Download as PDF, PPTX
More Related Content
Similar to Ch4 Reactions in Aqueous Solution (updated)
Ch4 Reactions in Aqueous Solution (updated)
- 1. Reactions in AqueousSolution Chapter 4 Dr. Sa’ib Khouri AUM- JORDAN Chemistry By Raymond Chang
- 2. 2 Solution is ahomogenous mixture of two or more substances. Solute is the substance present in a smaller amount. (In a solution may be more than one solute) Solvent is the substance present in a larger amount. Solution Solvent Solute Soft drink (l) Air (g) Soft Solder (s) H2O N2 Pb Sugar, CO2 O2, Ar, CH4 Sn Aqueous solutions: the solute is a liquid or a solid and the solvent is water General Properties of Aqueous Solutions A solution may be gaseous (air), solid (alloy), or liquid (seawater).
- 3. 3 Electrolyte is asubstance that, when dissolved in water, results in a solution that can conduct electricity. Nonelectrolyte does not conduct electricity when dissolved in water. nonelectrolyte weak electrolyte strong electrolyte Electrolytic Properties
- 4. 4 Examples of strongelectrolytes, weak electrolytes, and nonelectrolytes. Cations (+) and Anions (-) are responsible to conduct electricity in solution. Strong Electrolyte – 100% dissociation NaCl(s) Na + (aq) + Cl - (aq) H2O
- 5. 5 Hydration is theprocess in which an ion is surrounded by water molecules arranged in a specific manner. d+ d- H2O Hydration helps to stabilize ions in solution and prevents cations from combining with anions The Na+ and Cl- ions are separated from each other and undergo hydration
- 6. 6 A reversible reaction.The reaction can occur in both directions (chemical equilibrium)
- 7. 7 Precipitation Reactions Precipitate –insoluble solid that separates from solution Molecular equation Ionic equation Net ionic equation Pb2+ + 2NO3 - + 2K+ + 2I- PbI2 (s) + 2K+ + 2NO3 - Here K + and NO3 - are spectator ions Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq) Pb 2+ + 2I - PbI2 (s) e.g. Mixing of Pb(NO3)2 and KI solutions forms an insoluble yellow product, PbI2 through a double-displacement reaction:
- 8. 8 Precipitation of LeadIodide Pb 2+ + 2I - PbI2 (s)
- 9. 9 is the maximumamount of solute that will dissolve in a given quantity of solvent at a specific temperature. Solubility
- 10. 10 Examples of InsolubleCompounds CdS PbS Ni(OH)2 Al(OH)3 Practice Exercise Classify the following ionic compounds as soluble or insoluble: (a) CuS, (b) Ca(OH)2, (c) Zn(NO3)2, (d) Ag2SO4, (e) Na3PO4
- 11. 11 Writing Net IonicEquations 1. Write the balanced molecular equation. 2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3. Cancel the spectator ions on both sides of the ionic equation 4. Check that charges and number of atoms are balanced in the net ionic equation AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Ag+ + NO3 - + Na+ + Cl- AgCl (s) + Na+ + NO3 - Ag+ + Cl- AgCl (s) e.g. Write the net ionic equation for the reaction between silver nitrate and sodium chloride.
