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Types of Reactions and Solution Stoichiometry

Types of Reactions and Solution Stoichiometry

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Types of Reactions and Solution Stoichiometry

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  1. Types of Reactions and Solution Stoichiometry

  2. Water, the Common Solvent • Aqueous solutions – those solutions where water is the dissolving agent or solvent • Means something is dissolved in water • Water is a polar molecule • The electrons shared in the covalent H-O bond are not shared equally • The O is slightly negative and the H’s are slightly positive • This is represented by the lowercase delta (δ) which means partially followed by the correct sign • The polar nature of water causes hydration in ionic compounds • The cations are attracted to the slightly negative oxygen and the anions are attracted to the slightly positive hydrogen • The ionic compounds separate into anions and cations • NH4NO3H2O NH4+ (aq) + NO3-(aq)

  3. Water, the Common Solvent • Solubility – the ability of a substance to dissolve • Depends on the relative attraction of the ions for each other and for a water molecule • Polar molecules dissolve in water, especially those with an O-H bond such as alcohols • Water does not dissolve nonpolar molecules such as animal fat • “Like dissolves like” is a good rule of thumb for determining solubility • Polar substances dissolve ionic and polar compounds • Nonpolar substances dissolve nonpolar substances

  4. The Nature of Aqueous Solutions: Strong and Weak Electrolytes • Solute the part of the solution that is dissolved • Can change the concentration of the solution by changing the relative amounts of the solute and solvent • Electrical conductivity – ability to conduct electricity • Strong electrolyte – a substance, when dissolved in water, produces a solution that is a good conductor of electricity • Weak electrolyte - a substance, when dissolved in water, produces a solution that is a poor conductor of electricity • Nonelectrolytes - a substance, when dissolved in water, produces a solution that does not conduct electricity

  5. The Nature of Aqueous Solutions: Strong and Weak Electrolytes • Conductive properties of solutions were correctly identified by Svante Arrhenius in the early 1880s • Believed that they were based on the presence of ions • The idea was ridiculed at first, but later gained acceptance during the late 1890s • Strong electrolytes – ionize completely • Three classes of strong electrolytes: soluble salts, strong acids, and strong bases • Soluble salts are ionic compounds that ionize in water, such as NaCl or BaCl2 • Arrhenius first associated the sour taste of citrus fruits with acids • Found that certain acids acted like strong electrolytes when dissolved in water • Because of the ionization of acids in water

  6. The Nature of Aqueous Solutions: Strong and Weak Electrolytes • Arrhenius defined an acid as a substance that produces H+ ions (protons) when dissolved in water • Some acids completely ionize in water – strong acids • Include H2SO4, HNO3, HClO4, and HCl • Arrhenius defined a base as a substance that produces (OH)- ions when dissolved in water • Strong bases completely ionize • Have a bitter taste and a slippery texture • Include NaOH and KOH

  7. The Nature of Aqueous Solutions: Strong and Weak Electrolytes • Weak electrolytes do not ionize completely in water • Two classes: Weak acids and weak bases • Weak acids include HC2H3O2 • Weak bases include NH3 • Nonelectrolytes do not produce ions when dissolved in water • Include molecular compounds such as ethanol (C2H5OH) and sucrose (C12H22O11)

  8. The Composition of Solutions • Chemical reactions often take place when solutions are mixed • To perform stoichiometric calculations we must know two things • The nature of the reaction, which depends on the forms the chemicals take when dissolved • The amount of chemicals present in the solutions, usually expressed as concentrations

  9. The Composition of Solutions • Concentration is often expressed as molarity (M) • Moles of solute per volume of solution in liters • moles/L • To calculate the molarity of a solution • Convert the mass of solute into moles by dividing by the molar mass from the periodic table • Then divide the number of moles by the volume of solution • If the volume is not in liters we have to convert it to liters

  10. The Composition of Solutions • Example: Calculate the molarity of a solution prepared by dissolving 11.5g of solid NaOH in enough water to make 1.50L of solution 11.5g NaOH 1mol NaOH = .192M solution 40.00g NaOH 1.50L solution

  11. The Composition of Solutions • A solution’s concentration is given before the solute dissolves • Ionic compounds separate into their constituent ions • So a 1.0L solution of 1.0M NaCl doesn’t have 1.0mol NaCl. It has 1.0 mol Na+ ions and 1.0mol Cl- ions

