Transcript CHAPTER 4
CHAPTER 4 Reactions in Aqueous Solutions Aqueous Solutions aqueous solutions -solute dissolved in water nonelectrolytes - aqueous solutions do not conduct electricity C2H5OH - ethanol CH3 CH2 OH Aqueous Solutions C6H12O6 - glucose (blood sugar) C OH H C O H C H OH C H H C HO C HO OH Aqueous Solutions C12H22O11 - sucrose (table sugar) HO H C H C C H HO HO H 2 H O H C OH C H C H OH C C 2 O O H H C C HO OH C H Properties of Aqueous Solution Molecular compounds in water (e.g., CH3OH): no ions are formed. If there are no ions in solution, there is nothing to transport electric charge. Properties of Aqueous Solution Ionic Compounds in Water Ions dissociate in water. In solution, each ion is surrounded by water molecules. Transport of ions through solution causes flow of current. Properties of Aqueous Solution Properties of Aqueous Solution Strong electrolytes: completely dissociate in solution. For example: HCl(aq) H+(aq) + Cl-(aq) Weak electrolytes: produce a small concentration of ions when they dissolve. These ions exist in equilibrium with the unionized substance. For example: HC2H3O2(aq) H+(aq) + C2H3O2-(aq) Properties of Aqueous Solution strong electrolytes - extremely good conductors of electricity HCl, HNO3, etc. strong soluble acids NaOH, KOH, etc. strong soluble bases NaCl, KBr, etc. soluble ionic salts ionize in water essentially 100% Properties of Aqueous Solution weak electrolytes CH3COOH, (COOH)2 weak acids NH3, Fe(OH)3 weak bases some soluble covalent salts ionize in water much less than 100% Strong and Weak Acids acids generate H+ in aqueous solutions strong acids ionize 100% in water Strong and Weak Acids acids generate H+ in aqueous solutions strong acids ionize 100% in water 100 % HCl g H aq Cl aq - 100% HNO3 H 2O H3Oaq + NO3- aq or HNO3 H H 2O aq 3aq + NO Strong and Weak Acids entire list of strong water soluble acids HCl - hydrochloric acid HBr - hydrobromic acid HI - hydroiodic acid HNO3 - nitric acid H2SO4 - sulfuric acid HClO3 - chloric acid HClO4 - perchloric acid Strong and Weak Acids weak acids ionize less than 100% in water 10% or less! some common weak acids HF - hydrofluoric acid CH3COOH - acetic acid (vinegar) H2CO3 - carbonic acid (soda water) H2SO3 - sulfurous acid HNO2 - nitrous acid H3PO4 - phosphoric acid Strong and Weak Acids weak acids ionize as reversible or equilibrium reactions CH3COOH acetic acid why they ionize less than 100% O C H3C 7% OH CH 3COOH CH 3COO aq + H aq Strong Soluble Bases bases produce OH- ions in solution strong soluble bases ionize 100% in water KOH K (aq) + OH (aq) + - Ba(OH) 2 Ba (aq) + 2 OH (aq) 2+ - Strong Soluble Bases entire list of strong soluble bases LiOH - lithium hydroxide NaOH - sodium hydroxide KOH - potassium hydroxide RbOH - rubidium hydroxide CsOH - cesium hydroxide Ca(OH)2 - calcium hydroxide Sr (OH)2 - strontium hydroxide Ba (OH)2 - barium hydroxide Insoluble Bases ionic but insoluble in water not very basic Cu(OH)2 - copper (II) hydroxide Fe(OH)2 - iron (II) hydroxide Fe(OH)3 - iron (III) hydroxide Zn(OH)2 - zinc (II) hydroxide Mg(OH)2 - magnesium hydroxide Weak Bases covalent compounds that ionize slightly in water NH3 - ammonia is most common NH3g + H2O NH4aq + OH(aq) Solubility Rules strong acids completely water soluble strong bases completely water soluble soluble ionic salts Figure 4.3, page 80 of textbook