## Presentation on theme: "1 Chem 106, Prof. J.T. Spencer CHE 106: General Chemistry u CHAPTER FOUR Copyright © James T. Spencer 1995 - 1999 All Rights Reserved."— Presentation transcript:

2 Chem 106, Prof. J.T. Spencer Aqueous Reactions Aqueous Reactions and and Solution Chemistry CHAPTER 4 Chapt. 4.1

3 Chem 106, Prof. J.T. Spencer Aqueous Solution Stoichiometry Chapter 4 l Focus on reactions in water - the most common and “universal” solvent (aqueous solutions (aq) ) with unique properties (mp, bp, etc...) l Essential for life chemistry - biochemistry (i.e., cellular metabolism) and inorganic chemistry (i.e., rusting, precipitation). l Acids, bases and salts Chapt. 4.1

4 Chem 106, Prof. J.T. Spencer Solution Composition; Terms l Solution l Solution - homogeneous mixture of two or more substances l Solvent l Solvent - substance of greater amount in the homogeneous mixture (solution) l Solute l Solute - compounds “dissolved” in the solvent l Concentration l Concentration - the amount of solute dissolved in a solvent. Expressed in molarity (M) Chapt. 4.1 Molarity (M) = moles of solute volume of solution (in liters)

5 Chem 106, Prof. J.T. Spencer Molarity solution not L of solvent l Molarity - moles of solute per liter of solution. (not L of solvent!) l A 1.0 M solution contains 1.00 moles of solute in every liter of solution. l Calculate the molarity of a solution made by dissolving 0.0575 mol NH 4 Cl in 400. mL of water. Chapt. 4.1 Molarity (M) = moles of solute volume of solution (in liters) Molarity (M) = 0.0575 mol NH 4 Cl = 0.144 M NH 4 Cl 0.400 L of H 2 O Unit of Molarity is moles per liter

6 Chem 106, Prof. J.T. Spencer Molarity Calculate the molarity of a solution made by dissolving 9.93 g of Na 2 SO 4 in enough water to form 650. mL.[MW = Na 2 SO 4 = 142] Chapt. 4.1 Molarity (M) = moles of solute volume of solution (in liters) Moles of Na 2 SO 4 = 9.93 g (1 mol Na 2 SO 4 ) = 0.0699 mol 142 g M = 0.0699 mol Na 2 SO 4 = 0.108 M 0.650 L

7 Chem 106, Prof. J.T. Spencer Molarity l (A) Calculate the number of moles of HNO 3 in 2.0 L of a 0.200 M HNO 3 solution. l (B) What volume of a 0.30 M HNO 3 solution is required to supply 2.0 mol HNO 3 ? Chapt. 4.1 (A) Know: Molarity of solution Need: Moles of Volume of solution Solute (B) Know:Molarity of solutionNeed: Volume Moles of solute Molarity (M) = moles of solute volume of solution (in liters)

8 Chem 106, Prof. J.T. Spencer Molarity l (A) Calculate the number of moles of HNO 3 in 2.0 L of a 0.200 M HNO 3 solution. l (B) What volume of a 0.30 M HNO 3 solution is required to supply 2.0 mol HNO 3 ? Chapt. 4.1 (A) Moles of HNO 3 = 2.0 L soln (0.200 mol HNO 3 ) 1 L soln Moles of HNO 3 = 0.40 mol HNO 3 (B) L of HNO 3 Soln. = 2.0 mol HNO 3 (1 L soln) (0.30 mol HNO 3 ) L of HNO 3 Soln. = 6.7 L

9 Chem 106, Prof. J.T. Spencer Molarity Pure acetic acid is a liquid with a density of 1.049 g/mL at 25° C. Calculate the molarity of a solution of acetic acid prepared by dissolving 10.00 mL of acetic acid at 25° C in enough water to make 100.0 mL of solution. [Acetic acid = C 2 H 4 O 2, MW = 60.1 amu] Chapt. 4.1 (A) Know: Need: Density = 1.049g.mL Molarity of Vol. pure AA used final solution Vol. final solution Molarity (M) = moles of solute volume of solution (in liters)

10 Chem 106, Prof. J.T. Spencer Molarity Pure acetic acid is a liquid with a density of 1.049 g/mL at 25° C. Calculate the molarity of a solution of acetic acid prepared by dissolving 10.00 mL of acetic acid at 25° C in enough water to make 100.0 mL of solution. [Acetic acid = C 2 H 4 O 2, MW = 60.1 amu] Chapt. 4.1 Moles of AA = 10 mL AA (1.049 g AA) (1 mol AA) 1 mL AA 60.1 g AA = 0.175 mol AA M = moles AA = 0.175 mol AA = 1.75 M AA soln. V soln 0.100 L

11 Chem 106, Prof. J.T. Spencer Molarity and Dilution Volume of solution (L) = moles solute/molarity (M) Moles of solute = (Vol of soln in L)(Molarity) l Dilution - adding solvent to a “stock” solution to lower its concentration Chapt. 4.1 Molarity (M) = moles of solute volume of solution (in liters) M (initial) V (initial) = M (final) V (final) moles (initial) = moles (final)

