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Aqueous Reactions and Solution Stoichiometry (Chapter 4) Water has many unique chemical and physical properties. Possibly one of the most important is its ability to dissolve other substances to form solutions. Solutions are homogeneous mixtures of two or more substances. The solvent (usually the substance present in the greatest quantity) causes the other substance(s), the solute(s), to dissolve and enter into the solution. In an aqueous solution, water is the solvent, or dissolving medium. The symbol “(aq)” in a chemical equation indicates that the substance is dissolved in water. CaCO3 (s) + H2O (l) + CO2 (aq) Ca(HCO3)2 (aq) Some substances dissociate (split apart) in solution to form ions. Since such solutions can conduct an electric current, the dissolved substance is classified as an electrolyte. These substances consist of ions that are held together by the electrostatic forces of attraction between their opposite charges -- ionic bonding (see Chapter 8). H 2O NaCl (s) Na+(aq) + Cl–(aq) Nonelectrolytes do not dissociate in aqueous solution and thus their solutions do not conduct electricity. These substances consist of molecules where the atoms are covalently bonded (see Chapter 8). H 2O C12H22O11 (s) C12H22O11 (aq) Due to the strength of their ionic bonds, some ionic substances dissociate almost completely in aqueous solution. These substances, such as hydrobromic acid, are classified as strong electrolytes. HBr (aq) H+(aq) + Br–(aq) Some molecular substances, such as acetic acid, partially dissociate in aqueous solution and are classified as weak electrolytes. H+(aq) + CH3COO–(aq) CH3COOH (aq) The double arrows indicate that the reaction occurs in both directions. The longer arrow (pointing to the left) indicates that at chemical equilibrium, the concentration of the undissociated acetic acid is greater than that of its dissociated ions. NOTE: Please do not confuse concentration (mole per liter) with strength (degree of dissociation). Page 1 of 11 Water Solubility A substance’s solubility is usually expressed in terms of the grams of the solute that can dissolve in 100.00 cm3 of water. Sometimes is it more meaningful to discuss the solubility in terms of moles of solute per liter of solution. A substance’s solubility changes as the solution’s temperature varies. As a general rule, substances whose solubility is less than 0.01 mol/L at 25 oC are considered to be insoluble (a.k.a. slightly soluble). Some substances that are insoluble at 25 oC dissolve at slightly higher temperatures and are referred to as moderately soluble. The following solubility guidelines are taken from a variety of textbooks, but the primary source was pp. 134-135 of General Chemistry (6th Ed.), Whitten, Davis, and Peck, Saunders College Publishing. 1. 2. 3. 4. 5. 6. 7. 8. 9. The common inorganic acids (HCl, HNO3, etc.) are water soluble. Low-molecular-mass organic acids (HCOOH, CH3COOH, etc.) are also water soluble. All common compounds of the Group 1 metal ions (Li+, Na+, K+, Rb+, Cs+) and the ammonium ion, NH4+, are water soluble. The common nitrites, NO2–; nitrates, NO3–; acetates, CH3COO–; chlorates, ClO3–; and perchlorates, ClO4–, are water soluble. AgCH3COO, AgNO2 and KClO4 are moderately soluble. (a) The common chlorides, Cl–, are water soluble except for AgCl, Hg2Cl2, and PbCl2 which is soluble in hot water. (b) The common bromides, Br–, and iodides, I–, show approximately the same solubility behavior as chlorides, but there are some exceptions. As these halide ions (Cl–, Br–, I–) increase in size, the solubilities of their slightly soluble compounds decrease. HgBr2 is