  12. The Composition of Solutions • To calculate moles of solute, multiply volume of solution, in liters, by the molarity Example: Calculate the number of moles of Cl- ions in 1.75L of 1.0*10-3M ZnCl2 ZnCl2H2O Zn2+(aq) + 2Cl-(aq) 1.75L solution 1.0*10-3mol ZnCl2 2mol Cl- 3.5*10-3mol Cl- L solution 1mol ZnCl2

  13. The Composition of Solutions • To find volume of solution, divide by the molar mass of the solute to get moles of solute. Then divide by the molarity • Example: Typical blood serum is about .14M NaCl. What volume of blood contains 1.0mg NaCl? 1.0mgNaCl 1gNaCl 1mol NaCl L blood 1.2*10-4L blood 1000mg NaCl 58.45g NaCl .14mol NaCl

  14. The Composition of Solutions • Standard solution – a solution whose concentration is accurately known • To find the mass of solute in a standard solution • First, multiply the volume by the molarity to get moles of solute. Then convert to grams by multiplying by the molar mass

  15. The Composition of Solutions • Example: To analyze the alcohol content of a certain wine, a chemist needs 1.00L of an aqueous .200M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be weighed out to make this solution? 1.00L solution .200mol K2Cr2O7 294.20gK2Cr2O7 58.8g K2Cr2O7 1L solution 1molK2Cr2O7

  16. The Composition of Solutions • Dilution – when water is added to a solution in concentrated form to achieve a desired, lower molarity • To determine the volume of water needed to be added to a solution to achieve the desired molarity • First determine the moles of solute in the solution by multiplying the volume by desired molarity. Second, divide the moles of solute by the original molarity.

  17. The Composition of Solutions • Example: What volume of 16M sulfuric acid must be used to prepare 1.5L of a .10M H2SO4 solution? 1.5L solution .10mol H2SO4 1L solution 9.4*10-3L solution 1 L solution 16molH2SO4

  18. The Composition of Solutions • You can also use the equation M1V1 = M2V2 to solve solution problems. M1 is molarity of the first solution. V1 is the volume of the first solution. M2 is the molarity of the second solution and V2 is the volume of the second solution.

  19. Types of Chemical Reaction • Three types of solution reactions • Precipitation reactions • Acid-base reactions • Oxidation-reduction reactions

  20. Precipitation Reactions • Precipitation reactions – a reaction in which two solutions are mixed that produce an insoluble solid that settles out of the solution called the precipitate • In precipitation reactions, the ions of the reactants separate and recombine to form new ionic compounds. One of the products is insoluble and precipitates out

  21. Precipitation Reactions • Rules for the Solubility of Salts in Water • 1) Most nitrate (NO3-) salts are soluble. • 2) Most salts containing the alkali metal ions (Li+, Na+, K+, Cs+, and Rb+) and the ammonium ion (NH4+) are soluble. • 3) Most chloride, bromide, and iodide salts are soluble. Notable exceptions are salts containing the ions Ag+, Pb2+, and Hg22+. • 4) Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, Hg2SO4, and CaSO4. • 5) Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH and KOH. The compounds Ba(OH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble. • 6) Most sulfide (S2-), carbonate (CO32-), chromate (CrO42-), and phosphate (PO42-) are only slightly soluble.

  22. Describing Reactions in Solution • Three types of equations are used to describe reactions in solutions • Molecular equations give the overall reaction stoichiometry. K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq) • Complete ionic equations represents the actual forms of the products and reactants. All substances that are strong electrolytes are represented as ions. 2K+(aq) + CrO42-(aq) + Ba2+(aq) + 2NO3-(aq) BaCrO4(s) + 2K+(aq) + 2NO3-(aq)

  23. Describing Reactions in Solution • Net ionic equations include only those solution components directly involved in the reaction. Spectator ions are not included. • Ions that remain in solution and do not change their charge (under go reduction or oxidation) are removed when writing a net ionic equation. Ba2+(aq) + CrO42-(aq) BaCrO4(s)

  24. Stoichiometry of Precipitation Reactions • To do stoichiometry problems: • Write the balanced molecular equation. • Identify the species present in the combined solution and determine what reaction occurs. • Write the balanced net ionic equation for the reaction. • Calculate the moles of reactants. • Determine what reactant is limiting. • Calculate the moles of product or products, as required. • Convert to grams or other units, as required.