12 Chem 106, Prof. J.T. Spencer Molarity; Additional Examples l An experiment requires 150 mL of a 2 M HCl solution. All that is available in the laboratory is a 10.0 M HCl stock solution. How would you prepare the required solution? Chapt. 4.1 M (initial) V (initial) = M (final) V (final) M (initial) = 10 M HCl V (initial) = ? L M (final) = 2 M HCl V (final) = 0.150 L How?? V i = (2 M)(0.150 L) = 0.030 L of 10 M HCl (10 M)How?? 30 mL of the 10 M HCl solution is volumetrically pipetted into a 150 mL volumetric flask and diluted to mark

13 Chem 106, Prof. J.T. Spencer Molarity and Stiochiometry l How many mL of a 0.250 M HCl solution would be required to do each of the following operations; l prepare 50.0 mL of a 0.100 M HCl soln. l react with the OH- in 15.0 mL of a 0.200 M Ba(OH) 2 soln. l dissolve 0.500 g CaCO 3 according to the reaction; CaCO 3 + 2H + Ca +2 + H 2 O + CO 2 Chapt. 4.1 M (initial) V (initial) = M (final) V (final) Molarity (M) = moles of solute volume of solution (in liters) Equations:

14 Chem 106, Prof. J.T. Spencer Molarity and Dilution Chapt. 4.1 l How many mL of a 0.250 M HCl solution would be required to do each of the following operations; l prepare 50.0 mL of a 0.100 M HCl soln. M (initial) V (initial) = M (final) V (final) V i = M f V f M i M (initial) = 0.250 M HCl V (initial) = ? L M (final) = 0.100 M HCl V (final) = 0.050 L V i = (0.100 M)(0.050 L) = 0.020 L (0.250 M) 20 mL of the 0.250 M HCl solution is volumetrically pipeted into a 50 mL volumetric flask and diluted to mark

15 Chem 106, Prof. J.T. Spencer Molarity and Dilution Chapt. 4.1 l How many mL of a 0.250 M HCl solution would be required to do each of the following operations; l react with the OH- in 15.0 mL of a 0.200 M Ba(OH) 2 soln. 2 HCl + Ba(OH) 2 2 H 2 O + BaCl 2 moles OH - = (0.200 m Ba(OH) 2 ) (0.015L Ba(OH) 2 ) (2 m HCl) = 0.006mol liter 1 m Ba(OH) 2 vol HCl (L) = (0.006 mol HCl req.) = 0.024 L or 24 mL 0.250 M volume of solution (in liters)= moles of solute Molarity (M)

16 Chem 106, Prof. J.T. Spencer Molarity and Dilution Chapt. 4.1 volume of solution (in liters)= moles of solute Molarity (M) l How many mL of a 0.250 M HCl solution would be required to do each of the following operations; l dissolve 0.500 g CaCO 3 according to the reaction; CaCO 3 + 2H + Ca +2 + H 2 O + CO 2 moles CaCO 3 = (0.50 g CaCO 3 ) = 0.005 m CaCO 3 100.1 amu mol HCl req = (0.005 mol CaCO 3 ) (2 HCl ) = 0.010 mol HCl 1 CaCO 3 Vol HCl = 0.010 mol HCl = 0.040 L or 40 mL 0.250 M

17 Chem 106, Prof. J.T. Spencer Electrolytes Chapt. 4.2 Bulb glows brightly Strong Electrolyte Solution [many electrical current carriers] Bulb glows dimly Weak Electrolyte Solution [some current carriers] Bulb does not light Nonelectrolyte Solution [no electrical current carriers] solution The Ions complete the circuit by carrying electrical charge between the electrodes

18 Chem 106, Prof. J.T. Spencer Electrolytes Chapt. 4.2 l Electrolytes - solutes that exist as ions in solution and conduct electrical current. l Ionic compounds which dissociate into ions in solution are electrolytes; NaCl (s) Na +1 (aq) + Cl -1 (aq) l Other solutes which are not ionic but dissolve to form ions are electrolytes (HCl, H 2 SO 4, etc...) HCl (g) H +1 (aq) + Cl -1 (aq) l Non ionizing compounds form nonconducting solutions and are, therefore, nonelectrolytes. l Electrolyte solutes are characterized as either strong or weak depending on the extent of dissociation (and, therefore, current conduction). Most ionic compounds are strong electrolytes

19 Chem 106, Prof. J.T. Spencer Electrolytes Chapt. 4.2 l Strong electrolytes are completely dissociated in solution l Weak electrolytes are only partially dissociated into ions in solution [CHEMICAL EQUILIBRIUM] HC 2 H 3 O 2 (aq ) H + (aq) + C 2 H 3 O 2 - (aq) l Nonelectrolytes are not dissociated into ions in solution l Extent of dissolution does not dictate strong or weak electrolyte solution (i.e., HC 2 H 3 O 2 is very soluble but is a weak electrolyte while Ba(OH) 2 is only slightly soluble is a strong electrolyte)

20 Chem 106, Prof. J.T. Spencer Chemical Equilibria When a reaction proceeds to completion, its equation is written with a single arrow. HNO 3 (aq ) H + (aq) + NO 3 - (aq) When a reaction’s progress can be more accurately considered to be a balance between the forward and back reactions, then the two processes (forward and backward) are involved in an equilibrium which is written as a double arrow. HC 2 H 3 O 2 (aq ) H + (aq) + C 2 H 3 O 2 - (aq) Chapt. 4.2

