1 / 45

Chapter 4. Aqueous reactions and Solution Chemistry.

Chapter 4. Aqueous reactions and Solution Chemistry. δ +. Chemical reactions in water are important, as the chemistry of life occurs in water, and chemical reactions in the environment occur mainly in water. δ -. δ +. partial charges on a water molecule.

edena
Télécharger la présentation

Chapter 4. Aqueous reactions and Solution Chemistry.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 4. Aqueous reactions and Solution Chemistry. δ+ Chemical reactions in water are important, as the chemistry of life occurs in water, and chemical reactions in the environment occur mainly in water. δ - δ+ partial charges on a water molecule

  2. 4.1. General Properties of Aqueous Solutions. Electrolytic Properties: NaCl dissolved in water produces a solution that conducts electricity, whereas table sugar (sucrose, C12H22O11) does not. This can be demonstrated with an electric light-bulb that lights up only with ionic solutions.

  3. Demonstration of electrical conductivity of an electrolyte: Circuit passes through electrodes in solution in beaker bulb glows in presence of electrolyte but not of pure water Electrolyte Solution In beaker

  4. How circuit on previous slide works:

  5. Distribution of charge on the water molecule: The water molecule is polar, which in the case of water means that the hydrogens have partial positive charges, while the oxygen has a partial negative charge: δ+ The partial positive charges (δ+) are attracted to negative charges, while the partial negative charges (δ-) are attracted to positive charges H δ+ O H δ-

  6. The structure of water around anions and cations:Each anion is surrounded by partial positive charges from waters, and each cation is surrounded by partial negative charges from waters: δ+ δ+ δ- anion δ+ δ- cation δ+ δ+ δ+ δ- δ- δ+ δ- δ+ δ+ δ-

  7. - - + - + + H2O - + + + - - + - Ionic Compounds When ionic compounds dissolve in water, they tend to dissociate completely: water molecules Na+ cation Cl- anion NaCl (s) → Na+ (aq) + Cl- (aq) solid cation anion

  8. Molecular Compounds Most molecular compounds do not dissociate when they dissolve in water: Methanol molecule not ionized water + H2O Methanol (molecular compound) methanol dissolved in water

  9. Molecular Compounds that ionize in water: Some molecular compounds dissociate (ionize) in water (mainly acids) Strong acids, such as hydrochloric acid, dissociate completely (single arrow shows equilibrium lies well to the right): HCl (aq) → H+ (aq) + Cl- (aq) Weak acids, such as acetic acid, dissociate only partially (double arrow shows equilibrium does not lie completely to the right): HC2H3O2 (aq)  H+ (aq) + C2H3O2- (aq) acetic acid acetate ion

  10. Strong and Weak electrolytes: Strong electrolytes are completely ionized in water. Weak electrolyes only partially ionize in water. A prime example of a weak electrolyte is acetic acid. The fact that it only ionizes to a small extent is indicated by a double arrow: HC2H3O2(aq)  H+(aq) + C2H3O2-(aq) A single arrow is used to indicate complete ionization of strong electrolytes, e.g. NaOH: NaOH (aq) → Na+ (aq) + OH- (aq)

  11. 4.2 Precipitation reactions: A precipitate is an insoluble solid formed by a reaction in solution. mix clear solutions of AgNO3 and NaI to produce a precipitate of AgI(s) yellow precipitate of silver iodide (AgI) Bottom of Test-tube

  12. Soluble Ionic Compounds: Compds containing:Exceptions NO3- none C2H3O2- none Cl- Ag+, Hg22+, Pb2+ Br- “ I- “ SO42- Sr2+, Ba2+, Hg22+, Pb2+ (note: you will be given the information on this sheet in exams/tests)

  13. Insoluble Ionic Compounds: Compds containingexceptions S2- alkali metal cations, NH4+, Ca2+, Sr2+, Ba2+ OH- “ CO32- alkali metal cations, NH4+ PO43- “ (note: you will be given the information on this sheet in exams/tests)