- 13. Write balance completemolecular equation, ionic equation, and net ionic equations for the reaction that occurs when Pb(NO3)2 and Na2S solutions are mixed. Identify the spectator ions in the reaction. e.g. Molecular equation Ionic equation Net ionic equation
- 14. General Properties ofAcids and Bases Acids • Acids have a sour taste; e.g. Vinegar owes its sourness to acetic acid, and lemons and other citrus fruits contain citric acid • Acids cause color changes in plant dyes, e.g. acids change the color of litmus from blue to red. • Acids react with certain metals to produce hydrogen gas 2HCl (aq) + Mg (s) → MgCl2 (aq) + H2 (g) Acid-Base Reactions
- 15. Acids react withcarbonates and bicarbonates to produce carbon dioxide gas 2HCl (aq) + CaCO3 (s) → CaCl2 (aq) + CO2 (g) + H2O (l) HCl (aq) + NaHCO3 (s) → NaCl (aq) + CO2 (g) + H2O (l) • Aqueous acid solutions conduct electricity Bases • Have a bitter taste • Feel slippery, e.g. soaps, which contain bases, exhibit this property. • Cause color changes in plant dyes • Aqueous base solutions conduct electricity
- 16. 16 Arrhenius acid isa substance that produces H+ (H3O+) in water Arrhenius base is a substance that produces OH- in water Bronsted Acids and Bases Arrhenius’s definitions of acids and bases apply only to aqueous solutions
- 17. 17 A Brønsted acidis a proton donor A Brønsted base is a proton acceptor Johannes Brønsted (Danish chemist) proposed broader definitions: Note: Brønsted’s definition do not require acids and bases to be in aqueous solution NH3 (g) + HCl (g) → NH4Cl (s) Here HCl (g) is a Brønsted acid NH3 (g) is a Brønsted base
- 18. 18 Hydrochloric acid isa Brønsted acid because it donates a proton in water: HCl (aq) → H + (aq) + Cl - (aq) H+ ion is a hydrogen atom that has lost its electron; that is, it is just a bare proton and exists in the hydrated form: The hydrated proton, H3O + , is called the hydronium ion
- 19. 19 Monoprotic acids: yieldsone hydrogen ion upon ionization HCl H+ + Cl- HNO3 H+ + NO3 - CH3COOH H+ + CH3COO- Strong acid Strong acid Weak acid Diprotic acids: gives up two H + ions, in two separate steps H2SO4 H+ + HSO4 - HSO4 - H+ + SO4 2- Strong acid Weak acid Triprotic acids: yields three H+ ions, in three separate steps H3PO4 H+ + H2PO4 - H2PO4 - H+ + HPO4 2- HPO4 2- H+ + PO4 3- Weak acid Weak acid Weak acid
- 20.
- 21.
- 22. 22 Acid-Base Neutralization acid +base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H2O H+ + Cl - + Na+ + OH- Na + + Cl - + H2O H + + OH - H2O A neutralization reaction is a reaction between an acid and a base. Generally, aqueous acid-base reactions produce water and a salt (ionic compound) Acid-base reactions generally go to completion. Here only if we had started the reaction with equal molar amounts of the acid and the base.
- 23. 23 Neutralization reaction involvinga weak electrolyte weak acid + base salt + water HCN (aq) + NaOH (aq) NaCN (aq) + H2O HCN + Na + + OH - Na + + CN - + H2O HCN (aq) + OH - (aq) CN - (aq) + H2O The following are also examples of acid-base neutralization reactions, represented by molecular equations: HF(aq) + KOH(aq) ⟶ KF(aq) + H2O H2SO4(aq) + 2NaOH(aq) ⟶ Na2SO4(aq) + 2H2O HNO3(aq) + NH3(aq) ⟶ NH4NO3(aq) HNO3(aq) + NH+ 4 (aq) + OH− (aq) ⟶ NH4NO3(aq) + H2O
- 24. 24 Acid-Base Reactions Leadingto Gas Formation acid + base salt + water + CO2 2HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + H2O +CO2 2H + + 2Cl - + 2Na + + CO3 2- 2Na + + 2Cl - + H2O + CO2 2H+ (aq) + CO3 2- (aq) H2O + CO2 Certain salts like carbonates (containing the CO3 2- ion), bicarbonates (containing the HCO3 1- ion), sulfites (containing the SO3 2- ion), and sulfides (containing the S2- ion) react with acids to form gaseous products. Similar reactions involving other mentioned salts are
- 25.
- 26. 26 Oxidation-Reduction, or Redox,Reactions (electron transfer reactions) 2Mg 2Mg2+ + 4e- O2 + 4e- 2O2- Oxidation half-reaction (lose e-) Reduction half-reaction (gain e-) 2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e- 2Mg + O2 2MgO Example (Two elements combine to form a compound) The formation of magnesium oxide (MgO) from magnesium and oxygen. 2Mg(s) + O2(g) ⟶2MgO(s)
- 27. 27 Zn(s) + CuSO4(aq) ZnSO4 (aq) + Cu (s) (Zn is oxidized)Zn Zn2+ + 2e- (Cu2+ is reduced)Cu2+ + 2e- Cu Zn is the reducing agent Cu2+ is the oxidizing agent e.g. Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s) Cu Cu2+ + 2e- Ag+ + 1e- Ag (Ag+ is reduced) Ag+ is the oxidizing agent e.g. When metallic zinc is added to a solution containing copper(II) sulfate (CuSO4),
- 28.
