A strong electrolyte is an electrolyte that exists in solution almost entirely as ions PDF

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• A strong electrolyte is an electrolyte that exists in solution almost entirely as ions. • A weak electrolyte is an electrolyte that dissolves in water to give a relatively small percentage of ions. As a result, the light bulb shines weakly. • Compounds that dissolve readily are said to be soluble. • Compounds that dissolve very little are said to be insoluble. Solubility Rules • Molecular Equation • A chemical equation in which the reactants and products are written as if they were molecular

substances, even though they may actually exist in solution as ions. • State symbols are include: (s), (l), (g), (aq). • For example: • AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) • Although AgNO3, NaCl, and NaNO3 exist as ions in aqueous solutions, they are written as compounds in the molecular equation. • Complete Ionic Equation • A chemical equation in which strong electrolytes are written as separate ions in the solution. Other reactants and products are

written in molecular form. State symbols are included: (s), (l), (g), (aq). • For example: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) • In ionic form: Ag+(aq) + NO3-(aq) + Na+(aq)+Cl-(aq)  AgCl(s) + Na+(aq) + NO3-(aq) • Spectator Ion –An ion in an ionic equation that does not take part in the reaction. It appears as both a reactant and a product. • Net Ionic Equation • A chemical equation in which spectator ions are omitted. It

shows the reaction that actually occurs at the ionic level. • For example: Ag+(aq) + NO3-(aq) + Na+(aq) + Cl(aq)  AgCl(s) + Na+(aq) + NO3-(aq) • In net ionic form: Ag+(aq) + Cl-(aq)  AgCl(s) • Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. KBr + MgSO4  • Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. NaOH + MgCl2  • Molecular Equation

• (Balance the reaction and include state symbols) 2NaOH(aq) + MgCl2(aq)  2NaCl(aq) + Mg(OH)2(s) • Ionic Equation 2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + Mg(OH)2(s) • Net Ionic Equation 2OH-(aq) + Mg2+(aq)  Mg(OH)2(s) • Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. K3PO4 + CaCl2  • Molecular Equation • (Balance the reaction and include state symbols)

2K3PO4(aq) + 3CaCl2(aq)  6KCl(aq) + Ca3(PO4)2(s) • Ionic Equation 6K+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)  6K+(aq) + 6Cl-(aq) + Ca3(PO4)2(s) • Net Ionic Equation 2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s) Types of Chemical Reactions • Precipitation reactions: a solid ionic substance forms from the mixture of two solutions of ionic substances. •

Acid–base reactions: reactions that involve the transfer of a proton (H+) between reactants

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Oxidation–reduction reactions: reactions that involve the transfer of electrons between reactants. • A precipitate is an insoluble solid compound formed during a chemical reaction in solution.

• Predicting Precipitation Reactions 1. Predict the products (exchange of parts). 2. Determine the state of each product: (s), (l), (g), (aq). 3. If all products are aqueous (aq), no net reaction occurred. • Arrhenius Acid • A substance that produces hydrogen ions, H+, when it dissolves in water.

• Arrhenius Base • A substance that produces hydroxide ions, OH-, when it dissolves in water. • Brønsted–Lowry Acid • The species (molecule or ion) that donates a proton to another species in a proton-transfer reaction • Brønsted–Lowry Base • The species (molecule or ion) that accepts a proton from another species in a proton-transfer reaction • Household Acids and Bases • Acid-Base Indicator • A dye used to distinguish between an acidic and basic solution by

means of the color changes it undergoes in these solutions. • Strong Acid • An acid that ionizes completely in water. It is present entirely as ions; it is a strong electrolyte. • Common strong acids: HNO3 H2SO4 HClO4 HCl HBr HI • Weak Acid • An acid that only partly ionizes in water. It is present primarily as molecules and partly as ions; it is a weak electrolyte. • If an acid is not strong, it is weak. • In Figure A, a solution of HCl (a strong acid) illustrated on a

molecular/ionic level, shows the acid as all ions.

• Strong Base • A base that ionizes completely in water. It is present entirely as ions; it is a strong electrolyte. • Common strong bases: LiOH NaOH KOH Ca(OH)2 Sr(OH)2 Ba(OH)2 • Weak Base • A base that is only partly ionized in water. It is present primarily as molecules and partly as ions; it is a weak

electrolyte. These are often nitrogen bases such as NH3: • NH3(aq) + H2O(l) NH4+ (aq) + OH-(aq) • If a base is not strong, it is weak. • Classify the following as strong or weak acids or bases: a. KOH b. H2S c. CH3NH2 d. HClO4 • Polyprotic Acid

