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Topic 2. Solution Stoichiometry and Chemical 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. Water, the Common Solvent.
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Topic 2 Solution Stoichiometry and Chemical 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.
Water, the Common Solvent • The O-H bonds in water are covalent (electrons are shared) • 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 105ºC.
Hydration • The process of breaking the ions of salts apart. • Negative oxygen attracts cations (+) • Positive hydrogen attracts anions (-)
H H H H O O O H H H H O O H H O O H H H H H H O H O H Hydration
Solubility • How much of a substance will dissolve in a given amount of water. • Usually g/100 mL • Varies greatly, but if they dissolve, the ions are separated and they can move around. • Water can also dissolve non-ionic compounds if they have polar bonds, such as ethanol (C2H5OH).
The Nature of Aqueous SolutionsElectrolytes • Electricity is moving charges. • The ions that are dissolved can move. • Solutions of ionic compounds, such as NaCl, can conduct electricity. • These solutions are said to contain electrolytes. • Solutions are classified three ways.
Types of solutions • (1) Strong electrolytes- completely dissociate (break apart into ions). • Many ions, conduct well. • (2) Weak electrolytes- Partially dissociate. • Few ions, conduct electricity slightly. • (3) Non-electrolytes- Do not dissociate. • No ions, does not conduct electricity.
Strong/Weak - Acids/Bases • Acids- form H+ ions when dissolved. • Strong acids dissociate completely. • Produces many H+ ions • H2SO4 HNO3 HCl HBr HI HClO4 • Weak acids don’t dissociate completely. • Bases form OH- ions when dissolved. • Strong bases dissociate completely and form many ions. • KOH NaOH
The Composition of Solutions • Concentration - how much is dissolved. • Molarity = Moles of solute Liters of solution • abbreviated M • 1 M = 1 mol solute / 1 liter solution • Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution.
Calculation of Molarity I • Calculate the molarity of a solution perpared by dissolving 11.5g of solid NaOH in enough water to make 1.50 L of solution.
Calculation of Molarity II • Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.
Concentration of Ions • Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2.
Making solutions • Describe how to make 1.0 L of an aqueous 0.200 M K2Cr2O7 solution. • Describe how to make 250. mL of an aqueous 2.0 M copper (II) sulfate dihydrate solution.
Dilution • A stock solution is a concentrated solution of known concentration used to make more dilute solutions • If we adding more solvent to a known stock solution the moles of solute stay the same. • moles (before dilution) = moles (after dilution) • moles = M x V • therefore, M1V1 = M2V2
Dilution • What volume of a 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? • You have a 4.0 M stock solution. Describe how to make 1.0L of a .75 M solution.
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
Precipitation Reactions • Only happen if one of the products is insoluble • Otherwise all the ions stay in solution- nothing has happened. • Need to memorize the rules for solubility of salts in water (p. 144)
Solubility Rules • Most nitrate (NO3-) salts are soluble • Most salts containing the alkali metal ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium ion (NH4+) are soluble • Most chloride, bromide, and iodide salts are soluble. Exceptions are salts with Ag+, Pb+2, and Hg2+2 • Most sulfate salts are soluble, except salts with Pb+2, Ba+2, Hg+2,and Ca+2
Solubility Rules • Most hydroxides 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. • Sulfides (S2-), carbonates (CO32-), chromates (CrO42-), and phosphates (PO43-) are only slightly soluble (insoluble). • Lower number rules supersede so Na2S is soluble
Precipitation reaction • We can predict the products • Can only be certain by experimenting • The anion and cation switch partners • AgNO3 (aq) + KCl (aq) ® • Zn(NO3)2 (aq) + BaCr2O7 (aq) ® • CdCl2 (aq) + Na2S (aq) ®
Three Types of Equations • Formula (molecular) equation – gives the overall reaction stoichiometry • NaOH(aq) + FeCl3 (aq) ® NaCl(aq) + Fe(OH)3 (s) • Complete ionic equation – represents as ions all reactants and products that are strong electrolytes • Na+(aq) + OH-(aq)+ Fe+3 + Cl-(aq)® Na+(aq)+ Cl-(aq) + Fe(OH)3 (s) (spectator ions are those that don’t react) • Net ionic equation – includes only those components that are part of the reaction • OH- (aq) + Fe+3® Fe(OH)3 (s)
Writing Equations (Ex. 4.9) • Write the three types of equations for the reactions below: • A) Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate. • B) Aqueous potassium hydroxide is mixed with aqueous iron (III) nitrate to form a precipitate of iron (III) hydroxide and aqueous potassium nitrate.
Stoichiometry of Precipitation Reactions • Steps • Identify the species present and determine what reaction occurs • Write the balanced equation • Calculate moles of each reactant • Determine the limiting reactant • Calculate the moles of product • Convert to grams or other units
Determine Mass of Product formed (Ex. 4.11) • Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.