  25. Stoichiometry of Precipitation Reactions • Example: Calculate the mass of solid NaCl that must be added to 1.50L of a .100M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl. NaCl(s) + AgNO3(aq) NaNO3(aq) + AgCl(s)

  26. Stoichiometry of Precipitation Reactions Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) Na+(aq) + NO3-(aq) + AgCl(s) Cl-(aq) + Ag+(aq) AgCl(s) 1.50L solution .100mol Ag+ 1mol NaCl 58.45g NaCl 8.77g NaCl 1L solution 1mol Ag+ 1mol NaCl

  27. Acid-Base Reactions • Almost all substances can be classified as an acid or a base even those that do not contain H+ or OH- ions • To classify these substances Johannes Brønsted and Thomas Lowry published a new definition of acids and bases. • Brønsted-Lowry acids are proton donors • Brønsted-Lowry bases are proton acceptors

  28. Acid-Base Reactions • To determine if something is an acid or a base: • React the substance with a strong acid or strong base • Write the balanced molecular equation for the reaction. • Convert to an ionic equation and remove spectator ions. Remember the rules for soluble salts. • A weak acid will donate a H+ to the OH- ion of the strong base to form water • A weak base will receive a H+ from the strong acid to form water • Neutralization reactions – when just enough base is added to react with just enough acid in a solution to completely neutralize the reaction.

  29. Acid-Base Reactions • To perform acid-base reaction calculations: • 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 the moles of reactants. For reactions in solution, use the volumes of the original solution and their molarities • Determine the limiting reagent where appropriate • Calculate the moles of the required reactant or product • Convert to grams or volume of solution as required

  30. Acid-Base Reactions • Example: What volume of .100M HCl solution is needed to neutralize 25.0mL of .350M NaOH? HCl + NaOH HOH + NaCl H+ + Cl- + Na+ + OH- H2O + Na+ + Cl- H+ + OH- H2O 25.0mL NaOH 1L NaOH .350mol OH- 1mol H+ 1L solution 8.75*10-2L solution 1000mL NaOH 1L NaOH 1 mol OH- .100M H+

  31. Acid-Base Reactions • Example: In a certain experiment, 28.0mL of .250M HNO3 and 53.0mL of .320M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH- ions in excess after the reaction goes to completion? HNO3 +KOH H2O + KNO3 H+ + NO3- + K+ + OH- H2O + NO3- + K+ H+ + OH- H2O

  32. Acid-Base Reactions 28.0mL HNO3 1LHNO3 .250mol H+ 7.00*10-3mol H+ 1000mLHNO3 1L HNO3 53.0mL KOH 1L KOH .320mol OH- 1.70*10-2mol OH- 1000mL KOH 1L KOH H+ is the limiting reagent because there is less H+ produced thanOH- 1.70*10-2- 7.00*10-3 =1.00*10-2mol OH- in excess Total volume = sum of original volumes, so 28.0mL + 53.0mL = 81.0mL = .081L Molarity of excess OH- = mol OH- in excess / total volume of solution 1.00*10-2mol OH- / .081L solution = .123M OH-

  33. Acid-Base Reactions • Volumetric analysis – a technique for determining the amount of a certain substance by doing a titration • Titration – a technique in which one solution is used to analyze another • Involves the delivery from a buret of a measured volume of a known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). The titrant contains a substance that reacts in a known manner with the analyte • The point where enough titrant has been added to completely react with the analyte is known as the equivalence point or stoichiometric point • This point is marked by the presence of an indicator that changes color at or near the balance point • The point where the indicator actually changes color is known as the endpoint • Should be the same as the equivalence point

  34. Acid-Base Reactions • Three requirements for a successful titration • The exact reaction between titrant and analyte must be known and rapid • The stoichiometric endpoint must be marked accurately • The volume of titrant required to reach the stoichiometric point must be known accurately • Commonly used with strong acids and strong bases and is known as acid-base titrations • The indicator is usually phenolphthalein • Colorless in acidic solution and pink in basic solution