• An acid that results in two or more acidic hydrogens per molecule • For example: H2SO4, sulfuric acid • Neutralization Reaction • A reaction of an acid and a base that results in an ionic compound (a salt) and possibly water. • Write the molecular, ionic, and net ionic equations for the neutralization of sulfurous acid, H2SO3, by potassium hydroxide, KOH. • Molecular Equation

• (Balance the reaction and include state symbols) H2SO3(aq) + 2KOH(aq)  2H2O(l) + K2SO3(aq) • Ionic Equation H2SO3(aq) + 2K+(aq) + 2OH-(aq)  2H2O(l) + 2K+(aq) + SO32-(aq) • Net Ionic Equation H2SO3(aq) + 2OH-(aq)  2H2O(l) + SO32-(aq) • Acid-Base Reaction with Gas Formation • Some salts, when treated with an acid, produce a gas. Typically sulfides, sulfites, and carbonates behave in this way producing hydrogen sulfide, sulfur trioxide, and carbon dioxide, respectively.

• Gas-forming acid–base reactions: Na2S(aq) + 2HCl(aq)  2NaCl(aq) + H2S(g) Na2CO3(aq) + 2HCl(aq)  2NaCl(aq) + H2O(l) + CO2(g) Na2SO3(aq) + 2HCl(aq)  2NaCl(aq) + H2O(l) + SO2(g) • Write the molecular, ionic, and net ionic equations for the reaction of copper(II)

carbonate with hydrochloric acid. • Molecular Equation • (Balance the reaction and include state symbols) CuCO3(s) + 2HCl(aq)  CuCl2(aq) + H2O(l) + CO2(g) • Ionic Equation CuCO3(s) + 2H+(aq) + 2Cl-(aq)  Cu2+(aq) + 2Cl-(aq) + H2O(l) + CO2(g) • Net Ionic Equation CuCO3(s) + 2H+(aq)  Cu2+(aq) + H2O(l) + CO2(g)

• Oxidation Number • For a monatomic ion, the actual charge of the atom or a hypothetical charge assigned

to the atom in the substance using simple rules. Rules for Assigning Oxidation Numbers Rules for Assigning Oxidation Numbers • Elements: The oxidation number of an atom in an element is zero. • Monatomic ions: The oxidation number of an atom in a monatomic ion equals the charge on the ion. • Oxygen: The oxidation number of oxygen is -2 in most of its compounds. (An exception is O in H2O2 and other peroxides, where the oxidation number is -1.) • Hydrogen: The oxidation number of hydrogen is +1 in most of its

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compounds. (The oxidation number of hydrogen is -1 in binary compounds with a metal such as CaH2. Halogens: The oxidation number of fluorine is -1. Each of the other halogens (Cl, Br, I) has an oxidation number of -1 in binary compounds, except when the other element is another halogen above it in the periodic table or the other element is oxygen. Compounds and ions: The sum of the oxidation numbers of the atoms in a compound is zero. The sum of the oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion.

• Potassium permanganate, KMnO4, is a purple-colored compound; potassium manganate, K2MnO4, is a green-colored compound. Obtain the oxidation numbers of the manganese in these compounds. • K Mn O 2(+1) + 1(oxidation number of Mn) + 4(-2) = 0 2 + 1(oxidation number of Mn) + (-8) =0 (-6) + (oxidation number of Mn) = 0 Oxidation number of Mn = +6 In KMnO4, the oxidation number of Mn is +7.

In K2MnO4, the oxidation number of Mn is +6. • What is the oxidation number of Cr in dichromate, Cr2O72-?

• Half-reaction • One of two parts of an oxidation–reduction reaction, one part of which involves a loss of electrons (or increase in oxidation number) and the other part of which involves a gain of electrons (or decrease in oxidation number). • Oxidation • The half-reaction in which there is a loss of electrons by a species (or an increase in oxidation number).

• Reduction • The half-reaction in which there is a gain of electrons by a species (or a decrease in oxidation number). • Oxidizing Agent • A species that oxidizes another species; it is itself reduced. • Reducing Agent • A species that reduces another species; it is itself oxidized. Common Oxidation– Reduction Reactions

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Combination reaction Decomposition reaction Displacement reaction Combustion reaction • Combination Reaction • A reaction in which two substances combine to form a third substance. • For example: 2Na(s) + Cl2(g)  2NaCl(s) • Decomposition Reaction • A reaction in which a single compound reacts to give two or more substances. • For example: 2HgO(s)  2Hg(l) + O2(g)

• Displacement Reaction • A reaction in which an element reacts with a compound, displacing another element from it. • For example: Zn(s) + 2HCl(aq)  H2(g) + ZnCl2(aq) • Combustion Reaction • A reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a flame. • For example: 4Fe(s) + 3O2(g)  2Fe2O3(s)

• Balancing Simple Oxidation-Reduction Reactions: Half-Reaction Method • First, identify what is oxidized and what is reduced by determining oxidation numbers. • For the reaction Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s) 0 +1 +2 0 • •

Zn is oxidized from 0 to +2. Ag+ is reduced from +1 to 0.

