Which of the following solutions contains the greatest concentration of ions? 400.0 mL of 0.10 M NaCl. 300.0 mL of 0.10 M CaCl2. 200.0 mL of 0.10 M FeCl3. 800.0 mL of 0.10 M sucrose. a) contains mol of ions (0.400 L × 0.10 M × 2). b) contains mol of ions (0.300 L × 0.10 M × 3). c) contains mol of ions (0.200 L × 0.10 M × 4). d) does not contain any ions because sucrose does not break up into ions. Therefore, letter b) is correct. Copyright © Cengage Learning. All rights reserved

Notice The solution with the greatest number of ions is not necessarily the one in which: the volume of the solution is the largest. the formula unit has the greatest number of ions. Copyright © Cengage Learning. All rights reserved

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

Example 8 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. For letter a), adding water to the solution will increase the total volume of solution and therefore decrease the concentration. For letter b), pouring some of the solution down the drain will not change the concentration of the salt solution remaining. For letter c), adding more sodium chloride to the solution will increase the number of moles of salt ions and therefore increase the concentration. For letter d), water will evaporate from the solution and decrease the total volume of solution and therefore increase the concentration. Therefore, since only letter a) would decrease the concentration, letter e) cannot be correct. Copyright © Cengage Learning. All rights reserved

Example 9 What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a M NaOH solution? 60.0 mL The minimum volume needed is 60.0 mL. M1V1 = M2V2 (2.00 M)(V1) = (0.800 M)(150.0 mL) Copyright © Cengage Learning. All rights reserved

Colligative Properties
Depend only on the number, not on the identity, of the solute particles in an ideal solution: Vapor Pressure Reduction Boiling-point elevation Freezing-point depression Osmotic pressure Copyright © Cengage Learning. All rights reserved

Vapor Pressures of Solutions
Nonvolatile solute lowers the vapor pressure of a solvent. Raoult’s Law: Psoln = observed vapor pressure of solution solv = mole fraction of solvent = vapor pressure of pure solvent Copyright © Cengage Learning. All rights reserved

A Solution Obeying Raoult’s Law

Boiling-Point Elevation
Nonvolatile solute elevates the boiling point of the solvent. ΔT = Kbmsolute ΔT = boiling-point elevation Kb = molal boiling-point elevation constant msolute= molality of solute Copyright © Cengage Learning. All rights reserved

Freezing-Point Depression
When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = Kfmsolute ΔT = freezing-point depression Kf = molal freezing-point depression constant msolute= molality of solute Copyright © Cengage Learning. All rights reserved

Changes in Boiling Point and Freezing Point of Water

Example 1 A solution was prepared by dissolving g glucose in g water. The molar mass of glucose is g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution. °C The change in temperature is ΔT = Kbmsolute. Kb is 0.51 °C·kg/mol. To solve formsolute, use the equation m = moles of solute/kg of solvent. Moles of solute = (25.00 g glucose)(1 mol / g glucose) = mol glucose Kg of solvent = (200.0 g)(1 kg / 1000 g) = kg water msolute = ( mol glucose) / ( kg water) = mol/kg ΔT = (0.51 °C·kg/mol)( mol/kg) = 0.35 °C. The boiling point of the resulting solution is °C °C = °C. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Example 2 The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO3 i = 2 Na3PO4 i = 4 Copyright © Cengage Learning. All rights reserved

Ion Pairing Ion pairing is most important in concentrated solutions.
As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Ion pairing is most important for highly charged ions. Copyright © Cengage Learning. All rights reserved

Colloids A suspension of tiny particles in some medium.
Tyndall effect – scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm. Copyright © Cengage Learning. All rights reserved

Coagulation Destruction of a colloid.
Usually accomplished either by heating or by adding an electrolyte. Particles clump together and can be removed Naturally occurs at river deltas where fresh water meets salt water Copyright © Cengage Learning. All rights reserved

TYPES OF REACTIONS Precipitation Reactions Acid–Base Reactions
Oxidation–Reduction Reactions 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)
Ba(NO3)2 (aq) + K2CrO4(aq) → BaCrO4(s) + KNO3 (aq) Copyright © Cengage Learning. All rights reserved

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

Example 2 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+ a), b), and d) all form precipitates with Pb2+. A compound cannot form between only Pb2+ and 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

Break Up into Ions! Do NOT Break Up!
Strong Acids Strong Bases Soluble Salts Weak Acids Weak Bases Insoluble Salts Nonelectrolytes or Weak electrolytes

Example 3 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) + 2OH(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

Example 4 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 The balanced molecular equation is: 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s). mol Na3PO4 present to start and mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is g/mol, 1.1 g of Pb3(PO4)2 will form. Copyright © Cengage Learning. All rights reserved

Example 5 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 The concentration of nitrate ions left in solution after the reaction is complete is 0.27 M. Nitrate ions are spectator ions and do not participate directly in the chemical reaction. Since there were mol of Pb(NO3)2 present to start, then mol of nitrate ions are present. The total volume in solution is 30.0 mL. Therefore the concentration of nitrate ions = mol / L = 0.27 M. Copyright © Cengage Learning. All rights reserved