moderately soluble. (c) The common fluorides, F–, are water soluble except MgF2, CaF2, SrF2, BaF2, and PbF2. The common sulfates, SO42–, and sulfites, SO32–, are water soluble except PbSO4, BaSO4, HgSO4; CaSO4, SrSO4, and Ag2SO4 are moderately soluble. The common metal hydroxides, OH–, are insoluble in water except those of Group 1 metals and the heavier Group 2 metals: Ca(OH)2, Sr(OH)2, Ba(OH)2, and Ra(OH)2 are moderately soluble. The common carbonates, CO32–; chromates, CrO42-; oxalate, C2O42–; phosphates, PO43– ; and arsenates, AsO43–, are insoluble in water except those of Group 1 metals and NH4+. MgCO3 is moderately soluble. The common sulfides, S2–, are insoluble in water except for NH4+ and metal ions in Groups 1 and 2. Metal oxides, O2–, are generally insoluble. The following exceptions are soluble Li2O, Na2O, K2O, and BaO while CaO and SrO are moderately soluble. These exceptions dissolve to form hydroxides in an exothermic manner. Page 2 of 11 Ionic Equations During aqueous chemical reactions involving ionic substances, some ions are not involved in the actual reaction. These uninvolved ions are known as spectator ions. When writing the chemical equation to represent a reaction involving ions, we can choose to represent the reaction in one of three manners. The molecular equation shows all of the involved species and their phases: AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) The complete ionic equation shows all of the involved species with the strong electrolytes dissociated into aqueous ions. Solids, liquids, gases, and aqueous molecular species retain their molecular formulas: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) AgCl (s) + Na+(aq) ) + NO3–(aq) In the net ionic equation, the spectator ions are eliminated to show only the species that are actually involved in the chemical reaction: Ag+(aq) + Cl–(aq) AgCl (s) REACTION TYPES EXCHANGE (metathesis) reactions involve the cation from one solute exchanging anions with the cation from the other solute. Precipitation reactions occur when solute ions from two, or more, aqueous solutions combine to form a solid. Aqueous solutions of barium chloride and sodium sulfate react as follows: BaCl2 (aq) + Na2SO4 (aq) BaSO4 (s) + 2 NaCl (aq) The barium sulfate forms a solid precipitate that settles to the bottom of the solution. Acid-Base reactions: An acid dissociates in water to form H+(aq). A base dissociates in water to form OH–(aq). The neutralization reaction between the acid’s H+(aq) and the base’s OH–(aq) forms water. A salt is formed when the cation from the base combines with the anion from the acid. Soluble salts can be recovered by evaporating the water from the solution. Page 3 of 11 The reaction of nitric acid with potassium hydroxide (the base) produces water and potassium nitrate (the salt): HNO3 (aq) + KOH (aq) H2O (l) + KNO3 (aq) Some acid-base reactions produce salts that decompose to form water and a gas: 2 HNO3 (aq) + Na2CO3 (aq) 2 NaNO3 (aq) + H2O (l) + CO2 (g) Most acids and bases are weak… they are only partially dissociated in water. Strong acids and bases dissociate almost completely in water. Since there are so few strong acids and bases, it is easier to remember which common acids and base are strong rather than trying to learn rules to determine whether or not an acid or base is weak or strong. Common Strong Acids Hydrochloric Hydrobromic Hydroiodic Chloric Perchloric Nitric Sulfuric ** HCl HBr HI HClO3 HClO4 HNO3 H2SO4** Common Strong Bases Lithium hydroxide Sodium hydroxide Potassium hydroxide Rubidium hydroxide Cesium hydroxide Calcium hydroxide Strontium hydroxide Barium hydroxide LiOH NaOH KOH RbOH CsOH Ca(OH)2 