21 Chem 106, Prof. J.T. Spencer Chemical Equilibria Chapt. 4.2 Backward ReactionForward Reaction

22 Chem 106, Prof. J.T. Spencer l Precipitation Reactions l Precipitation Reactions - formation of an insoluble solid (less than 0.01 mol/L) AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) l Solubility Rules l Solubility Rules - allow predictions of whether a ppt will form. Rules not based upon simple physical properties (i.e., charge, conductivity, etc...) – Rules developed by observations, but many generalizations are possible Metathesis Reactions Chapt. 4.2

23 Chem 106, Prof. J.T. Spencer l Soluble – all Nitrates (NO 3 - ) and acetates (C 2 H 3 O 2 - ) and ammoniums – all Group 1 metals and fluorides – all chlorides, bromides and iodides (except Ag, Hg, and Pb) – all sulfates (except (Ca, Sr, Ba, Pb, Hg, and Ag) Solubility Chapt. 4.2

24 Chem 106, Prof. J.T. Spencer l Soluble – all Nitrates (NO 3 - ) and acetates (C 2 H 3 O 2 - ) and ammoniums – all Group 1 metals and fluorides – all chlorides, bromides and iodides (except Ag, Hg, and Pb) – all sulfates (except (Ca, Sr, Ba, Pb, Hg, and Ag) l Insoluble – all sulfides (S 2- ) except those of groups 1 (1A) and 2 (2A) and ammonium – all carbonates (CO 3 2- ) except those of group 1 (1A) and ammonium – all phosphates except those of group 1 (1A) and ammonium – all hydroxides except those of group 1 (1A) and Ba, Sr and Ca l Use Rules to Predict Metathesis Reactions Solubility Chapt. 4.2

25 Chem 106, Prof. J.T. Spencer l Metathesis Reactions l Metathesis Reactions - a reaction of ionic species in which partners exchange (metathesis is Greek for “transpose”). AX + BYAY + BX l These reactions require a “driving force” by removing ions from solution by; – formation of an insoluble solid – formation of either a soluble weak electrolyte or soluble nonelectrolyte – formation of a gas which escapes from solution Metathesis Reactions Chapt. 4.2

26 Chem 106, Prof. J.T. Spencer l Driving force can also be the formation of weak electrolytes or nonelectrolytes – acid base neutralizations (even water insoluble hydroxides can act as bases in these reactions) Mg(OH) 2 ( s ) + 2 HCl MgCl 2 ( aq ) + 2 H 2 O – metal oxides with acids (oxide can react with 2H + to yield water) FeO( s ) + H 2 SO 4 ( aq ) Fe(SO 4 )( aq ) + H 2 O( l ) FeO( s ) + 2H + ( aq ) Fe 2+ ( aq ) + H 2 O( l ) – ionic reactions occur when ions are removed to form a weak electrolyte HCl( aq ) + NaC 2 H 3 O 3 ( aq ) HC 2 H 3 O 3 ( aq ) + NaCl( aq ) Metathesis Reactions Chapt. 4.2

27 Chem 106, Prof. J.T. Spencer l Driving force may also be the formation of a gas as a product which has a low water solubility. – 2HCl( aq ) + Na 2 S( aq )H 2 S( g ) + 2 NaCl( aq ) 2H + (aq) + S 2- ( aq )H 2 S( g ) (net ionic) – HCl( aq ) + NaHCO 3 ( aq )NaCl( aq ) + [H 2 CO 3 ( aq )] H 2 O( l ) + CO 2 ( g ) Metathesis Reactions Chapt. 4.2 UNSTABLE INITIAL PRODUCT gas formation to drive reaction

28 Chem 106, Prof. J.T. Spencer l Reaction of PbS with HCl l Reaction of Cr(C 2 H 3 O 2 ) 2 with HCl l Reaction of Co(CO 3 ) with HCl Metathesis Reactions Chapt. 4.2 Pb +2 ( aq ) + S 2- ( aq ) + 2H + ( aq ) + 2Cl - ( aq ) PbCl 2 ( s ) + H 2 S( g ) complete equation, net ionic equation driving force = precipitation and gas evol. Cr +2 ( aq ) + 2C 2 H 3 O 3 - ( aq ) + 2H + ( aq ) + 2Cl - ( aq ) Cr +2 ( aq ) + 2HC 2 H 3 O 3 ( aq ) + 2Cl - ( aq ) 2C 2 H 3 O 3 - ( aq ) + 2H+( aq ) 2HC 2 H 3 O 3 ( aq ) (net ionic) Co +2 ( aq ) + CO 3 -2 ( aq ) + 2H + ( aq ) + 2Cl - ( aq ) Co +2 ( aq ) +CO 2 ( g ) + H 2 O( l ) +2Cl - ( aq )

29 Chem 106, Prof. J.T. Spencer l Molecular equation l Molecular equation - shows the complete chemical formulas. Ba(OH) 2 (s) + 2 HNO 3 (aq) Ba(NO 3 ) 2 (aq) + 2 H 2 O (l) l Complete Ionic Equation l Complete Ionic Equation - in solutions, where the reactants and products are ionized, it is more useful (and accurate) to indicate ions. Ba 2+ (aq) + 2OH - (aq) + 2H + (aq) + 2NO 3 - (aq) Ba 2+ (aq) + 2NO 3 - (aq) + 2 H 2 O (l) l Net Ionic Equation l Net Ionic Equation - omits ions which appear unchanged ( spectator ions ) on both sides of eqn. 2OH - (aq) + 2H + (aq) 2 H 2 O (l) Ionic Equations Chapt. 4.2

