Transcript Solutions
Chapter 4 Aqueous solutions Types of reactions Parts of Solutions Solution- homogeneous mixture. Solute- what gets dissolved. Solvent- what does the dissolving. Soluble- can be dissolved. Miscible- liquids dissolve in each other. Aqueous solutions Dissolved in water. Water is a good solvent because the molecules are polar. The oxygen atoms have a partial negative charge. The hydrogen atoms have a partial positive charge. The angle is 105º. Hydration The process of breaking the ions of salts apart. Ions have charges and attract the opposite charges on the water molecules. Hydration H H H H H Solubility How much of a substance will dissolve in a given amount of water. • Usually listed in g/100 mL Solubilities vary greatly, but if they do dissolve the ions are separated, and they can move around. Water can also dissolve non-ionic compounds if they have polar bonds. Electrolytes Electricity is moving charges. The ions that are dissolved can move. Solutions of ionic compounds can conduct electricity and are therefore called “electrolytes.” Types of solutions Solutions are classified three ways: • Strong electrolytes- completely dissociate (fall apart into ions). • Many ions- Conduct well. • Weak electrolytes- Partially fall apart into ions. • Few ions -Conduct electricity slightly. • Non-electrolytes- Don’t fall apart. • No ions- these sol’ns don’t conduct. Types of solutions Acids- form H+ ions when dissolved. • Strong acids fall apart completely. • many ions • There are just 6 strong acids – MEMORIZE THEM! • Sulfuric acid: H2SO4 • Nitric acid: HNO3 • Hydrochloric acid: HCl • Hydrobromic acid: HBr • Hydroiodic acid: HI • Perchloric acid: HClO4 • Weak acids- don’t dissociate completely. Types of solutions Bases - form OH- ions when dissolved. • Strong bases- many ions. Most common: • Lithium hydroxide: LiOH • Sodium hydroxide: NaOH • Potassium hydroxide: KOH • Rubidium hydroxide: RbOH • Calcium hydroxide: Ca(OH)2 • Strontium hydroxide: Sr(OH)2 • Barium hydroxide: Ba(OH)2 MOM to the rescue! Milk of magnesia = aqueous suspension of magnesium hydroxide Is Mg(OH)2 a strong base? Yes, technically… since all of the dissolved magnesium hydroxide dissociates. However, it has a very low solubility in water. Measuring Solutions Concentration- how much is dissolved. mol solute Molarity M L sol'n 1 M = 1 mol solute / 1 liter solution What is this??? A one molar solution!!! Ha ha ha ha ha ha! 1L Molarity Practice Problems Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution. How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution? What would the concentration be if you used 27g of CaCl2 to make 500. mL of solution? • What is the concentration of each ion? Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution. Molarity mol solute L sol'n 34.6g NaCl 1 mol 58.44 g Molarity 0.125L Molarity 4.74 M How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution? Molarity mol solute L sol'n mol HCl 2.7 M 0.0500L mol HCl 0.135 mol 36.46g HCl 4.92 g HCl 1 mol HCl What would the concentration be if you used 27g of CaCl2 to make 500. mL of solution? What is the concentration of each ion? mol solute Molarity L sol'n Why? 27 g CaCl2 1 mol 110.98g Molarity 0.500L MolarityCaCl2 0.49 M [Ca2+]=0.49M [Cl-]=0.98M Because for every one CaCl2 there are two Cl- and one Ca2+… thus for every 1 mol CaCl2, 1 mole of Ca2+ and 2 moles of Cl- form. Molarity Practice Problems Calculate the concentration of a solution made by dissolving 45.6 g of Fe2(SO4)3 to 475 mL. • What is the concentration of each ion? 45.6g Fe2 (SO4 )3 1 mol 399.88g Molarity of Fe2 (SO4 )3 0.475L [Fe2+]= 2(.240 M)=.480 M [SO42-]= 3(.240 M)=.720 M 0.240M This is because for every 1 Fe2(SO4)3 particle there are 2 iron ions and 3 sulfate ions Practice Problems – Making Solutions Describe how to make 100.0 mL of a 1.0 M K2Cr2O4 solution. mol K2Cr2O4 1.0 M 0.1000L mol K2Cr2O4 0.10 mol 246.2g 1 mol 24 g Answer: Measure out 24 g of K2Cr2O4 in a volumetric flask. Add enough distilled water to dissolve the solid. Dilute to a total solution volume of 100.0 mL. Practice Problems – Making Solutions Describe how to make 