- 29. 2 9 Oxidation number (or oxidationstate) 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, H2, O2, P4 = 0 2. In monatomic ions (composed of only one atom), the oxidation number is equal to the charge on the ion. Li+ = +1; Fe3+ = +3; O2- = -2 3. The oxidation number of oxygen is usually –2. In H2O2 and O2 2- it is –1. 4.4 Signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely. Rules to assign oxidation numbers
- 30. 30 4. The oxidationnumber of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases (e.g. NaH), its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5. Group IA metals are +1, IIA metals are +2 HCO3 - O = - 2 H = +1 For C: 3(- 2) + 1 + x = -1 x=+4 C = +4 e.g. What are the oxidation numbers of all the elements in HCO3 - ? 7. Oxidation numbers do not have to be integers. e.g. Oxidation number of oxygen in the superoxide ion, O2 -, is –½.
- 31. 31 The Oxidation Numbersof Elements in their Compounds The more common oxidation numbers are in color
- 32. 32 NaIO3 Na = +1O = -2 3(-2) + 1 + x = 0 x=+5, I = +5 IF7 F = -1 7(-1) + x = 0 x=+7, I = +7 K2Cr2O7 O = -2 K = +1 7(-2) + 2(+1) + 2x = 0 x=+6, Cr = +6 e.g. What are the oxidation numbers of all the elements in each of these compounds? NaIO3 IF7 K2Cr2O7
- 33. 33 Types of RedoxReactions Combination Reactions Decomposition Reactions A combination reaction is a reaction in which two or more substances combine to form a single product. Decomposition reactions are the opposite of combination reactions, or the breakdown of a compound into two or more components.
- 34. 3 4 Combustion Reactions A combustionreaction is a reaction in which a substance reacts with oxygen, usually with the release of heat and light to produce a flame Example: Burning of propane (C3H8), a component of natural gas that is used for domestic heating and cooking: Here, we focus only on the oxidation number of O atoms, which changes from 0 to -2. Displacement Reactions In a displacement reaction, an ion (or atom) in a compound is replaced by an ion (or atom) of another element. 1. Hydrogen Displacement All alkali metals and some alkaline earth metals (Ca, Sr, and Ba), which are the most reactive of the metallic elements, will displace hydrogen from cold water C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(l) Example next slide
- 35. 35 Example for combustionrxn in the last slide
- 36. Zinc (Zn) andmagnesium (Mg) do not react with cold water but do react with hydrochloric acid, as follows: 2. Metal Displacement A metal in a compound can be displaced by another metal in the elemental state Example: zinc replacing copper ions and copper replacing silver ions Metal displacement reactions find many applications, the goal of which is to separate pure metals from their ores e.g. vanadium is obtained by treating vanadium(V) oxide with metallic calcium V2O5(s) + 5Ca(l) ⟶2V(l) + 5CaO(s)
- 37. 37 The metals arearranged according to their ability to displace hydrogen from an acid or water The activity series for metals According to this series, any metal above hydrogen will displace it from water or from an acid, but metals below hydrogen will not react with either water or an acid Any metal listed in the series will react with any metal (in a compound) below it. For example, Zn is above Cu, so zinc metal will displace copper ions from copper sulfate.
- 38. 38 3. Halogen Displacement.Another activity series summarizes the halogens’ behavior in halogen displacement reactions: F2 > Cl2 > Br2 > I2 The power of these elements as oxidizing agents decreases as we move down Group 7A from fluorine to iodine, so molecular fluorine can replace chloride, bromide, and iodide ions in solution Molecular bromine, in turn, can displace iodide ion in solution:
- 39. 39 Disproportionation Reaction an elementin one oxidation state is simultaneously oxidized and reduced. Here the oxidation number of oxygen in the reactant (-1) both increases to zero in O2 and decreases to -2 in H2O This reaction describes the formation of household bleaching agents One reactant in a disproportionation reaction always contains an element that can have at least three oxidation states. The element itself is in an intermediate.
- 40.
- 41.