30 Chem 106, Prof. J.T. Spencer l Net Ionic Equations greatly simplify equations and points out similarities and differences between reactions more easily. (i.e., ANY neutralization reaction appears the same); 2OH - (aq) + 2H + (aq) 2 H 2 O (l) Ionic Equations Chapt. 4.2

31 Chem 106, Prof. J.T. Spencer l Net Ionic Equations greatly simplify equations and points out similarities and differences between reactions more easily. (i.e., EVERY neutralization reaction appears the same); 2OH - (aq) + 2H + (aq) 2 H 2 O (l) l To write an ionic equation; – is the compound soluble? – if its is soluble, is it a strong electrolyte? » Only if both answers are yes then an ionic equation is good. (otherwise write molecular equation) Ionic Equations Chapt. 4.2

32 Chem 106, Prof. J.T. Spencer l Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) Ionic Equations Chapt. 4.2

33 Chem 106, Prof. J.T. Spencer l Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Ionic Equations Chapt. 4.2

34 Chem 106, Prof. J.T. Spencer l Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Pb +2 + 2NO 3 -1 + 2Na +1 + SO 4 -2 Ionic Equations Chapt. 4.2

35 Chem 106, Prof. J.T. Spencer l Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Pb +2 + 2NO 3 -1 + 2Na +1 + SO 4 -2 Double Replacement… Ionic Equations Chapt. 4.2

36 Chem 106, Prof. J.T. Spencer l Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Pb +2 + 2NO 3 -1 + 2Na +1 + SO 4 -2 PbSO 4 + NaNO 3 Ionic Equations Chapt. 4.2

37 Chem 106, Prof. J.T. Spencer l Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Pb +2 + 2NO 3 -1 + 2Na +1 + SO 4 -2 PbSO 4 + 2NaNO 3 Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Ionic Equations Chapt. 4.2

38 Chem 106, Prof. J.T. Spencer l Pb +2 + 2NO 3 -1 + Na +1 + SO 4 -2 PbSO 4 + NaNO 3 Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Pb +2 + 2NO 3 -1 + 2Na +1 + SO 4 -2 PbSO 4 + 2Na + + 2NO 3 - Ionic Equations Chapt. 4.2

39 Chem 106, Prof. J.T. Spencer l Pb +2 + 2NO 3 -1 + Na +1 + SO 4 -2 PbSO 4 + NaNO 3 Guidelines: nitrates are ALWAYS SOLUBLE, sulfates are soluble except Hg, Ag, Pb, and heavy Group II Pb +2 + 2NO 3 -1 + 2Na +1 + SO 4 -2 PbSO 4 + 2Na + + 2NO 3 - Net Ionic: Pb +2 + SO 4 -2 PbSO 4 Ionic Equations Chapt. 4.2

40 Chem 106, Prof. J.T. Spencer l CuBr 2 (aq) + NaOH(aq) Ionic Equations Chapt. 4.2

41 Chem 106, Prof. J.T. Spencer l CuBr 2 (aq) + NaOH(aq) Guidelines: halides are soluble except Hg, Ag, Pb and hydroxides are always insoluble except group I, group II and ammonium Ionic Equations Chapt. 4.2

42 Chem 106, Prof. J.T. Spencer l CuBr 2 (aq) + NaOH(aq) Guidelines: halides are soluble except Hg, Ag, Pb and hydroxides are always insoluble except group I, group II and ammonium Cu +2 + 2Br - + Na + + OH - Ionic Equations Chapt. 4.2

43 Chem 106, Prof. J.T. Spencer l CuBr 2 (aq) + NaOH(aq) Guidelines: halides are soluble except Hg, Ag, Pb and hydroxides are always insoluble except group I, group II and ammonium Cu +2 + 2Br - + Na + + OH - Double Replacement Ionic Equations Chapt. 4.2

44 Chem 106, Prof. J.T. Spencer l CuBr 2 (aq) + NaOH(aq) Guidelines: halides are soluble except Hg, Ag, Pb and hydroxides are always insoluble except group I, group II and ammonium Cu +2 + 2Br - + Na + + OH - Cu(OH) 2 + NaBr Ionic Equations Chapt. 4.2

45 Chem 106, Prof. J.T. Spencer l Cu +2 + 2Br - + Na + + OH - Cu(OH) 2 + NaBr Guidelines: halides are soluble except Hg, Ag, Pb and hydroxides are always insoluble except group I, group II and ammonium Ionic Equations Chapt. 4.2

46 Chem 106, Prof. J.T. Spencer l Cu +2 + 2Br - + Na + + OH - Cu(OH) 2 + NaBr Guidelines: halides are soluble except Hg, Ag, Pb and hydroxides are always insoluble except group I, group II and ammonium Cu +2 + 2Br - + Na + + OH - Cu(OH) 2 + Na + + Br - Ionic Equations Chapt. 4.2

