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Chapter 5 Molecular View of Reactions in Aqueous Solutions: Part 2
Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop
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Predicting Precipitation Reactions
Metathesis Reaction Reactions where anions and cations exchange partners. Also called Double replacement reaction Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq) Precipitation reactions Metathesis reactions where precipitate forms How can we predict if compounds are insoluble? Must know Solubility Rules
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Table 5.1 Solubility Rules
Soluble Compounds All salts of the alkali metals (Group IA) are soluble. All salts containing NH4+, NO3, ClO4, ClO3, and C2H3O2 are soluble. All chlorides, bromides, and iodides (salts containing Cl, Br, or I) are soluble except when combined with Ag+, Pb2+, and Hg22+ (note the subscript 2). All salts containing SO42 are soluble except those of Pb2+, Ca2+, Sr2+, Ba2+, and Hg22+.
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Table 5.1 Solubility Rules
Insoluble Compounds 5. All metal hydroxides (ionic compounds containing OH) and all metal oxides (ionic compounds containing O2 are insoluble except those of Group IA and those of Ca2+, Sr2+, and Ba2+. When metal oxides do dissolve, they react with water to form hydroxides. The oxide ion, O2, does not exist in water. For example: Na2O(s) + H2O 2NaOH(aq) 6. All salts containing PO43, CO32, SO32 and S2 are insoluble except those of Group IA and NH4+
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Learning Check: Solubility Rules
Which of the following compounds are expected to be soluble in water? Ca(C2H3O2)2 FeCO3 AgCl Yes No soluble : the substance will be aqueous (aq) if water is available. insoluble : substance will remain a solid (s). even in water.
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Your Turn Which of the following will be the solid product of the reaction of Ca(NO3)2 (aq) + Na2CO3 (aq) ? CaCO3 NaNO3 Na(NO3)2 Na2(NO3)2 H2O
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Metathesis (Double Replacement) Reaction
AB + CD AD + CB Cations and anions change partners Charges on each ion don’t change Formulas of products are determined by charges of reactants Occur only if solid, gas, weak electrolyte or non-electrolyte product forms Otherwise, all ions are spectator ions What is a metathesis reaction?
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Predicting Products of Double Replacement Reactions
Identify the ions involved: Distinguish between subscripts that count and those that are characteristic of a polyatomic ion. Swap partners and make neutral with appropriate subscripts Assign states using solubility rules Balance equation counting subscript Characteristic subscripts are the only subscripts that transfer to the product side. 2 HCl(aq)+ Ca(OH)2(aq) CaCl H2O (aq) ions: H+, Cl– Ca2+, 2OH –
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Predict if Ionic Reaction Occurs
Write molecular equation for metathesis reaction Determine which ion combinations form insoluble salt, water, weak electrolyte, or gas. Translate molecular equation into ionic equation Cancel spectator ions, to give net ionic equation Check for driving force: formation of weak electrolyte, solid, gas or non-electrolyte
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Learning Check: Predict Products
Pb(NO3)2(aq) + Ca(OH)2(aq) BaCl2(aq) + Na2CO3(aq) 2Na3PO4(aq) + 3Hg2(NO3)2(aq) 2 NaCl(aq) + Ca(NO3)2(aq) Pb(OH)2(s) + Ca(NO3)2(aq) BaCO3(s) + 2 NaCl (aq) 6 NaNO3(aq) + (Hg2)3(PO4)2(s) Note that Hg22+ is a polyatomic ion. NR (No reaction) CaCl2(aq) + 2 NaNO3(aq)
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Learning Check Predict the reaction that will occur when aqueous solutions of Cd(NO3)2 and Na2S are mixed. Write molecular, ionic and net ionic equations. Molecular Equation: Cd(NO3)2(aq) + Na2S(aq) Ionic Equation: Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq) Net Ionic Equation: Cd2+(aq) + S2–(aq) CdS(s) CdS(s) + 2NaNO3(aq) Dialog: Net ionic eqn. Let’s us generalize. Any soluble compound of Cd2+ and any soluble compound of S2– will react to form CdS. Cd(C2H3O2)2(aq) + (NH4)2S(aq) CdS(s) + 2NH4C2H3O2(aq) CdS(s) + 2NO3–(aq) Na+(aq)
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Learning Check Write molecular, ionic and net ionic equations for the reaction that occurs when Pb(NO3)2 and Fe2(SO4)3 are mixed in solution. Molecular Equation 3Pb(NO3)2(aq) + Fe2(SO4)3(aq) PbSO4(s) + 2Fe(NO3)3(aq) Ionic Equation 3Pb2+(aq) + 6NO3–(aq) + 2Fe3+(aq) + 6SO42–(aq) Fe3+(aq) + 6NO3–(aq) + PbSO4(s) Net Ionic Equation 3Pb2+(aq) + 6SO42–(aq) 3PbSO4(s) Pb2+(aq) + 2SO42–(aq) PbSO4(s) Must make lowest possible whole number coefficients 1 2 1
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Your Turn Will mixing aqueous solutions of Mg(C2H3O2)2 and CsCl yield a precipitate? Yes No Molecular Equation: Mg(C2H3O2)2(aq) + 2CsCl(aq) MgCl2(aq) CsC2H3O2(aq) Ionic Equation: Mg2+(aq) + 2C2H3O2–(aq) + 2Cs+(aq) + 2Cl–(aq) Mg2+(aq) + 2Cl–(aq) + 2Cs+(aq) + 2C2H3O2– (aq) All ions cancel. No Net Ionic Equation. No precipitate formed.