  14. To use the Tables on the two preceding slides, consider the following examples: Which of the following compounds are insoluble in water? a) BaSO4 b) K2SO4 c) PbCl2 d) (NH4)2CO3 a) If we look at the tables, we see that all sulfates are soluble except Ba2+ (plus others) insoluble b) Tables show all sulfates are soluble, and alkali metal ions are not exceptions, so is soluble c) All chlorides are soluble, except Pb2+ (plus others) so is insoluble d) All carbonates are insoluble, except NH4+ (ammonium ion) plus others, so is soluble

  15. Ionic Equations Molecular equation: (do not show as ions) Pb(NO3)2 + 2 KI → PbI2(s) + 2 KNO3 Complete Ionic equation (show all ions): Pb2+ + 2 NO3-+ 2 K+ + 2 I- → PbI2(s) + 2 K+ + 2 NO3- Net ionic equation (omit ions that do not change during the reaction, NO3- and K+): Pb2+(aq) + 2 I-(aq) → PbI2(s)

  16. Note: Ca(NO3)2 (aq) + NaC2H3O2 (aq) → ? Spectator ions: ALL ! If all salts are soluble, no precipitation reaction will take place

  17. Example: What is the net ionic reaction that corresponds to the molecular reaction below? Ca(NO3)2 (aq) + Na2CO3 (aq) → CaCO3(s) + 2 NaNO3(aq) Write out the complete ionic equation and then remove spectator ions (ions that do not change): Ca2+ + 2 NO3- + 2 Na+ + CO32- → CaCO3(s) + 2 Na+ + 2 NO3- Spectator ions: 2 Na+, 2 NO3- Net ionic equation: Ca2+ (aq) + CO32- (aq) → CaCO3 (s)

  18. 4.3 Acid-Base Reactions. Acids are substances that ionize in aqueous solutions to form protons, i.e. H+(aq) ions. HCl, HNO3 are monoprotic acids. i.e. they ionize to give only one proton per acid molecule: HNO3(aq)  H+(aq) + NO3- (aq) H2SO4 is a diprotic acid, and gives two protons. H2SO4(aq)  2 H+(aq) + SO42-(aq) (Note that for H2SO4 only the first ionization is complete.)

  19. Bases: Bases are substances that accept protons. A base is a proton acceptor. Common bases are OH- compounds of group 1 cations (NaOH, KOH) or heavier Group 2 (Ba(OH)2). These produce the hydroxide (OH-(aq)) ion in solution. NaOH (aq)  Na+ (aq) + OH-(aq) bases will react with a proton (this is the net ionic reaction below): H+ (aq) + OH- (aq)  H2O (l) acidbase water

  20. Ammonia – a weak base Many bases do not contain OH-, such as NH3 (ammonia). NH3(aq) + H+(aq)  NH4+(aq) + H2O or NH3(aq) + H2O NH4+(aq) + OH-(aq) ammoniaammonium The latter equilibrium lies well to the left, so NH3 is a weak base. A solution of NH3 in water contains only a tiny fraction as NH4+ ions (< 1%). bigger arrow shows equilibrium lies to the left

  21. The difference between ammonia and ammonium: To remember the difference between ammonia and ammonium, remember that ammonium is a cation like sodium or potassium. ammonia ammonium NH3(neutral NH4+ (a cation) molecule)

  22. Strong and Weak Acids and Bases. Acids and bases that are STRONG ELECTROLYTES. i.e. completely ionize in aqueous solution, are strong acids and bases. Those that are WEAK ELECTROLYTES i.e. only partly ionize, are weak acids and bases. For example, HF is a weak acid, since it is only partly ionized. However, it is very reactive since the F-(aq) ion will attack glass!