47 Chem 106, Prof. J.T. Spencer l Cu +2 + 2Br - + Na + + OH - Cu(OH) 2 + NaBr Guidelines: halides are soluble except Hg, Ag, Pb and hydroxides are always insoluble except group I, group II and ammonium Cu +2 + 2Br - + Na + + OH - Cu(OH) 2 + Na + + Br - Net Ionic: Cu +2 + 2OH - Cu(OH) 2 Ionic Equations Chapt. 4.2

48 Chem 106, Prof. J.T. Spencer Acids Chapt. 4.3 H 3 PO 3 (aq) H + (aq) + H 2 PO 3 -1 (aq) H 2 PO 3 -1 (aq) H + (aq) + HPO 3 -2 (aq) HPO 3 -2 (aq) H + (aq) + PO 3 -3 (aq) l Acid l Acid - a substance able to ionize to form H + ion, “proton”, thereby increasing the H + concentration in solution. ACIDS ARE PROTON DONORS l Type of acid named by how many protons it can donate (i.e., monoprotic, diprotic,....).

49 Chem 106, Prof. J.T. Spencer Bases Chapt. 4.3 l Base - l Base - a substance able to ionize to form OH - ions, thereby increasing the OH - concentration in aqueous solutions. l Base - Neutralization Reactions l Base - a substance which reacts with or accepts H + ions [Neutralization Reactions]: H + (aq) + OH - (aq) H 2 O (l ) BASES ARE PROTON ACCEPTORS

50 Chem 106, Prof. J.T. Spencer Bases Chapt. 4.3 l Kinds of Bases - l Ionic Hydroxides - dissociate to form OH - and M + l Proton Acceptors w/out Hydroxide (OH - ) NH 3 (aq) + H 2 O (l) NH 4 + (aq ) + OH - (aq) only a small portion of the ammonia (ca. 1%) reacts with the water to form the ammonium ion, it is a weak electrolyte (weak base). Note: Compounds, such as CH 3 OH, which contain an OH group, are essentially not dissociated, are not electrolytes, and are not bases.

51 Chem 106, Prof. J.T. Spencer l Increasing concentration of acid (H + ) decreases concentration of base (OH - ). l Increasing concentration of base (OH - ) decreases the concentration of acid (H + ) l Strong and Weak Acids and Bases l Strong and Weak Acids and Bases - Strong indicates complete dissociation (strong electrolyte) and weak refers to only partial dissociation into ions (weak electrolyte) l Strong acids and bases are generally more reactive than weak acids and bases when the reactivity depends only on H + or OH - concentration Acids and Bases Chapt. 4.3

52 Chem 106, Prof. J.T. Spencer A cids and Bases Chapt. 4.3 H + with a halide ion l Strong Acids l Strong Acids - HClhydrochloric acid HBrhydrobromic acid HIhydroiodic acid HNO 3 nitric acid H 2 SO 4 sulfuric acid HClO 3 chloric acid HClO 4 perchloric acid l Strong Bases Group 1 (or 1A) metal hydroxides (i.e., LiOH) Group 2 (or 2A) heavy metal hydroxides (i.e., Ca(OH) 2 ) Most Acids (others) Are Weak

53 Chem 106, Prof. J.T. Spencer l Salts l Salts are ionic compounds formed by replacing one or more H + of an acid by a different cation. l HCl becomes NaCl by replacing H + with Na + l H 2 SO 4 becomes CuSO 4 by replacing 2H + with Cu +2 l HNO 3 becomes CoNO 3 by replacing H + with Co +1 l HClO 4 becomes KClO 4 by replacing H + with K +1 l Salts l Salts are typically strong electrolytes [with the exception of salts of heavy metal cations such as mercury and lead]. l A neutralization reaction between an acid and a metal hydroxide produces water and a salt; HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq) Salts Chapt. 4.3

54 Chem 106, Prof. J.T. Spencer EXCEPT l Most salts are strong electrolytes[EXCEPT heavy metal salts, i.e., HgCl 2 ]. EXCEPT l Most acids are weak electrolytes [EXCEPT HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3 and HClO 4 ]. l The common strong bases are the hydroxides of the alkali metals and the heavy alkaline earths. l Ammonia is a weak electrolyte. l Most other compounds are nonelectrolytes. Acids and Bases Chapt. 4.3 Identifying Strong and Weak Acids and Bases

55 Chem 106, Prof. J.T. Spencer Identify each of the following as a strong electrolyte, a weak electrolyte or a non electrolyte AND (where possible) as a weak or strong acid, base or salt (in aqueous solution); l strong base (strong electrolyte) l salt (strong electrolyte) l weak acid (weak electrolyte) l strong acid (strong electrolyte) l nonelectrolyte l weak base (weak electrolyte) l salt (strong electrolyte) l nonelectrolyte l weak base (weak electrolyte) Acids, Bases, and Salts Chapt. 4.3 LiOH BaBr 2 H 3 PO 4 H 2 SO 4 CH 3 CN Ba(OH) 2 CuCl 2 C 6 H 12 O 6 NH 3

56 Chem 106, Prof. J.T. Spencer l Ba(OH) 2 (s) + 2 HNO 3 (aq) Ba(NO 3 ) 2 (aq) + 2 H 2 O (l) l H 3 PO 4 (aq) + 3 LiOH (aq) Li 3 PO 4 (aq) + 3 H 2 O (l) l CH 3 CO 2 H (aq) + NaOH (aq) CH 3 CO 2 Na (aq) + H 2 O (l) l NH 3 (aq) + HCl (aq) NH 4 + + Cl - (aq) Acids, Bases, and Salts Chapt. 4.3 Complete and balance each of the following equations (1) Barium hydroxide reacting with nitric acid (2) Phosphoric acid reacting with lithium hydroxide (3) Acetic acid is neutralized with sodium hydroxide (4) Ammonia (aq) is added to hydrochloric acid (aq )