Acids, Bases, and Salts Dissociation = pre-formed ions in solid move apart in solution. Ionization = neutral substance forms ions in solution. Acid = substances that ionizes to form H+ in solution (e.g. HCl, HNO3, CH3CO2H, lemon, lime, vitamin C). Bases = substances that react with the H+ ions formed by acids (e.g. NH3, Drano™, Milk of Magnesia™). Acids, Bases, and Salts Reactions in Aqueous Solutions molecular equations all reactants & products in molecular or ionic form Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) total ionic equation ions as they exist in solution Zn(s) + Cu 2 aq + SO 24aq Zn 2 aq + SO 24aq + Cu (s) Reactions in Aqueous Solutions net ionic equation shows ions that participate in reaction and removes spectator ions spectator ions do not participate in the reaction Zn (s) + Cu 2 aq Zn 2 aq + Cu (s) Displacement Reactions Metathesis reactions involve swapping ions in solution: AX + BY AY + BX. Metathesis reactions will lead to a change in solution if one of three things occurs: an insoluble solid is formed (precipitate), weak or nonelectrolytes are formed, or an insoluble gas is formed. Displacement Reactions Displacement Reactions Precipitation Reactions form an insoluble compound crystals molecular equation Ca(NO 3 ) 2 (aq) + K 2 CO 3 ( aq) 2 KNO 3 ( aq ) + CaCO 3 (s) Precipitation Reactions form an insoluble compound crystals molecular equation Ca(NO 3 ) 2 (aq) + K 2 CO 3 ( aq) 2 KNO 3 ( aq ) + CaCO total ionic reaction 3 (s) Precipitation Reactions total ionic reaction Ca 2aq 2 NO 3- aq 2 K aq CO 32-aq 2K aq 2 NO 3 aq CaCO 3 s Precipitation Reactions total ionic reaction Ca 2aq 2 NO 3- aq 2 K aq CO 32-aq 2K net ionic reaction aq 2 NO 3 aq CaCO 3 s Precipitation Reactions net ionic reaction Ca 2 aq + CO 23aq CaCO3(s) Acids, Bases, and Salts Neutralization occurs when a solution of an acid and a base are mixed: HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Notice we form a salt (NaCl) and water. Salt = ionic compound whose cation comes from a base and anion from an acid. Neutralization between acid and metal hydroxide produces water and a salt. Acid-Base Reactions acid + base salt + water recall the acids and bases from before molecular equation Acid-Base Reactions acid + base salt + water recall the acids and bases from before molecular equation HBr(aq) + KOH (aq) KBr(aq) + H 2 O ( ) total ionic equation Acid-Base Reactions total ionic equation aq H aq + Br +K aq aq + OH net ionic equation K aq aq + Br + H2O() Acid-Base Reactions net ionic equation H aq + OH aq H 2 O ( ) - Acid-Base Reactions molecular equation Ca(OH)2 (aq) + 2 HNO3(aq) Ca(NO3 ) 2 (aq) + 2 H 2O( ) Acid-Base Reactions molecular equation Ca(OH)2 (aq) + 2 HNO3(aq) Ca(NO3 ) 2 (aq) + 2 H 2O( ) total ionic equation Acid-Base Reactions total ionic equation Ca 2aq + 2 OH - aq + 2 H aq + 2 NO 3- aq Ca 2aq + 2 NO 3- aq + 2 H 2 O ( ) net ionic equation Acid-Base Reactions net ionic equation - 2 OH aq + 2 H aq 2 H 2O () or better - OH aq + H aq H 2O () Solution Composition Solution = solute dissolved in solvent. Solute: present in smallest amount. Water as solvent = aqueous solutions. Change concentration by using different amounts of solute and solvent. Molarity: Moles of solute per liter of solution. If we know: molarity and liters of solution, we can calculate moles (and mass) of solute. Solution Composition Molarity: Moles of solute per liter of solution. Concentration of Solutions solution - one substance dissolved in another, commonly one is a liquid concentration - amount of solute dissolved in a solvent • “sweet tea” • mixed drinks units of concentration m a ss o f so lu te % b y m a ss o f so lu te = 100% m a ss o f so lu tio n m a ss o f so lu tio n = m a ss o f s o lu te + m a ss o f s o lv e n t Molarity (Molar Concentration) molarity = mol solute/L of solution = M Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. Molarity (Molar Concentration) molarity = mol solute/L of solution = M Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. ? mol H 2 SO 4 12.5 g H 2 SO 4 1 mol H 2 SO 4 L sol' n 1.75 L sol' n 98.1 g H 2 SO 4 0.0728 mol H 2 SO 4 0.0728 M H 2 SO 4 L Molarity (Molar Concentration) Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . Molarity (Molar Concentration) Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . 0.800 mol Ca(NO 3 ) 2 ? g Ca(NO 3 ) 2 3.50 L L 164 g Ca(NO 3 ) 2 459 g Ca(NO 3 ) 2 1 mol Ca(NO 3 ) 2 Dilution of Solutions take a concentrated solution and add water to it • make tea “less sweet” number of moles of solute remains constant M1V1 = M2V2 works because # of moles is constant If 10 mL of 12 M HCl is added to enough water to give 100 mL of solution, what is concentration of solution? Dilution of Solutions take a concentrated solution and add water to it • make tea “less sweet” number of moles of solute remains constant M1V1 = M2V2 works because # of moles is constant If 10 mL of 12 M HCl is added to enough water to give 100 mL of solution, what is concentration of solution? (10 mL)(12 M) = M2(100 mL) M2 = 1.2 M HCl Dilution of Solutions What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? Dilution of Solutions What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? M 2 x V2 M1 x V1 M 2 x V2 V1 M1 2.50 L x 2.40 M V1 0.333 L or 333 mL 18.0 M Stoichiometry with Solutions There are two different types of units: •laboratory units (macroscopic units: measure in lab); •chemical units (microscopic units: relate to moles). Always convert the laboratory units into chemical units first. •Grams are converted to moles using molar mass. •Volume or molarity are converted into moles using M = mol/L. Use the stoichiometric coefficients to move between reactants and product. Stoichiometry with Solutions Stoichiometry with Solutions What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4? Na 2SO 4 + BaCl 2 BaSO ? L BaCl 2 4 + 2 NaCl 1 mol Na 2SO 4 4.32 gNa 2SO 4 142 g Na 2SO 4 1 mol BaCl 2 1 L BaCl 2 1 mol Na 2SO 4 0.500 mol BaCl 0.0608 L 2 Stoichiometry with Solutions What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate? What mass of aluminum hydroxide precipitates? Al(NO3)3 + 3 NaOH Al(OH)3 + 3 NaNO3 Stoichiometry with Solutions What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate? What mass of aluminum hydroxide precipitates? Al(NO3)3 + 3 NaOH Al(OH)3 + 3 NaNO3 ? mL NaOH = 50.0 mL Al(NO 3 ) 3 sol' n 0.200 mol Al(NO 3 ) 3 sol' n 3 mol NaOH 1 L Al(NO 3 ) 3 sol' n 1 mol Al(NO 3 ) 3 1 L NaOH 150 mL NaOH sol' n 0.200 mol NaOH Stoichiometry in Solution What mass of Al(OH)3 precipitates in (a)? Stoichiometry in Solution What mass of Al(OH)3 precipitates in (a)? ? g Al(OH) 3 50.0 mL Al(NO 3 ) 3 sol'n 0.200 mol Al(NO 3 ) 3 1 mol Al(OH) 3 78.0 g Al(OH) 3 1 L Al(NO 3 ) 3 sol'n 1 