2. Acid Base Reactions Arrhenius: • Acid – substance produces H+ ions • HCl (aq) → H+(aq) + Cl-(aq) • Base – substance produces OH- ions • NaOH (aq) → Na+(aq) + OH-(aq) • What is the net ionic equation for the reaction of HCl (aq) and NaOH (aq) ? • Acid + Base ® salt + water • H+ + OH-® H2O
Acid Base Reactions Bronsted-Lowry: • Acid – is a proton donor • Base – is a proton accept • NH3(aq) + H2O (l) → NH4+(aq) + OH-(aq)
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.
Steps to Calc Acid Base Rxns • List the Species. • Write the balanced net ionic equation. • Change the given quantities of reactants to moles. • Determine the limiting reactant • Calculate the moles of reactant or product • Convert to grams or volume of solution
Neutralization Reaction • What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? • In a certain experiment, 28.0 mL of 0.250 M HNO3 and 53.0 mL of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction.
Acid-BaseTitrations • A titration involves delivery of a measured volume of solution of known concentration (titrant) into a solution of unknown concentration (analyte). • Equivalence point – where the titrant reacts exactly with the analyte. • An indicator is used which changes color at or near the equivalence point. • The point where the indicator actually changes color is the end point.
Neutralization Titration • A student weighs out a 1.3009-g sample of potassium hydrogen phthalate (KHC8H4O4 or KHP). KHP (molar mass 204.22 g/mol) has one acidic hydrogen. • The student dissolves the KHP in distilled water, adds phenolphthalein (phen) as an indicator, and titrates the resulting solution with an NaOH solution to the endpoint. • A volume of 41.20 mL of NaOH solution is required to react exactly with the 1.3009 g KHP. Calculate the concentration of the NaOH solution.
3. Oxidation-Reduction reaction • Called a redox reaction • An Oxidation-reduction reaction involves the transfer of electrons. • How do we determine if electrons are being transferred?
Oxidation States • Provide a way of keeping track of the electrons in redox rxns. • Not necessarily true of what is in nature, but it works. • Memorize the rules for assigning: • An atom in an element (in its standard state) is zero. Na(s), O2(g), Hg(l) • A monatomic ion is the same as its charge. Na+, Cl-
Oxidation states • Fluorine is -1 in its compounds. • Oxygen is usually -2 in its compounds. (Except peroxides containing O22- in which oxygen is -1). • Hydrogen is +1 in its covalent compounds. • The sum of the oxidation states must be zero in compounds or equal to the charge of the ion.
Oxidation States • Assign the oxidation states to each element in the following. • CO2 • SF6 • NO3- • H2SO4 • KMnO4
Oxidation-Reduction • Transfer electrons, so the oxidation states change. • 2Na + Cl2® 2NaCl • CH4 + 2O2® CO2 + 2H2O • Oxidation is the loss of electrons. • Reduction is the gain of electrons. • OIL RIG • LEO GER
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.
Identify the • Substance oxidized • Substance reduced • in the following reactions • Fe(s) + O2(g)® Fe2O3(s) • Fe2O3(s)+ 3 CO(g)® 2 Fe(l) + 3 CO2(g) • SO32-(aq) + H+(aq) + MnO4- (aq) ® SO42- (aq) + H2O (l) + Mn+2(aq)
Half-Reactions • All redox reactions can be thought of as happening in two halves. • One produces electrons - Oxidation half. • The other requires electrons - Reduction half. • Write the half reactions for the following. • Na + Cl2® Na+ + Cl- • SO32- + H+ + MnO4- ® SO42- + H2O + Mn+2
Balancing Redox Equations • In aqueous solutions the key is the number of electrons produced must be the same as those required. • For reactions in acidic solution an 8 step procedure. • Write separate half reactions • For each half reaction balance all reactants except hydrogen and oxygen • Balance oxygen using H2O
Balancing Redox Equations • Balance hydrogen using H+ • Balance the charge using electrons ( e- ) • Multiply half-reactions by an integer to make electrons equal • Add half-reactions and cancel identical species • Check that charges and elements are balanced.
Practice • The following reactions occur in aqueous solution. Balance them • MnO4- + Fe+2® Mn+2 + Fe+3 • H++ Cr2O7 + C2H5OH → Cr3++ CO2 + H2O • Cr(OH)3 + OCl- + OH-® CrO4-2 + Cl- + H2O • Cu + NO3- ® Cu+2 + NO(g) • Pb + PbO2 + SO4-2® PbSO4 • Mn+2 + NaBiO3® Bi+3 + MnO4-
Basic Solution • Do everything you would with acid, but add one more step. • Add enough OH- to both sides to neutralize the H+ • CrI3 + Cl2® CrO4-+ IO4- + Cl- • Fe(OH)2 + H2O2® Fe(OH)-