57 Chem 106, Prof. J.T. Spencer l Metals react with many compounds - corrosion reaction is specifically the reaction of a metal with an “environmental” (naturally occurring l Oxidation Reactions l Oxidation Reactions - loss of electrons by a substance (the substance is oxidized). 4 Na( s ) + O 2 ( g )2 Na 2 O( s ) Na goes from 0 charge to +1 charge ( oxidized )O goes from 0 charge to -2 charge ( reduced ) Reactions with Metals Chapt. 4.4 O -2 Na +1 Na 0 O0O0 O0O0 4 + 2

58 Chem 106, Prof. J.T. Spencer l 4 Na( s ) + O 2 ( g )2 Na 2 O( s ) Na goes from 0 charge to +1 charge ( oxidized ) O goes from 0 charge to -2 charge ( reduced ) l ReductionReactions l Reduction Reactions - gain of electrons by a substance (the substance is reduced). l When one reactant loses electrons (oxidized) some other reactant must gain the same number of electrons (reduced) [net electron transfer] Reactions with Metals Chapt. 4.4

59 Chem 106, Prof. J.T. Spencer l May formally assign charges to atoms in polar covalent compounds (bookkeeping purposes). l Charges assigned are called Oxidation Numbers. l H ox. no. = +1 and F ox. no. = -1 l Oxidation numbers DO NOT correspond with actual charges within a molecule (except for ionic compounds). Oxidation Numbers HFHFHFHF  EN = 4.0 - 2.1 = 1.9 Polar Covalent Bond   partial + charge partial - charge

60 Chem 106, Prof. J.T. Spencer Oxidation Numbers l Rules to determining Oxidation Number: – Ox. No. of an element is 0. – Ox. No. of a monatomic ion is the same as its charge. – For binary compounds, the element with the greater EN is assigned the negative Ox. No. (same charge as found in ionic compounds for the anion). – The sum of the oxidation numbers (total) must equal the charge of the molecule (including the neutral case).

61 Chem 106, Prof. J.T. Spencer Oxidation Numbers l Rules to determining Oxidation Number: – Ox. No. of an element is 0. – Ox. No. of a monatomic ion is the same as its charge. – Assign the following oxidation states to compounds IN ORDER until one element is left » Group 1 metals = +1 » Group 2 metal = +2 » Aluminum = +3 » Fluorine = -1 » Hydrogen = +1 » Oxygen = -2 » Halides = -1 » Assign the top oxidation number to elements starting from right

62 Chem 106, Prof. J.T. Spencer Oxidation Numbers l Determine Oxidation Numbers: Cl ClClCH H H H NH+1 OONH ClClS F FFCl

63 Chem 106, Prof. J.T. Spencer Oxidation Numbers l Determine Oxidation Numbers: Cl ClClCH H H H NH+1 OONH ClClSI F FFCl H = +1H = +1H = +1 C = +2N = -3N = +3 Cl = -1O = -2 S = +2F = -1 Cl = -1Cl = +3

64 Chem 106, Prof. J.T. Spencer the metal is oxidized l Many metals react with acids to form salts and hydrogen gas (the metal is oxidized!) metal + acidsalt + hydrogen Zn( s ) + 2HCl( aq )ZnCl 2 ( aq ) + H 2 ( g ) Fe( s ) + H 2 SO 4 ( aq )FeSO 4 ( aq ) + H 2 ( g ) (H+ is reduced and the metal is oxidized) l Can we predict when reactions involving oxidation and reduction will occur? – Different metals vary in the ease with which they are oxidized – Activity Series Oxidation of Metals Chapt. 4.4

65 Chem 106, Prof. J.T. Spencer l Activity Series lists metals in decreasing order of ease of oxidation (how easily a metal loses electrons). l Metals at the top are most easily oxidized and are referred active metals (i.e., alkali and alkaline earth metals) l Metals at the bottom are most stable WRT oxidation (i.e., noble metals) l Series can be used to predict reactions; ANY METAL ON THE LIST CAN BE OXIDIZED BY IONS OF THE ELEMENT BELOW IT l Only those metals ABOVE hydrogen on the list can react with acids to form hydrogen gas. Activity Series of Metals Chapt. 4.4

66 Chem 106, Prof. J.T. Spencer Activity Series Chapt. 4.4 Lithium Potassium Barium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron NickelHydrogen Copper Silver Platinum Gold EasierOxidationMoreDifficultOxidation M(s) M +n + n e - Alkali Metals Noble Metals

67 Chem 106, Prof. J.T. Spencer l Iron ions cannot oxidize Copper l Copper ions can oxidize Iron l Copper does not react with H+ l Nickel reacts with H+ to form hydrogen l Nickel is oxidized by Copper ions l Gold is not oxidized by other metal ions Activity Series Chapt. 4.4 Lithium Potassium Barium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron NickelHydrogen Copper Silver Platinum Gold EasierOxidationMoreDifficultOxidation EXAMPLES ANY METAL ON THE LIST CAN BE OXIDIZED BY IONS OF THE ELEMENT BELOW IT