Example 6 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 The concentration of phosphate ions left in solution after the reaction is complete is M. Phosphate ions directly participate in the chemical reaction to make the precipitate. There were mol of Na3PO4 present to start, therefore there was mol of phosphate ions present to start mol of phosphate ions were used up in the chemical reaction, therefore mol of phosphate ions is leftover ( – mol). The total volume in solution is 30.0 mL. Therefore the concentration of phosphate ions = mol / L = M. Copyright © Cengage Learning. All rights reserved

Redox Reactions Reactions in which one or more electrons are transferred. Transfer may occur to form ions Oxidation – increase in oxidation state (loss of electrons); reducing agent Reduction – decrease in oxidation state (gain of electrons); oxidizing agent Note: Combustion reactions are redox! Double replacement reactions are NEVER redox! Copyright © Cengage Learning. All rights reserved

Reaction of Sodium and Chlorine

Rules for Assigning Oxidation Numbers
FIMHO 1. F—free elements and fluorine. Free elements always have an oxidation number of 0. Fluorine is always -1.

2. I—ions. The oxidation number of ions is the charge of the ion.
3. M—metals. Use the Periodic Table to get the charge of a metal. If the metal has more than one charge, try to determine the charge from the second element. Ex—ZnCl2

4. H—hydrogen. H is always +1 EXCEPT when H is bonded to a Group 1 metal. When H is bonded to a Group 1 metal it is -1. Ex.—NaH vs. HCl 5. O—oxygen. O is always -2 EXCEPT when it is a peroxide (-1) or bonded to F (+2). Ex—Na2O2

ALL OXIDATION NUMBERS MUST ADD TO THE TOTAL CHARGE ON THE COMPOUND OR ION!!

Example 1 Find the oxidation states for each of the elements in each of the following compounds: K2Cr2O7 CO32- MnO2 PCl5 SF4 K = +1; Cr = +6; O = –2 C = +4; O = –2 Mn = +4; O = –2 P = +5; Cl = –1 S = +4; F = –1 K2Cr2O7; K = +1; Cr = +6; O = -2 CO32-; C = +4; O = -2 MnO2; Mn = +4; O = -2 PCl5; P = +5; Cl = -1 SF4; S = +4; F = -1 Copyright © Cengage Learning. All rights reserved

Example 2 Which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent. Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l) 2CuCl(aq)  CuCl2(aq) + Cu(s) Zn – reducing agent; HCl – oxidizing agent c) CuCl acts as the reducing and oxidizing agent Copyright © Cengage Learning. All rights reserved

Balancing Oxidation–Reduction Reactions by Oxidation States
Write the unbalanced equation. Determine the oxidation states of all atoms in the reactants and products. Show electrons gained and lost. Use coefficients to equalize the electrons gained and lost. Balance the rest of the equation by inspection. Add appropriate states. Copyright © Cengage Learning. All rights reserved

Gas-Forming Reactions
Some metathesis (double replacement) reactions do not give the product expected. In this reaction, the expected product (H2CO3) decomposes to give a gaseous product (CO2). CaCO3 (s) + HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l) © 2009, Prentice-Hall, Inc.``

When a carbonate or bicarbonate reacts with an acid, the products are a salt, carbon dioxide, and water. CaCO3 (s) + HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l) NaHCO3 (aq) + HBr (aq) NaBr (aq) + CO2 (g) + H2O (l) © 2009, Prentice-Hall, Inc.

Performing Calculations for Acid–Base Reactions
List the species present in the combined solution before any reaction occurs, and decide what reaction will occur. Write the balanced net ionic equation for this reaction. Calculate moles of reactants. Determine the limiting reactant, where appropriate. Calculate the moles of the required reactant or product. Convert to grams or volume (of solution), as required. Copyright © Cengage Learning. All rights reserved

Acid–Base Titrations Titration – delivery of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). Equivalence point – enough titrant added to react exactly with the analyte. Endpoint – the indicator changes color so you can tell the equivalence point has been reached. Copyright © Cengage Learning. All rights reserved

Example 1 For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of M sulfuric acid to reach the endpoint? 1.00 mol NaOH The balanced equation is: H2SO4 + 2NaOH → 2H2O + Na2SO moles of sulfuric acid is present to start. Due to the 1:2 ratio in the equation, 1.00 mol of NaOH would be required to exactly react with the sulfuric acid. 1.00 mol of sodium hydroxide would be required. Copyright © Cengage Learning. All rights reserved

Complex Ion Formation Usually formed from a transition metal and ligand (polar molecules or anions) “Rule of thumb”: place twice the number of ligands around a cation as the charge on the ion Ex: Cu(NH3)42+

Ag+ (aq) + 2Cl- (aq) → [AgCl2]-
A precipitate can become soluble if a complex ion is formed Keywords to look for: “excess”, “concentrated” Example 2: Write the balanced net ionic equation for the reaction of aqueous silver nitrate and excess hydrochloric acid. Ag+ (aq) + 2Cl- (aq) → [AgCl2]-

Chapter 4 & 11—Reactions & Solutions

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