Sr(OH)2 Ba(OH)2 Note: H2SO4’s first ionization is strong while its second ionization is weak H2SO4 (aq) H+(aq) + HSO4–(aq) H+(aq) + SO42–(aq) HSO4–(aq) OXIDATION-REDUCTION reactions involve the transfer of electrons between atoms. Oxidation involves the loss of electrons which makes the oxidized substance more positive, thus increasing its ability to combine with oxygen. Reduction involves the gain of electrons which makes the reduced substance more negative thus reducing its ability to combine with oxygen. The substance that is oxidized gives up electrons – allowing another substance to gain electrons (reducing it). Thus the oxidized substance is classified as the reducing agent. The substance that is reduced gains electrons – allowing another substance to lose electrons (oxidizing it). Thus the reduced substance is classified as the oxidizing agent. In a chemical reaction, oxidation cannot occur without an accompanying reduction. Page 4 of 11 Oxidation numbers (a.k.a. oxidation states) are a “bookkeeping” tool to help determine whether or not electrons have been transferred during a chemical reaction. Oxidation numbers are assigned on a per atom basis. The following general “rules” are adapted mainly from pp. 138-139 of General Chemistry (6th Ed.), Whitten, Davis, and Peck, Saunders College Publishing will help with the assignment of oxidation numbers. In order of precedence the rules are: 1. 2. 3. 4. 5. 6. 7. 8. 9. The atoms of free, uncombined elements have oxidation numbers of zero. This includes the diatomic elements: H2, N2, O2, F2, Cl2, Br2, and I2 as well as O3, P4, S8, and C60. The oxidation number of an element in a monatomic ion is equal to the charge on the ion. The sum of the oxidation numbers of all the atoms in a compound is zero. The sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion. Fluorine’s oxidation number is 1- in all of its compounds. Hydrogen’s oxidation number is 1+ in most of its compounds, except its oxidation number is 1- in metallic hydrides. Oxygen’s oxidation number is 2- in its compounds, except for: (a) peroxides, O22–, where each O has a 1-. Examples: H2O2, Na2O2, and CaO2. (b) superoxides, O2–, where each O has a 1/2–. Examples: KO2 and RbO2. (c) In OF2, O has a 2+. The element’s position on the periodic table often helps with the assignment of oxidation numbers: (a) Group 1 metals have an oxidation number of 1+ in all of their compounds. (b) Group 2 metals have an oxidation number of 2+ in all of their compounds. (c) Group 3 metals have an oxidation number of 3+ in all of their compounds with a few rare exceptions. (d) Group 14 can an oxidation number of 2+ or 4+, except that C and Si often have other charges. (e) Group 15 elements have an oxidation number of 3– in binary compounds with metals, hydrogen, or in ammonium. They can also have an oxidation number of 3+ or 5+ in compounds with more electronegative elements. (f) Group 16 (except O) elements have 2– in binary compounds with metals or hydrogen, or with ammonium. They can also have an oxidation number of 4+ or 6+ with O and the lighter halogens. (g) Group 17 elements have 1– in binary compounds with metals or hydrogen, or with ammonium. Cl, Br, and I can have an oxidation number of 1+, 3+, 5+, or 7+ with O or a lighter halogen. The oxidation number for the transition (Groups 3-12) and the inner-transition metals must be determined based on what they are combined with – or from the Roman numeral in their compound name. Exceptions: Ag forms 1+, Cd and Zn form 2+, Page 5 of 11 Example: Zn(s) + 2 HBr(aq) ZnBr2(aq) + H2(g) Zn(s) and H2(g) have oxidation numbers of zero H+(aq) has a 1+ oxidation number Br–(aq) is a