  35. Acid-Base Reactions • Neutralization Titration Example: A student carries out an experiment to standardize (determine the exact concentration of) a sodium hydroxide solution. To do this, the student weighs out a 1.3009g sample of potassium hydrogen phthalate (KHC8H4O4, often abbreviated KHP). KHP (molar mass 204.22g/mol) has one acidic hydrogen. The student dissolves the KHP in distilled water, adds phenolphthalein as an indicator, and titrates the resulting solution with the sodium hydroxide to the phenolphthalein endpoint. The difference between the final and initial buret readings indicate that 41.20mL of the sodium hydroxide solution is required to react exactly with the 1.3009g KHP. Calculate the concentration of the sodium hydroxide solution. KHC8H4O4(aq)+ NaOH(aq) H2O(l) + NaKC8H4O4(aq)

  36. Acid-Base Reactions HC8H4O4-(aq)+ OH-(aq) H2O(l) + C8H4O42-(aq) 1.3009g KHC8H4O4 1 molKHC8H4O4 1mol NaOH .1546 M NaOH 204.22g KHC8H4O4 1 mol KHC8H4O4 4.120*10-2L

  37. Acid-Base Reactions • Neutralization Analysis Example: An environmental chemist analyzed the effluent (waste material) from an industrial process known to produce the compound carbon tetrachloride (CCl4) and benzoic acid (HC7H5O2), weak acid that has one acidic hydrogen atom per molecule. A sample of this effluent weighing .3518g was shaken with water, and the resulting aqueous solution required 10.59mL of .1546M NaOH for neutralization. Calculate the percent of HC7H5O2 in the original sample.

  38. Acid-Base Reactions • HC7H5O2 (aq) + NaOH (aq) H2O(l) +NaC7H5O2 (aq) .01059L NaOH .1546mol NaOH 1mol HC7H5O2 122.12g HC7H5O2 1L solution 1mol NaOH 1mol HC7H5O2 .1999gHC7H5O2 .1999gHC7H5O2 / .3518geffluent * 100 = 56.82%

  39. Oxidation-Reduction Reactions • Oxidation-reduction or redox reactions – reactions in which one or more electrons are transferred • Reactions in which ions form are redox reactions • More commonly, we think of redox reactions as reactions where the oxidation state changes • Oxidation states or oxidation numbers – a concept that provides a way to keep track of electrons in redox reactions, especially those involving covalent substances • Based on the locations of shared electrons • Oxidation numbers are usually integers, but fractions are possible • Example: Fe3O4 • Each iron has an oxidation state of 8/3 and each oxygen has one of -2

  40. Oxidation-Reduction Reactions • Rules for Assigning Oxidation States • 1. The oxidation state of an atom in an element is 0. • Examples: the substances Na(s), O2(g), O3(g), and Hg(l) is 0. • 2. The oxidation state of a monatomic ion is the same as its charge. • Examples: the Na+ ion is +1 and of the Cl- ion is -1. • 3. In its compounds, flourine is always assigned an oxidation state of -1. • 4. Oxygen is usually assigned an oxidation state of -2 in its covalent compounds • Examples: CO, CO2, SO2, and SO3. • Exceptions to this rule include peroxides (compounds containing the O22- group), where each oxygen is assigned an oxidation state of -1 • Examples hydrogen peroxide (H2O2), and OF2 in which oxygen is assigned a +2 state. • 5. In its covlaent compounds with nonmetals, hydrogen is assigned an oxidation state of +1. • Examples: HCl, NH3, H2O, and CH4 • 6. The sum of the oxidation states must equal the charge. For neutral compounds, they must equal zero. For polyatomic ions, they must equal the charge • Examples: In H2O, the oxidations numbers equal 0. In CO32-, the oxidation numbers equal -2 (Carbon is 4 and each oxygen is -2)

  41. Oxidation-Reduction Reactions • Redox reactions are characterized by the transfer of electrons • Ion Formation: • 2Na(s) + Cl2(g) 2NaCl(s) • Reactions involving an element, such as combustion • CH4(g) + 2O2(g) CO2(g) + 2H2O • Oxidation – an increase in oxidation state or a loss of electrons • Reduction – a decrease ion oxidation state or a gain of electrons • The part of the reaction that is oxidized is also known as the reducing agent or electron donor • The part of the reaction that is reduced is known as the oxidizing agent or electron acceptor • Carbon is oxidized because it has lost electrons • Oxygen is reduced because it gained electrons • CH4 is the reducing agent • O2 is the oxidizing agent