68 Chem 106, Prof. J.T. Spencer l Note: Copper should not be oxidized by acid (H + ) to form hydrogen (from activity series table data) l Copper does not react with HCl( aq ) l HOWEVER, copper does react with nitric acid. WHY? Activity Series Chapt. 4.4 Copper reacts with the nitrate ion of the acid NOT the H+ (in agreement with the activity series) Cu( s ) + 4HNO 3 ( aq ) Cu(NO 3 ) 2 ( aq ) + 2H 2 O( l ) + 2NO 2 ( g )

69 Chem 106, Prof. J.T. Spencer l Lithium l Silver l Chromium l Nickel l Hydrogen l Magnesium Activity Series Chapt. 4.4 Lithium Potassium Barium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron NickelHydrogen Copper Silver Platinum Gold M(s) M +n + n e - EXAMPLES l Which of the following metals will be oxidized by Zn(SO 4 )? l Yes l No l Yes ANY METAL ON THE LIST CAN BE OXIDIZED BY IONS OF THE ELEMENT BELOW IT

70 Chem 106, Prof. J.T. Spencer l How do we determine (experimentally) the concentration of an unknown solution? – USE CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY INVOLVING ONE SOLUTION OF KNOWN CONCENTRATION BUT WHAT REACTIONS? Acid Base Reactions (previously discussed) Metathesis Reactions Oxidation-Reduction Reactions Solution Stoichiometry

71 Chem 106, Prof. J.T. Spencer l How do we determine (experimentally) the concentration of an unknown solution? KNOWN CONCENTRATION TITRATION – use a second solution ( standard solution ) of KNOWN CONCENTRATION which reacts with the first solution, then use molar/molarity conversions and the balanced equation to determine the concentration of the unknown solution (TITRATION). Solution Stoichiometry Chapt. 4.5 Grams of Compound A Grams of Compound B Moles of Compound A Moles of Compound B Volume or Molarity of Compound A of Compound A Volume or Molarity of Compound B of Compound B use MW use MW use M=mol/V use M=mol/V use bal. eqn.

72 Chem 106, Prof. J.T. Spencer l Add volume of a known conc. standard solution until equivalence point [reaction of first “excess” of standard solution with “indicator” to give a visible color change (i.e., litmus)]. Titrations Chapt. 4.5 Equivalence Point 1 drop add

73 Chem 106, Prof. J.T. Spencer Titrations: Indicators Indicator - A compound that changes color and is used to mark the end point of a titration. The color change occurs when either there is an excess of acid or base in solution. COLORLESS RED Phenolphthalein

74 Chem 106, Prof. J.T. Spencer l Acid-Base Indicator from Red Cabbage l Aqueous Carbon Dioxide Titrations: Indicators Indicator - A compound that changes color and is used to mark the end point of a titration. The color change occurs when either there is an excess of acid or base in solution. S Tape 1:1 (5:44 + 3:28)

75 Chem 106, Prof. J.T. Spencer Titrations Chapt. 4.5 M (strd) V (strd) = M (unk) V (unk) M (strd) = Molarity of strd V (strd) = vol. to equiv. point M (unk) = determine V (unk) = vol. of unknown solution volume of solution (in liters)= moles of solute Molarity (M) known Five Example Problems Follow To Illustrate

76 Chem 106, Prof. J.T. Spencer Titrations l What volume of 0.827 M KOH solution is required to completely neutralize 35.00 mL of 0.737 H 2 SO 4 solution? M (strd) V (strd) = M (unk) V (unk) M (unk) = 0.8270 M KOH V (unk) = ? L KOH M (strd) = 0.737 M H 2 SO 4 V (strd) = 0.0350 L H 2 SO 4 V(KOH) = (0.737 M)(0.0350 L) 2 = 0.0624 L (0.827 M) V(KOH) = 62.4 mL l 2 KOH + H 2 SO 4 2 H 2 O + K 2 SO 4

77 Chem 106, Prof. J.T. Spencer Titrations l What volume of 0.827 M KOH solution is required to completely neutralize 35.00 mL of 0.737 H 2 SO 4 solution? V(KOH) = (0.737 M)(0.0350 L) 2 = 0.0624 L (0.827 M) V(KOH) = (0.737 mol H 2 SO 4 )(0.0350 L H 2 SO 4 )(2 mol KOH) 1 L H 2 SO 4 soln 1 mol H 2 SO 4 0.827 mol KOH 1 L KOH soln l 2 KOH + H 2 SO 4 2 H 2 O + K 2 SO 4

78 Chem 106, Prof. J.T. Spencer l A 17.5 mL sample of acetic acid (CH 3 CO 2 H) required 29.6 mL of 0.250 M NaOH to neutralize it. What was the molarity of the acetic acid solution? Titrations Chapt. 4.5

79 Chem 106, Prof. J.T. Spencer l A 17.5 mL sample of acetic acid (CH 3 CO 2 H) required 29.6 mL of 0.250 M NaOH to neutralize it. What was the molarity of the acetic acid solution? Titrations M (strd) V (strd) = M (unk) V (unk) M (strd) = 0.250 M NaOH V (strd) = 0.0296 L NaOH M (unk) = ? V (unk) = 0.0175 L acetic acid M(aa) = (0.250 M)(0.0296 L) (0.0175 L) M(aa) = 0.423 M Chapt. 4.5