spectator ion and has an oxidation number of 1–. Zn2+(aq) has an oxidation number of 2+. Zn(s) Zn2+(aq) + 2 e– 2 H+(aq) + 2 e– H2(g) is the oxidation, Zn becomes more + is the reduction, H+ becomes less + Zn(s) is the reducing agent and H+(aq) is the oxidizing agent REPLACEMENT The Activity Series (Table 4.5, p. 124) can be used to predict whether or not a metal will “replace” another in an aqueous compound. Metals listed higher on the activity series list are more easily oxidized and thus reduce the lower metal: Ni(s) + Cu2+(aq) Ni2+(aq) + Cu(s) A similar series can be created for the halogens: F would replace Cl–, Br–, or I– Cl would replace Br–, or I– Br would replace I– I would not replace F–, Cl–, or Br– Cl2 (aq) + 2 NaI (aq) 2 NaCl (aq) + I2 (aq) I2 (aq) + 2 NaF (aq) No Reaction These “activities” are directly related to the elements’ standard reduction potentials (Chapter 20: Electrochemistry). Since most of the chemical reaction they we study are in aqueous solution, we need to be able to determine the amount of reactants and products present in a given volume of solution. There are several methods of expressing the concentration of the solutes; we will mainly be using molarity (M). Molarity = moles of solute liters of solution mol = M x L L = Page 6 of 11 mol M Example Problem #1: What is the molarity of 27.5 g of potassium nitrate in 235 mL of solution? mol KNO3 = L solution M = 1 mol KNO3 101.10324 g KNO3 = 1.16 M KNO3 1 L 235 mL x 1000 mL 27.5 g KNO3 x Example Problem #2 What is molar concentration for the ions formed when 125 g Ca(NO3)2 is dissolved in 437 mL of solution? M Ca(NO3)2 = M Ca 2+ mol Ca(NO3) 2 = L solution 1 mol Ca(NO3)2 164.08788 g Ca(NO3) 2 = 1.74 M Ca(NO3)2 1 L 437 mL x 1000 mL 125 g Ca(NO3) 2 x 1 mol Ca 2+ = 1.74 M Ca(NO3)2 x = 1.74 M Ca2+ 1 mol Ca(NO3) 2 Note: Round-off error accounts for the fact that the NO3– concentration exceeds double that for Ca2+ In the laboratory, reagent solutions are typically prepared by dissolving solid, liquid, or gaseous substances with deionized water. Some solutions are prepared by diluting “stock” solutions with deionized water. NOTE: The addition of water during the dilution process has an insignificant impact on the number of moles of solute present in the solution. Example Problem #3 What mass of solid barium chloride must be dissolved to prepare 1250 mL of a 0.245 M 0.245 mol solution? Note: 0.245 M is equivalent to 1 L 0.245 M BaCl2 x 1250 mL x 1 L 208.2324 g BaCl2 x = 63.8 g BaCl2 1000 mL 1 mol BaCl 2 Page 7 of 11 Example Problem #4 What volume of concentrated hydrochloric acid (11.7 M) must be diluted with water to prepare 2250.0 mL of 0.347 M solution? Method 1: 2250.0 mL solution x 0.347 M HCl x 1 L solution 1 L HCl x = 0.0667 L HCl 1000 mL solution 11.7 M HCl We would probably convert the volume to 66.7 mL. Method 2: Since the moles of solute is unchanged by the dilution process, and moles = molarity x liters, we can use the relationship: M1V1 = M2V2. Since we are solving for a volume, V1, we do not necessarily need to change the 2230.0 mL to liters. M 2V 2 M1 (0.347 M HCl)(2250.0 mL solution) V1 = = 66.7 mL solution 11.7 M HCl M1V 1 = M 2V 2 V 1 = SOLUTION STOICHIOMETRY When we are performing stoichiometric calculations involving aqueous solutions, we must follow pretty much the same steps as we did in Chapter 3: 1. Write the balanced chemical equation for the reaction. 2. Convert the mass, or volume, of each reactant to moles. 3. Use the coefficient ratios from the balanced chemical equation to determine the moles of product(s) formed. 4. Convert the moles of products to grams to obtain the theoretical yield. 