  23. Common Strong Acids and Bases (Table 4.2) Acids: HCl, HBr, HI, HClO3, HClO4, HNO3, H2SO4. Bases: Group 1. LiOH, NaOH, KOH, RbOH, CsOH Heavy group 2: Ca(OH)2, Sr(OH)2, Ba(OH)2.

  24. Weak acids and bases: If not listed as strong, it is probably weak. Some examples: Weak acids: HF, H3PO4, H2SO3, HC2H3O2, (hydrofluoric, phosphoric, sulfurous, acetic acids) Weak bases: NH3(ammonia)

  25. Neutralization reactions: (p 134) Acids = sour taste (e.g. vinegar), Bases = bitter (e.g. soap). HCl(aq) + NaOH(aq) → NaCl (aq) + H2O acidbase salt water Net ionic reaction: H+ (aq) + OH- (aq)  H2O The Na+ and Cl- are spectator ions acidbase water

  26. Balancing acid-base reactions Note: to balance these equations, just make the OH- and H+ balance out 2 HCl(aq) + Ca(OH)2(aq) → CaCl2 (aq) + 2 H2O 3 HCl(aq) + Al(OH)3(aq) → AlCl3 (aq) + 3 H2O Two protons needed for two hydroxides

  27. 4.5. Concentration of Solutions Molarity. This is used to express the concentrations of solutions: Molarity = moles solute_______ Volume of solution in liters Units = mol/liter (moles per liter)

  28. A 1.00 molar (written 1.00 M) solution has one mol of substance dissolved in every liter of solution: fill the flask to the 1 liter mark with water stopper 1 liter flask 1 mole 1.00 M (empty) CuSO4 (159.5 g)CuSO4 Dissolve the 1 mole of CuSO4 in the flask  line on neck of flask shows 1 liter mark

  29. Examples: What strength solution do we obtain if we dissolve: a) 0.25 mol CuSO4 in a 250 ml flask: Conv.: 1 liter = 1000 ml 250 ml = 250 ml x 1liter = 0.25 liter 1000 ml Molarity = ____ moles of solute_____ volume of solution in liters = _______0.25 moles_______ 0.25 liters = 1.00 M (or moles/liter)

  30. Examples: What molarity solution do we obtain if we dissolve: b) 4.82 g CuSO4 in a 200 ml flask: Note: 200 ml = 0.200 liter. Molecular mass CuSO4 = 159.5 g/mol moles CuSO4 = 4.82 g x 1mole = 0.0302 moles 159.5 g molarity = 0.0320 moles = 0.151 M 0.200 liter

  31. Expressing the concentration of an Electrolyte: A 1.00 M solution of NaOH is also 1.00 M in Na+ ions and OH- ions. NaOH (aq)  Na+ (aq) + OH-(aq) 1 molar 1 molar 1 molar A 1.00 M solution of H2SO4 is 1.00 M in SO42-, but 2.00 M in H+. H2SO4 (aq)  2 H+ (aq) + SO42- (aq) 1 molar 2 molar 1 molar

  32. Just pay attention to the coefficents in the balanced equation A 1.00 M solution of Na3PO4 is 1.00 M in PO43-, but 3.00 M in Na+. What about a 0.1 M solution? Na3PO4 (aq)  3 Na+ (aq) + PO43- (aq) 1 3 1 1 M3M 1 M or 0.1 M0.3M 0.1 M coefficients in balanced equation You notice again the power of the balanced equation. You can multiply all the coefficients by 1.0 M or 0.1 M.