• Next, write the unbalanced half-reactions. Zn(s)  Zn2+(aq) (oxidation) Ag+(aq)  Ag(s) (reduction) • Now, balance the charge in each half reaction by adding electrons. Zn(s)  Zn2+(aq) + 2e(oxidation) e- + Ag+(aq)  Ag(s) (reduction) • Since the electrons lost in oxidation are the same as those gained in reduction, we need each half-reaction to have the same number of

electrons. To do this, multiply each half-reaction by a factor so that when the halfreactions are added, the electrons cancel. Zn(s)  Zn2+(aq) + 2e(oxidation) 2e- + 2Ag+(aq)  2Ag(s) (reduction) • Lastly, add the two halfreactions together. Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)

• Balance the following oxidationreduction reaction: FeI3(aq) + Mg(s)  Fe(s) + MgI2(aq)

• Balancing the half-reactions: Mg(s)  Mg2+(aq) + 2e(oxidation) Fe3+(aq) + 3e-  Fe(s) (reduction) • Multiply the oxidation halfreaction by 3 and the reduction half-reaction by 2. 3Mg(s)  3Mg2+(aq) + 6e(oxidation) 2Fe3+(aq) + 6e-  2Fe(s) (reduction) • Add the half-reactions together.

2Fe3+(aq) + 3Mg(s)  2Fe(s) + 3Mg2+(aq) • Now, return the spectator ion, I-. 2FeI3(aq) + 3Mg(s)  2Fe(s) + 3MgI2(aq)

• Molar Concentration, Molarity, (M) • Moles of solute per liter of solution

• To prepare a solution, add the measured amount of solute to a volumetric flask, then add water to bring the solution to the mark on the flask.

• Dilution • When a higher concentration solution is used to make a lessconcentration solution, the moles of solute are determined by the amount of the higherconcentration solution. The number of moles of solute remains constant. MiVi = MfVf Note: The units on Vi and Vf must match. • Diluting a solution quantitatively requires specific glassware. • The photo at the right shows a volumetric flask used in dilution.

• You place a 1.62-g of potassium dichromate,

K2Cr2O7, into a 50.0-mL volumetric flask. You then add water to bring the solution up to the mark on the neck of the flask. What is the molarity of K2Cr2O7 in the solution? • A solution of sodium chloride used for intravenous transfusion (physiological saline solution) has a concentration of 0.154 M NaCl. How many moles of NaCl are contained in 500.mL of physiological saline? How many grams of NaCl are in the 500.-mL of solution?

• A saturated stock solution of NaCl is 6.00 M. How much of this stock solution is needed to prepare 1.00-L of physiological saline soluiton (0.154 M)? • Quantitative Analysis • The determination of the amount of a substance or species present in a material. • Gravimetric Analysis • A type of quantitative analysis in which the amount of a species in a material is determined by converting the species to a product that can

be isolated completely and weighed. • The figure on the right shows the reaction of Ba(NO3)2 with K2CrO4 forming the yellow BaCrO4 precipitate. • The BaCrO4 precipitate is being filtered in the figure on the right. It can then be dried and weighed. • A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compound gave 1.788 g

of silver chloride, what is the mass percent of silver in the compound? • Volumetric Analysis • A type of quantitative analysis based on titration • Titration • A procedure for determining the amount of substance A by adding a carefully measured volume with a known concentration of B until the reaction of A and B is just complete. • In the titration above, the indicator changes color to indicate when the reaction is just complete.

• Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas: • ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g) • How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS? • Molar mass of ZnS = 97.45 g

= 0.157 L = 157 mL HCl solution

• A dilute solution of hydrogen peroxide is sold in drugstores as a mild antiseptic. A typical solution was analyzed for the percentage of hydrogen peroxide by titrating it with potassium permanganate: • A solution is composed of two parts: the solute and the solvent. • Solute : The gas (or solid) in a solution of gases (or solids), or the component present in the smaller amount. • Solvent :The liquid in the case of a solution of gases or solids, or the component present in the larger amount.

• Fluids that mix with or dissolve in each other in all proportions are said to be miscible (left). • A saturated solution is in equilibrium with respect to the amount of dissolved solute. • The rate at which the solute leaves the solid state equals the rate at which the solute returns to the solid state. • The solubility of a solute is the amount that dissolves in a given quantity of solvent at a given temperature.