  42. Balancing Oxidation-Reduction Reactions • To balance redox reactions, you must balance the charge or the electrons • split the reaction into two half-reactions, one involving the oxidation and one involving the reduction • Ce4+(aq) + Sn2+(aq) Ce3+(aq) + Sn4+(aq) • Reduction: Ce4+(aq) Ce3+(aq) • Oxidation: Sn2+(aq) Sn4+(aq)

  43. Balancing Oxidation-Reduction Reactions • Balance all of the elements, except hydrogen and oxygen • Balance using H2O • Balance hydrogen using H+ • Balance the charge using electrons (e-) • If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half reactions • Add the half-reactions and cancel identical species • Check that charge and elements are still balanced

  44. Balancing Oxidation-Reduction Reactions • Reduction: Ce4+(aq) Ce3+(aq) • 1e- + Ce4+(aq) Ce3+(aq) • Oxidation: Sn2+(aq) Sn4+(aq) • Sn2+(aq) Sn4+(aq) + 2e- • 2(1e- + Ce4+(aq) Ce3+(aq)) • 2e- + 2Ce4+(aq) 2Ce3+(aq) • 2e- +2Ce4+(aq) + Sn2+(aq) 2Ce3+(aq) + Sn4+(aq) + 2e- • 2Ce4+(aq) + Sn2+(aq) 2Ce3+(aq) + Sn4+(aq)

  45. Balancing Oxidation-Reduction Reactions • MnO4-(aq) + Fe2+(aq) Acid Fe3+(aq) + Mn2+(aq) • MnO4-(aq) Acid Mn2+(aq) • Fe2+(aq) Acid Fe3+(aq) • MnO4-(aq) Acid Mn2+(aq) + 4H2O • 8H+(aq) + MnO4-(aq) Acid Mn2+(aq) + 4H2O • 8H+(aq) + MnO4-(aq) Acid Mn2+(aq) + 4H2O has a charge of 7+ on the left and 2+ on the right, so we add five electrons to the left • 5e- +8H+(aq) + MnO4-(aq) Acid Mn2+(aq) + 4H2O

  46. Balancing Oxidation-Reduction Reactions • Fe2+(aq) Acid Fe3+(aq) • Fe2+(aq) Acid Fe3+(aq) + 1e- • Multiply by 5 to balance electrons • 5(Fe2+(aq) Acid Fe3+(aq) + 1e-) • 5Fe2+(aq) Acid 5Fe3+(aq) + 5e- • 5Fe2+(aq) + 5e- + 8H+(aq) + MnO4-(aq) Acid Mn2+(aq) + 4H2O + 5Fe3+(aq) + 5e- • 5Fe2+(aq) + 8H+(aq) + MnO4-(aq) Acid Mn2+(aq) + 4H2O + 5Fe3+(aq)

  47. Balancing Oxidation-Reduction Reactions • For basic solutions, balance as if H+ ions were present • Then add OH- ions to form water with H+ ions • Ag(s) + CN-(aq) + O2Basic Ag(CN)2-(aq) • Ag(s) + CN-(aq) Basic Ag(CN)2-(aq) • Ag(s) + 2CN-(aq) Basic Ag(CN)2-(aq) • Ag(s) + 2CN-(aq) Basic Ag(CN)2-(aq) + 1e- • O2Basic • O2Basic 2H2O

  48. Balancing Oxidation-Reduction Reactions • O2 + 4H+Basic 2H2O • 4e- + O2 + 4H+Basic 2H2O • 4(Ag(s) + 2CN-(aq) Basic Ag(CN)2-(aq) + 1e-) • 4Ag(s) + 8CN-(aq) Basic 4Ag(CN)2-(aq) + 4e- • 4e- + O2 + 4H+ + 4Ag(s) + 8CN-(aq) Basic 4Ag(CN)2-(aq) + 4e- +2H2O • O2 + 4H+ + 4Ag(s) + 8CN-(aq) Basic 4Ag(CN)2-(aq) +2H2O • 4OH- + O2 + 4H+ + 4Ag(s) + 8CN-(aq) Basic 4Ag(CN)2-(aq) + 2H2O + 4OH- • O2 + 4H2O + 4Ag(s) + 8CN-(aq) Basic 4Ag(CN)2-(aq) + 2H2O + 4OH- • O2 + 2H2O + 4Ag(s) + 8CN-(aq) Basic 4Ag(CN)2-(aq) + 4OH-