- 42. 42 Concentration of solutions Theconcentration of a solution is the amount of solute present in a given quantity of solvent or solution. Molarity (M) = moles of solute (n) liters of solution(V) e.g. What mass of KI is required to make 500 mL of a 2.80 M (mol/L) KI solution? volume of KI solution (L) moles KI grams KI 500. mL = 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x Or directly: mass = V (L) X M (mol/L) X MM (g/mol)
- 43. 43 Preparing a solutionof known molarity (a) known amount of a solid solute is transferred into the volumetric flask; then water is added through a funnel. (b) The solid is slowly dissolved by gently swirling the flask. (c) After the solid has completely dissolved, more water is added to bring the level of solution to the mark. Knowing the volume of the solution and the amount of solute dissolved in it, we can calculate the molarity of the prepared solution.
- 44.
- 45. 45 Dilution of solutions:is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f)= Mi Vi Mf Vf=
- 46. 46 e.g. How wouldyou prepare 100.0 mL of 0.200 M HCl from a stock solution of 4.00 M HCl? MiVi = MfVf Mi = 4.00 M Mf = 0.200 M Vf = 0.1000 L Vi = ? L Vi = MfVf Mi = 0.200 M x 0.1000 L 4.00 M = 0.00500 L = 5.00 mL Dilute 5.00 mL of acid with water to a total volume of 100.0 mL.
- 47. 47 Gravimetric Analysis an analyticaltechnique based on the measurement of mass. (a) A solution containing a known amount of NaCl in a beaker. (b) The precipitation of AgCl upon the addition of AgNO3 solution from a measuring cylinder. (c) AgCl precipitate is filtered through a vacuum filtration, which allows the liquid (but not the precipitate) to pass through. Basic steps for gravimetric analysis e.g. Solution Stoichiometry
- 48. Examples A Gravimetric Calculation A10.00 mL solution containing Cl- was treated with excess AgNO3 to precipitate 0.4368 g of AgCl. What was the molarity of Cl- in the unknown? Solution The formula mass of AgCl is 143.321. Precipitate weighing 0.4368 g contains Because 1 mol of AgCl contains 1 mol of Cl- , there must have been 3.0483 x 10-3 mol of Cl- in the unknown.
- 49.
- 50. Introduction to Titrations Ina titration, increments of reagent solution (the titrant and it is usually delivered from a buret) are added to analyte until their reaction is complete. Requirements -Reaction must be quick -Reaction must proceed according to a defined equation -Availability of indicator -Reactions must be complete or have a large equilibrium constant From the quantity of titrant required, the quantity of analyte that must have been present can be calculated Common titrations rely on acid-base, precipitation, oxidation-reduction, or complex formation reactions.
- 51. 51 Acid-Base Titrations In titration,a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete. If we know the volumes of the standard and unknown solutions, we can calculate the concentration of the unknown one. The acid often chosen for standardazation of bases is potassium hydrogen phthalate (KHP), for which the molecular formula is KHC8H4O4 (molar mass = 204.2 g) The reaction between KHP and sodium hydroxide is KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O
- 52. Indicator: substance thatchanges its color at/or near the equivalence point. Slowly add base to unknown acid UNTIL the indicator changes color (End point) Equivalence point: the point at which the acid has completely reacted with or been neutralized by the base. End point: is marked by a sudden change in a physical property of the solution, e.g. color change of a reactant or an external indicator added.
- 53. 53 e.g. What volumeof a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution? The chemical equation volume acid moles acid moles base volume base H2SO4 + 2NaOH 2H2O + Na2SO4 4.50 mol H2SO4 1000 mL soln x 2 mol NaOH 1 mol H2SO4 x 1000 ml soln 1.420 mol NaOH x25.00 mL = 158 mL M acid rxn coef. M base
- 54. Other method At theequivalence point moles of NaOH = 2 x moles of H2SO4 (M x V)NaOH 2= (M x V)H2SO4 1.420 x VNaOH = 2 x 4.50 x 25.00 VNaOH 158= mL
- 55. 55 Example: The acidicsubstance in vinegar is acetic acid, HC2H3O2. When 4.00 g of a certain vinegar sample was titrated with 0.200 M NaOH, 25.56 mL of the NaOH solution were required to reach the equivalence point. What is the percent composition by mass of HC2H3O2 in the vinegar? A) 7.7% B) 30.7% C) 76.8% D) 3.8% E) 15.6%