250. mL of an 2.0 M copper (II) sulfate dihydrate solution. mol CuSO4 2H2O 2.0 M 0.250L 195.61g mol CuSO4 2H2O 0.50 mol 1 mol 98 g Answer: Measure out 98 g of CuSO4•2H2O in a volumetric flask. Add enough distilled water to dissolve the solid. Dilute to a total solution volume of 250. mL. Dilution Adding more solvent to a known solution. The moles of solute stay the same. • M1 V1 = M2 V2 Stock solution is a solution of known concentration used to make more dilute solutions Dilution Practice Problems What volume of a 1.7 M solutions is needed to make 250 mL of a 0.50 M solution? M1V1 = M2V2 (1.7 M)(V1) = (0.50 M)(250 mL) V1 = 74 mL Dilution Practice Problems 18.5 mL of 2.3 M HCl is added to 250 mL of water. What is the concentration of the resulting solution? M1V1 = M2V2 (2.3 M)(18.5 mL) = (M2)(250 mL+18.5 mL) M2 = 0.16 M Dilution Practice Problems You have a 4.0 M stock solution. Describe how to make 1.0L of a .75 M solution. M1V1 = M2V2 (4.0 M)(V1) = (0.75 M)(1.0 L) V1 = 0.19 L = 190 mL Measure out 190 mL of 4.0 M stock solution. Dilute with enough water to make 1.0 L solution Dilution Practice Problems 25 mL 0.67 M of H2SO4 is added to 35 mL of 0.40 M CaCl2 . What mass CaSO4 is formed? H2SO4 + CaCl2 CaSO4 + 2HCl 0.67mol mol H2SO4 0.025L 0.017 mol L 0.40mol mol CaCl2 0.035L 0.014 mol L Dilution Practice Problems 25 mL 0.67 M of H2SO4 is added to 35 mL of 0.40 M CaCl2 . What mass CaSO4 is formed? LR H2SO4 + CaCl2 CaSO4 + 2HCl 0.017mol 0.014mol 0.014 mol CaCl2 1 mol CaSO4 136.14 g CaSO4 1.9 g CaSO4 1 mol CaCl2 1 mol CaSO4 Types of Reactions Precipitation reactions • When aqueous solutions of • • ionic compounds are poured together a solid forms. A solid that forms from mixed solutions is a precipitate If you’re not a part of the solution, your part of the precipitate! LOL! :-D Precipitation reactions Sample reaction… Overall (molecular) eq’n: 3NaOH(aq) + FeCl3(aq) NaCl(aq) + Fe(OH)3(s) The above rxn is really… Total ionic eq’n: 3Na+(aq)+3OH-(aq)+Fe+3(aq) + 3Cl-(aq) Na+(aq) + 3Cl-(aq) +Fe(OH)3(s) So all that really happens is… Net ionic eq’n: 3OH-(aq) + Fe+3(aq) Fe(OH)3(s) (Double replacement reaction) Precipitation reaction We can predict the products Can only be certain by experimenting The anion and cation switch partners Practice Problems – Predicting Products of Precipitation Rxns AgNO3(aq) + KCl(aq) KNO3( Ag+ + AgCl( ) K+ Cl- Zn(NO3)2(aq) + BaCr2O7(aq) Zn2+ NO3- ) NO3- Ba2+ Cr2O72- ZnCr2O7( CdCl2(aq) + Na2S(aq) 2NaCl( Cd2+ Cl- Na+ S2- ) ) + Ba(NO3)2( + CdS( ) ) Precipitations Reactions Only happen if one of the products is insoluble Otherwise all the ions stay in solutionnothing has happened. Need to memorize the rules for solubility! • See Solubility Rules Handout! Solubility Rules 1) All nitrates are soluble 2) Alkali metals ions and NH4+ ions are soluble 3) Halides are soluble except Ag+, Pb+2, and Hg2+2 4) Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2 5) Most hydroxides are slightly soluble (insoluble) except NaOH and KOH 6) Sulfides, carbonates, chromates, and phosphates are insoluble Lower number rules supersede so Na2S is soluble Practice Problems – Predicting Products of Precipitation Rxns AgNO3(aq) + KCl(aq) KNO3(aq ) + AgCl( s Ag+ K+ Cl- Zn(NO3)2(aq) + BaCr2O7(aq) Zn2+ NO3- ) NO3- Ba2+ Cr2O72- ZnCr2O7( s ) + Ba(NO3)2( aq) CdCl2(aq) + Na2S(aq) 2NaCl(aq ) + CdS( s Cd2+ Cl- Na+ S2- ) Three Types of Equations Molecular Equation- written as whole formulas, not the ions. • Complete Ionic Equation shows dissolved electrolytes as the ions. • • K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq) 2K+ + CrO4-2 + Ba+2 + 2 NO3- BaCrO4(s) + 2K+ + 2 NO3Spectator ions are those that don’t react. Net Ionic equation shows only those ions that react, not the spectator ions • Ba+2 + CrO4-2 BaCrO4(s) Practice Problem Write the three types of equations for the reactions when these solutions are mixed. • iron (III) sulfate and potassium sulfide • lead (II) nitrate and sulfuric acid iron (III) sulfate and potassium sulfide Fe2(SO4)3(aq) + 3K2S(aq) Fe2S3(s) + 3K2SO4(aq) 2Fe3+(aq) + 3SO42-(aq) + 6K+(aq) + 3S2-(aq) Fe2S3(s) + 6K+(aq) + 3SO42-(aq) 2Fe3+(aq) + 3S2-(aq) Fe2S3(s) lead (II) nitrate and sulfuric acid Pb(NO3)2(aq) + H2SO4(aq) PbSO4(s) + 2HNO3(aq) Pb2+(aq) + 2NO3-(aq) + 2H+ + SO42-(aq) PbSO4(s) + 2H+(aq) + 2NO3-(aq) Pb2+(aq) + SO42-(aq) PbSO4(s) Stoichiometry of PrecipitationPractice Problems Exactly the same, except you may have to figure out what the pieces are. • What mass of solid is formed when 100.00 mL of 0.100 M barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide? • What volume of 0.204 M HCl is needed to precipitate the silver from 50.ml of 0.0500 M silver nitrate solution ? Stoichiometry of PrecipitationPractice Problems What mass of solid is formed when 100.00 mL of 0.100 M barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide? BaCl2(aq) + 2NaOH(aq) Ba(OH)2(s) + 2NaCl(aq) 0.01 mol 0.01 mol LR 0.01mol NaOH 1 mol Ba(OH)2 1 71 .35g Ba(OH)2 2 mol NaOH 1 mol Ba(OH)2 0.857g Ba(OH)2 Stoichiometry of PrecipitationPractice Problems • What volume of 0.204 M HCl is needed to precipitate the silver from 50.ml of 0.0500 M silver nitrate solution ? HCl(aq) + AgNO3(aq) HNO3(aq) + AgCl(s) ? mL 0.0025 mol 0.0025mol AgNO3 1 mol HCl 0.0025 mol HCl 1 mol AgNO3 0.0025mol 0.204 M xL volume .012L 12mL Types of Reactions Acid-Base For our purposes an acid is a proton donor. a base is a proton acceptor usually OHWhat is the net ionic equation for the reaction of HCl(aq) and KOH(aq)? Acid + Base salt + water H+ + OH- H2O Acid - Base Reactions Often called a neutralization reaction Because the acid neutralizes the base. Often titrate to determine concentrations. Solution of known concentration (titrant) is added to the unknown (analyte) until the equivalence point is reached where enough titrant has been added to neutralize it. Titration Where the indicator changes color is the endpoint. Not always exactly at the equivalence point, but should pick an indicator that changes color very close to the endpoint. # of H+ x MA x VA = # of OH- x MB x VB or NA x VA = NB x VB Titration Practice Problems A 50.00 mL sample of aqueous Ca(OH)2 requires 34.66 mL of 0.0980 M Nitric acid for neutralization. What is [Ca(OH)2 ]? NaVa = NbVb (0.0980N)(34.66mL)=(Nb)(50.00mL) Nb=0.06793 NCa(OH)2=(M)(2) [Ca(OH)2]=0.03397M Titration Practice Problems 75 mL of 0.25M HCl is mixed with 225 mL of 0.055 M Ba(OH)2. What is the concentration of the excess H+ or OH-? [H+] = 0.25 M [OH-] = 0.110 M 0.25mol 0.075L 0.01875mol H L 0.110mol 0.225L 0.02475mol OHL Excess [OH-] = 0.02475 mol – 0.01875 mol = 0.00600 mol [OH ]excess 0.00600mol 0.000200 M (.075L .225L) Types of Reactions Oxidation-Reduction (“Redox”) Ionic compounds are formed through the transfer of electrons An oxidation-reduction rxn involves the transfer of electrons We need a way of keeping track – oxidation states Oxidation States A way of keeping track of the electrons. Not necessarily true of what is in nature, but it works. need the rules for assigning • memorize these! Rules for assigning oxidation states The oxidation state of an atom in an element is zero (i.e. Na(s), O2(g), O3(g), Hg(l)) Oxidation state for monoatomic ions are the same as their charge. (i.e. Na+, Cl-) Oxygen is assigned an oxidation state of -2 in its covalent compounds except as a peroxide (such as H2O2). In compounds with nonmetals hydrogen is assigned the oxidation state +1. In its compounds fluorine is always –1. The sum of the oxidation states must be zero in compounds or equal the charge of the ion. Oxidation States Practice Assign the oxidation states to each element in the following. • CO2 +4 -2 • NO3+5 -2 -2 • H2SO4 +1 +6 -2 • Fe2O3 +3 -2 • Fe3O4 +8/3 -2 Oxidation-Reduction Electrons are transferred, so the oxidation states change. • 2Na + Cl2 2NaCl • CH4 + 2O2 CO2 + 2H2O 0 -4 +1 0 +1 -1 0 +4 -2 +1 -2 OIL RIG • • Oxidation is the loss of electrons. Reduction is the gain of electrons. LEO GER • • Losing electrons - oxidation Gaining electrons - reduction Oxidation-Reduction Oxidation means an increase in oxidation state - lose electrons. Reduction means a decrease in oxidation state - gain electrons. The substance that is oxidized is called the reducing agent. The substance that is reduced is called the oxidizing agent.