  33. Interconverting Molarity, Moles, and Volume: If we know any two of the above, we can calculate the third quantity. e.g. calculate no. of moles HNO3 in 2.0 L of 0.200 M HNO3. Molarity = moles/liters 0.200 M = x moles/2.0 L x moles = 0.200 mol x 2.0 liter 1 liter = 0.4 moles

  34. Sample exercise: How many grams of Na2SO4 are required to make 0.35 L of 0.500 M Na2SO4? Moles = 0.35 L x 0.500 moles/liter = 0.175 moles. O.175 moles Na2SO4 (MM = 142.0 g/mol) Grams = 0.175 moles x 142.0 g/mol = 24.9 g

  35. Dilution (p. 148). When we dilute something, we take a volume of solution and place it in a larger container and make it more dilute by adding water: make up to mark with water transfer solution to larger flask 0.5 M CuSO4 solution 1 M CuSO4 solution 1 liter flask 2 liter flask 2 liters of solution

  36. Dilution (p. 148). Molarity = moles liter So moles = molarity x liters In diluting a solution, the number of moles of dissolved substance stays the same. Moles before dilution = moles after dilution, so: before dilution after dilution M(conc) x V(conc) = M(dil) x V(dil)

  37. Exercise 4.14: How many milliliters of 3.0 M H2SO4 are needed to make 450 ml of 0.10 M H2SO4? 3.0 M x V(conc) = 0.10 M x 450 ml V(conc) = 0.10 M x 450 ml 3.0 M = 15.0 ml Note: make sure units of volume are the same on both sides: liters and liters, or ml and ml. M(conc) XV(conc) =M(dil) xV(dil)

  38. Problem: What volume of 2.50 M Pb(NO3)2 contains 0.0500 mol of Pb? M = moles/volume So volume = moles/M = 0.0500 mol/2.50 M = 0.0500 mol x 1 liter 2.50 mol = 0.02 L or 20 ml

  39. 4.6. Solution stoichiometry. Here we need to be able to convert from grams to moles, and then use M = moles/volume to convert from moles to molarity. Thus for example: 79.8 g x 1 mole 159.5 g 0.5 mol x 1 liter 1 liter 0.5 mol 79.8 g of CuSO4 1 liter of 0.5 M CuSO4

  40. Problem: How many grams of Ca(OH)2 are needed to neutralize 25.0 ml of 0.100 M HCl? Balanced equation: 2HCl + Ca(OH)2 → CaCl2 + 2 H2O 2 moles 1 mole 2 moles 2 moles so will need 1 mole calcium hydroxide to neutralize 2 moles HCl.

  41. Moles HCl = 0.100 M x 0.025 L = 0.0025 moles 2HCl + Ca(OH)2 → CaCl2 + 2 H2O 2 moles 1 mole 2 moles 2 moles 0.0025 moles Factor = moles we have/moles in equation = 0.0025/2 = 0.00125 So moles of Ca(OH)2 = 0.00125 x 1 = 0.00125 mol

  42. Convert moles to grams: Mol. Mass Ca(OH)2 = 74.1 g/mol Therefore grams = 74.1 g x 0.00125 mol 1 mole = 0.0926 g

  43. Solution Stoichiometry and Chemical Analysis What volume of a 0.30 M HCl solution is needed to completely react 3.5 g of Ca(OH)2 ? 2 HCl (aq) + Ca(OH)2 (aq) → 2 H2O (aq) + CaCl2 (aq) 21 The question is how many moles of Ca(OH)2 do we have, and how many moles of HCl will we need to react with it? No. of moles Ca(OH)2 = 3.5 g x 1 mole = 0.035 mol 100.0 g

  44. From the balanced equation we see that we need 2 mols of HCl for each mol Ca(OH)2: 2 HCl (aq) + Ca(OH)2 (aq) → 2 H2O (aq) + CaCl2 (aq) 21 0.070 mol 0.035 mol Problem then is what volume of 0.30 M HCl will provide 0.070 mol? Molarity = moles/volume 0.30 M = 0.070 volume Volume = 0.070 mol x 1 liter = 0.23 L 0.30 mol or 230 mL

  45. What is the concentration of a solution that is made by adding 0.3L of water to 15mL of a 0.65M solution? Mconc= 0.65M Vconc = 15mL = 0.015L Vdil = 0.3L + 15mL = 0.3L + 0.015L = 0.315L rearrange

More Related