• An unsaturated solution is a solution not in equilibrium with respect to a given dissolved substance and in which more of the substance can be dissolved. • A supersaturated solution is a solution that contains more dissolved substance than a saturated solution does. This occurs when a solution is prepared at a higher temperature and is then slowly cooled. This is a very unstable situation, so any disturbance causes precipitation. • Solubility can be understood in terms of two factors:

• The natural tendency toward disorder favors dissolving. • The relative forces between and within species must be considered. • Stronger forces within solute species oppose dissolving. • Stronger forces between species favor dissolving. • For molecular solutions, this can be summarized as “Like dissolves like.” In other words, solutes dissolve in solvents that have the same type of intermolecular forces. • An immiscible solute and solvent are illustrated at right.

• When considering ionic solutes in water, we need to examine the hydration energy and the lattice energy. • The stronger ion-dipole force between the ion and the solvent— that is, hydration energy—favors dissolving. • A stronger force between ions— that is, lattice energy—opposes dissolving.

• The force of attraction between water and both a cation and an anion is illustrated to the left with lithium fluoride, LiF.

• The process of dissolving occurs at the surfaces of the solid. Here we see water hydrating (dissolving) ions. • The hydration energy for AB2 must be greater than the hydration energy for CB2. • In general, solubility depends on temperature. • In most cases, solubility increases with increasing temperature. However, for a number of compounds, solubility decreases with increasing temperature.

• The difference is explained by differences in the heat of solution. • When dissolving absorbs heat (is endothermic), the temperature of the solution decreases as the solute dissolves. The solubility will increase as temperature increases. • When dissolving releases heat (is exothermic), the temperature of the solution increases as the solute dissolves. The solubility will decrease as temperature increases.

• Hot packs use an exothermic solution process. • Henry’s law describes the effect of pressure on gas solubility: The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution. • This is expressed mathematically in the equation • S = kH P • where • S = gas solubility • kH = Henry’s law constant for the gas • P = partial pressure of the gas over the solution

• In general, pressure has little or no effect on the solubility of solids or liquids in water. • The solubility of a gas increases as pressure increases, as illustrated at right. • Helium–oxygen mixtures are sometimes used as the breathing gas in deep-sea diving. At sea level (where the pressure is 1.0 atm), the solubility of pure helium in blood is 0.94 g/mL. What is the solubility of pure helium at a depth of 1500 feet?

• Pressure increases by 1.0 atm for every 33 feet of depth, so at 1500 feet the pressure is 46 atm. (For a helium–oxygen mixture, the solubility of helium will depend on its initial partial pressure, which will be less than 1.0 atm.)

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P1 = 1.0 atm P2 = 46 atm S1 = 0.94 g/mL S2 = ?

• At high elevations, the partial pressure of oxygen decreases, decreasing the solubility of oxygen in water. Fish require a certain minimum level of dissolved oxygen to survive.

• The concentration of a solute can be quantitatively expressed in several ways: • • • •

Molarity Mass percentage of solute Molality Mole fraction

• Molarity is the moles of solute per liter of solution. It is abbreviated as M. • Mass percentage of solute is the percentage by mass of solute in a solution.

• An experiment calls for 36.0 g of a 5.00% aqueous solution of potassium bromide. Describe how you would make up such a solution. • A 5.00% aqueous solution of KBr has 5.00 g KBr per 100. g solution. The remainder of the 100. g is water: 95 g. • We can use this ratio to determine the mass of KBr in 36.0 g solution:

• Since 1.8 g KBr is required for 36.0 g of solution, the remainder consists of 34.2 g water.

• We make the solution by mixing 1.8 g KBr in 34.2 g water.

• Molality is the moles of solute per kilogram of solvent. It is abbreviated as m. • Iodine dissolves in a variety of organic solvents. For example, in methylene chloride, it forms an orange solution. What is the molality of a solution of 5.00 g iodine, I2, in 30.0 g of methylene chloride, CH2Cl2? • Mass of solute = 5.00 g I2

• Mass of solvent = 30.0 g CH2Cl2

• Mole fraction is the moles of component over the total moles of solution. It is abbreviated .

• A solution of iodine, I2, in methylene chloride, CH2Cl2,

contains 5.00 g I2 and 56.0 g CH2Cl2. What is the mole fraction of each component in this solution? • Mass of solute = 5.00 g I2 • Mass of solvent = 56.0 g CH2Cl2 • • A bottle of bourbon is labeled 94 proof, meaning that it is 47% by volume of alcohol in water. What is the mole fraction of ethyl alcohol, C2H5OH, in the bourbon? The density of ethyl alcohol is 0.80 g/mL.

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