mol Al(NO 3 ) 3 1mol Al(OH) 3 0.780 g Al(OH) 3 Titrations Titrations Suppose we know the molarity of a NaOH solution and we want to find the molarity of an HCl solution. We know: molarity of NaOH, volume of HCl. What do we want? Molarity of HCl. What do we do? Take a known volume of the HCl solution, measure the mL of NaOH required to react completely with the HCl. Titrations What do we get? Volume of NaOH. We know molarity of the NaOH, we can calculate moles of NaOH. Next step? We also know HCl + NaOH NaCl + H2O. Therefore, we know moles of HCl. Can we finish? Knowing mol(HCl) and volume of HCl (20.0 mL above), we can calculate the molarity. Titrations What is the molarity of a KOH solution if 38.7 mL of KOH solution is required to react with 43.2 mL of 0.223 M HCl? KOH + HCl KCl + H 2 O 43.2 mL 0.223 M HCl = 9.63 mmol HCl 1 mol KOH 9.63 mmol HCl 9.63 mmol KOH 1 mol HCl 9.63 mmol KOH 0.249 M KOH 38.7 mL KOH Titrations What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? Titrations What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? Ba(OH) 2 + 2 HCl BaCl 2 + 2 H 2 O 1L 0.103 mol (44. 1 mL HCl)( )( HCl) = 4.54x10 -3 mol HCl 1000mL 1L 1 mol Ba(OH) 2 -3 4.54 x10 mol HCl 2 mol HCl 2.27x10 -3 mol Ba(OH) 2 Titrations What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? Ba(OH) 2 + 2 HCl BaCl 2 + 2 H 2 O 2.27x10 -3 mol Ba(OH) 2 0.0593 M Ba(OH) 2 0.0383 mL Ba(OH) 2 Oxidation Number Rules The oxidation number of any free, uncombined element is zero. The oxidation number of an element in a simple (monatomic) ion is the charge on the ion. In the formula for any compound, the sum of the oxidation numbers of all elements in the compound is zero. In a polyatomic ion, the sum of the oxidation numbers of the constituent elements is equal to the charge on the ion. Oxidation Number Rules Hydrogen, H, in combined form, has the oxidation number +1, usually. Oxygen, O, in combined form, has the oxidation number -2, usually. Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 K2Sn(OH)6 H3PO4 Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 +1 +5 -2 K2Sn(OH)6 H3PO4 Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 +1 +5 -2 K2Sn(OH)6 +1 +4 -2 +1 H3PO4 Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 +1 +5 -2 K2Sn(OH)6 H3PO4 +1 +1 +5 -2 +4 -2 +1 Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32- HCO3- Cr2O72- Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32+4 -2 HCO3- Cr2O72- Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32- HCO3- +4 -2 +1+4 -2 Cr2O72- Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32- HCO3- +4 -2 +1+4 -2 Cr2O72+6 -2 Redox Reactions oxidation - loss of electrons • oxidation number increases reduction - gain of electrons • oxidation number decreases or reduces! oxidizing agents - chemical species that gain electrons and oxidize some other species • electron thieves • they are reduced Redox Reactions reducing agents - chemical species that lose electrons and reduce some other species • victims of electron thieves • they are oxidized How do we balance redox reactions? Balancing Redox Reactions Balancing redox reactions. • Half-Reaction Method Half Reaction Method Balancing redox reactions by half reaction method rules: Write the unbalanced reaction Break into 2 half reactions • 1 for oxidation • 1 for reduction Mass balance each half reaction by adding appropriate