5. If an actual yield is given, determine the reaction’s percentage yield. Page 8 of 11 Example Problem #5: What is the percent yield for a reaction that produces 17.9 g of lead (II) sulfate when 125 mL of 0.500 M lead (II) nitrate is combined with 75.5 mL of 0.295 M aluminum sulfate? 3 Pb(NO3)2 (aq) + Al2(SO4)3 (aq) 3 PbSO4 (s) + 2 Al(NO3)3 (aq) 125 mL Pb(NO 3)2 x 1L 1000 mL 75.5 mL Al 2( SO 4)3 x x 0.500 M Pb(NO3)2 x 3 mol PbSO4 3 mol Pb(NO3)2 1 L x 303.2636 g PbSO 4 1 mol PbSO 4 3 mol PbSO 4 303.2636 g PbSO 4 x 0.295M Al 2(SO 4) 3 x x 1 mol Al 2(SO 4) 3 1 mol PbSO 4 1000 mL = 19.0 g PbSO 4 = 20.3 g PbSO 4 The Pb(NO3)2 is the limiting reactant and the theoretical yield for the lead (II) sulfate is 19.0 g. % yield= actual yield x 100 theoretical yield % yield = 17.9 g x 100 = 94.2 % 19.0 g TITRATIONS Titration is an analytical procedure where the amount of an unknown substance is determined by the addition of a standard solution (a solution with a known concentration) until the equivalence point is reached. The equivalence point is reached when stoichiometrically equal amounts of each reactant have been added to the solution. In most titrations, if a suitable indicator (a substance that changes color at a specific color) is used, the indicator will change color (the end-point) at approximately the equivalence point. There are several indicators to choose from for a particular titration. The choice of indicator is determined by the pH at which the equivalence point occurs (see Chapter 17 for additional details). Indicators such as phenolphthalein, a weak organic acid, is often used in acid-base titrations involving a strong base (such as NaOH) and a weak acid (such as HC2H3O2). Phenolphthalein is colorless in acidic solution and turns to pink/red in base (around pH = 8.2). Methyl red or methyl orange might be used for titrations involving strong acids, such as HCl (aq), and weak bases, such as NH3 (aq). The equivalence point for this titration would occur at a pH of about 4 … well below where phenolphthalein changes color. Indicators such as bromthymol blue or litmus change color at a pH of about 7, which would be appropriate for use during a strong acid-strong base titration. Page 9 of 11 Example Problem #6 What is the molarity of formic acid, HCOOH, a monoprotic acid with one ionizible H+, if 45.73 mL of 0.1025 M strontium hydroxide was required to titrate a 20.0 mL sample of the formic acid solution? 2 HCOOH (aq) + Sr(OH)2 (aq) 2 H2O (l) + Sr(COOH)2 (aq) Method 1: M HCOOH = mol HCOOH L HCOOH 45.73 mL Sr (OH ) 2 x = 1 L Sr (OH) 2 1000 mL Sr (OH) 2 20.0 mL HCOOH x x 0.1025 M Sr (OH ) 2 x 2mol HCOOH 1 mol Sr (OH ) 2 1 L HCOOH = 0.469 M HCOOH 1000 mL HCOOH Method 2: At the equivalence point, the mol H+(aq) = mol OH–(aq), thus: mol ionized H + ions mol ionized OH ions x Macid x Vacid = x M base x Vbasse mol acid molecules mol base molecules HCOOH donates one H+ per molecule, and Sr(OH)2 donates two OH– for each of its molecules Thus we have: 1 x M acid x Vacid = 2 x M base x Vbase M acid = M acid = 2 x Mbase x Vbase 1 x Vacid 1 L 1000 mL = 0.469 M HCOOH 1 L 1 x 20.0 mL x 1000 mL 2 x 0.1025 M x 45.73 mL x Page 10 of 11 Example Problem #7 What is the mass percent of HCOOH in the solution if the density is 1.03 g/cm3? Assume we have 1.0000 L of the solution. The mass of HCOOH in one liter of the solution would be found by multiplying the molarity by the volume to get moles and then multiplying by HCOOH’s molar mass. The mass of the solution is determined by multiplying its volume by its density. mass % = g HCOOH g solution 0.469 mol HCOOH x x 100 = 46.02538 g HCOOH 1 mol HCOOH 1.03 g 1000 mL x 1.0000 L x 1 mL 1 L Page 11 of 11 x 100 = 2.10 % HCOOH