80 Chem 106, Prof. J.T. Spencer l Example - l Example - Acetic acid (HC 2 H 3 O 2 ) reacts with sodium hydroxide in a neutralization reaction. If 8.30 mL of an acetic acid soln. requires 42.4 mL of 0.1050 M NaOH to reach the equivalence point, how many grams of acetic acid are in a 500 mL sample? HC 2 H 3 O 2 + NaOH H 2 O + NaC 2 H 3 O 2 Titrations Chapt. 4.5

81 Chem 106, Prof. J.T. Spencer l Example - l Example - Acetic acid (HC 2 H 3 O 2 ) reacts with sodium hydroxide in a neutralization reaction. If 8.30 mL of an acetic acid soln. requires 42.4 mL of 0.1050 M NaOH to reach the equivalence point, how many grams of acetic acid are in a 500 mL sample? HC 2 H 3 O 2 + NaOH H 2 O + NaC 2 H 3 O 2 Titrations Chapt. 4.5 M (strd) V (strd) = M (unk) V (unk) M (strd) = 0.1050 M NaOH V (strd) = 0.0424 L NaOH M (unk) = ? V (unk) = 0.00830 L acetic acid M(aa) = (0.1050 M)(0.0424 L) (0.00830 L) M(aa) = 0.536 M mol(aa) in 500 mL = (0.537 M aa)(0.5 L) = 0.268 mol aa g(aa) = (0.268 mol aa)(60 amu) = 16.1 g aa in 500 mL of soln

82 Chem 106, Prof. J.T. Spencer l Example - l Example - In the laboratory, 6.67 g of Sr(NO 3 ) 2 was dissolved to make 750 mL of solution. A 100 mL sample of this solution was titrated with a 0.0460 M Na 2 CrO 4 solution. What volume of the Na 2 CrO 4 solution is required to precipitate all the Sr 2+ as SrCrO 4 ? (MW Sr(NO 3 ) 2 =211.6) Sr(NO 3 ) 2 (aq) + Na 2 CrO 4 (aq) SrCrO 4 (s) + Na(NO 3 ) (aq) Titrations Chapt. 4.5 (6.67 g Sr(NO 3 ) 2 ) = 0.0315 mol Sr +2 (211.6 g/mol) and (0.0315 mol Sr +2 ) = 0.042 M and (0.042 M)(0.100 L) = 0.0042 mol (0.750 L) 0.0042 Sr +2 = 0.0042 mol CrO 4 -2 0.042 M (0.100) = 0.0460(V) V = 0.042(0.100) 0.0460 V = 0.0913 L

83 Chem 106, Prof. J.T. Spencer Titrations Chapt. 4.5 l Example - l Example - The arsenic in a 1.22 gram sample of pesticide was chemically converted to AsO 4 3-. It was then titrated with Ag +1 to form Ag 3 AsO 4 as a precipitate. If it took 25.0 mL of a 0.102 M Ag +1 solution to reach the equivalence point, what is the percentage arsenic is the pesticide? l What is the balanced equation? AsO 4 3- (aq) + 3Ag +1 (aq) Ag 3 AsO 4 (s) l How many moles of AsO 4 3- are in the sample? l How many moles As in sample ? l How many grams of As in sample?

84 Chem 106, Prof. J.T. Spencer l Example - l Example - The arsenic in a 1.22 gram sample of pesticide was chemically converted to AsO 4 3-. It was then titrated with Ag +1 to form Ag 3 AsO 4 as a precip. If it took 25.0 mL of a 0.102 M Ag +1 solution to reach the equivalence point, what is the percentage arsenic is the pesticide? l How many moles of AsO 4 3- are in the sample? AsO 4 3- (aq) + 3Ag +1 (aq) Ag 3 AsO 4 (s) Titrations Chapt. 4.5 therefore Molarity = moles therefore moles = (M)(V) V moles Ag 1+ = (0.102M)(0.025L) = 0.00244 mol Ag 1+ mol AsO 4 3- = (0.00255 mol Ag 1+ )(1 AsO 4 3- ) = 0.00085 mol (3 Ag 1+ ) AsO 4 3-

85 Chem 106, Prof. J.T. Spencer l Example - l Example - The arsenic in a 1.22 gram sample of pesticide was chemically converted to AsO 4 3-. It was then titrated with Ag +1 to form Ag 3 AsO 4 as a precip. If it took 25.0 mL of a 0.102 M Ag +1 solution to reach the equivalence point, what is the percentage arsenic is the pesticide? l How many moles and grams As in sample ? Titrations Chapt. 4.5 moles AsO 4 3- = moles As g As = (mole As)(AW of As) = (0.00085 mol As)(74.9 amu) = 0.0637 grams As in sample % As = g As (100) = (0.0637 g As) (100) = 5.22% As g total (1.22 g sample)

86 Chem 106, Prof. J.T. Spencer l Molarity l Strong Electrolytes, Weak Electrolytes and Nonelectrolytes l Acids, Bases and Salts l Neutralization Reactions l Metathesis Reaction l Solubility rules l Reactions of Metals l Activity Series and oxidations and reductions l Solution Stoichiometry l Titrations End Chapter Four