stoichiometric coefficents • in acidic solutions we can add H+ or H2O • in basic solutions we can add OH- or H2O Half Reaction Method Charge balance the half reactions by adding appropriate numbers of electrons Multiply each half reaction by a number to make the number of electrons added equal to the number of electrons consumed Add the two half reactions Eliminate any common terms Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. Sn Br2 Sn Br starting reaction 2+ 4+ - Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. Sn Br2 Sn Br starting reaction 2+ 4+ - Sn Sn 2e oxidation half reaction 2+ 4+ - Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. Sn Br2 Sn Br starting reaction 2+ Sn 2+ 4+ - Sn 2e oxidation half reaction 4+ - Br2 2e 2 Br reduction half reaction - - Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. Sn 2+ Br2 Sn Sn 2+ Sn 4+ Br 4+ - starting reaction 2e oxidation half reaction - Br2 2e 2 Br reduction half reaction - Sn 2+ Br2 Sn - 4+ 2 Br balanced reaction - Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. Cr2 O 7 2 Fe 2 + Cr 3+ Fe 3+ starting reaction Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. Cr2 O 7 Fe 2+ 2 Fe Fe 3+ 2+ Cr - 3+ Fe 3+ starting reaction + 1e oxidation half reaction Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 2 Cr2O7 Fe Cr Fe start rxn Fe2+ Fe3+ +1e- ox. 2 2+ 3+ 3+ Cr2O7 14 H 6 e 2 Cr 7 H2O + - 3+ red. Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 2 Cr2O7 Fe Cr Fe startrxn 6 Fe Fe +1e ox. 2+ 2 2+ 3+ 3+ - 3+ Cr2O7 14H 6 e 2 Cr 7 H2O + - 3+ red. Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 2 Cr2O7 Fe2+ Cr 3+ Fe3+ start rxn 6 Fe Fe + 1e ox. 2+ 2 3+ - Cr2O7 14 H 6 e 2 Cr 7 H 2O + 2 - 3+ red. 6 Fe Cr2O7 14 H 6 Fe 2 Cr 7 H 2O 2+ + 3+ 3+ Half Reaction Method Example: In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH)4-, to chromate ions, CrO42-. Write and balance the net ionic equation for this reaction. Half Reaction Method Cr(OH) 4 H 2 O 2 CrO 4 2 (basic solution) 2 2 Cr(OH) 4 4OH - CrO 4 4H 2 O + 3e - 3 H 2 O 2 2e - 2OH 2 2 Cr(OH) 4 8OH 3 H 2 O 2 2 CrO 4 8H 2 O + 6OH - 2 2 Cr(OH) 4 2OH - 3 H 2 O 2 2 CrO 4 8H 2 O - Half Reaction Method Example: When chlorine is bubbled into basic solution, it forms chlorate ions and chloride ions. Write and balance the net ionic equation. Half Reaction Method Cl 2 ClO 3 Cl (basic solution) 1 Cl 2 12 OH 2 ClO 3 6 H 2 O + 10e - 5 Cl 2 2 e 2 Cl - - 6 Cl 2 12 OH 2 ClO 3 6 H 2 O + 10 Cl - 3 Cl 2 6 OH ClO 3 3 H 2 O + 5 Cl - Synthesis Question Nylon is made by the reaction of hexamethylene diammine CH2 CH2 CH2 NH2 H2N CH2 CH2 CH2 with adipic acid HO C O H2 C C H2 H2 C C H2 O C OH Synthesis Question * in a 1 to 1 mole ratio. The structure of nylon is: O H2 H2 H2 H2 H2 C C C C C C N C C C C N C H H H2 H2 H2 H2 O where the value of n is typically 450,000. On a daily basis, a DuPont factory makes 1.5 million pounds of nylon. How many pounds of hexamethylene diamine and adipic acid must they have available in the plant each day? n * Group Activity Manganese dioxide, potassium hydroxide and oxygen react in the following fashion: 4 MnO 2 + 4 KOH + 3 O 2 4 KMnO 4 2 H 2 O A mixture of 272.9 g of MnO2, 26.6 L of 0.250 M KOH, and 41.92 g of O2 is allowed to react as shown above. After the reaction is finished, 347.6 g of K2MnO4 is separated from the reaction mixture. What is the per cent yield of this reaction?