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2. Predicting Acid−Base Reactions
Neutralization reaction When mixed in 1:1 molar ratio, acid and base solutions lose their acidic and basic properties Combination of H+ and OH– to form salt and water Salt Ionic compound without H+, OH–, or O2– Acid Base Salt Water HClO4(aq) + NaOH(aq) NaClO4(aq) + H2O
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Neutralization Between Strong Acid and Strong Base
Molecular Equation 2HCl(aq) + Ca(OH)2(aq) → Ionic Equation 2H+(aq) + 2Cl–(aq) + Ca2+(aq) + 2OH–(aq) → H2O + Ca2+(aq) + 2Cl–(aq) Net Ionic Equation 2H+(aq) + 2OH–(aq) → 2H2O H+(aq) + OH–(aq) → H2O True for any strong acid and strong base 2H2O + CaCl2(aq)
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Weak Acid with Strong Base
Molecular Equation: HC2H3O2(aq) + NaOH(aq) → Ionic Equation: HC2H3O2(aq) + Na+(aq) + OH–(aq) → H2O + Na+(aq) + C2H3O2–(aq) Net Ionic Equation: HC2H3O2(aq) + OH–(aq) → H2O + C2H3O2–(aq) H2O + NaC2H3O2(aq) Write weak electrolytes in “molecular form”
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Neutralization of Strong Acid with Insoluble Base
Insoluble Hydroxides Molecular Equation Mg(OH)2(s) + 2HCl(aq) Ionic Equation Mg(OH)2(s) + 2H+(aq) + 2Cl–(aq) Mg2+(aq) + 2Cl–(aq) + 2H2O Net Ionic Equation Mg(OH)2(s) + 2H+(aq) Mg2+(aq) + 2H2O MgCl2(aq) + 2H2O Fig. 5.19
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Neutralization of Strong Acid with Insoluble Base
Insoluble Oxides – Basic Anhydrides Molecular Equation Al2O3(s) + 6HCl(aq) Ionic Equation Al2O3(s) + 6H+(aq) + 6Cl–(aq) 2Al3+(aq) + 6Cl–(aq) + 3H2O Net Ionic Equation Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O 2AlCl3(aq) + 3H2O
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Strong Acid with Weak Base
Molecular Equation: NH3(aq) + HCl(aq) Ionic Equation : NH3(aq) + H+(aq) + Cl–(aq) NH4+(aq) + Cl–(aq) Net Ionic Equation : NH3(aq) + H+(aq) NH4+(aq) NH4Cl(aq) Neutralization doesn’t always form water. Fig 5.20 A hydronium ion transfers a hydrogen ion to an ammonia molecule. The products are ammonium ion and water NH3(aq) + H3O+(aq) NH4+ (aq) + H2O
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Challenge Problem What is the Net Ionic Equation for reaction of an insoluble hydroxide and a weak acid? Molecular Equation Mg(OH)2(s) + 2HC2H3O2(aq) Ionic Equation Mg(OH)2(s) + 2HC2H3O2(aq) Mg2+(aq) + 2H2O C2H3O2–(aq) There are NO spectator ions! So Net Ionic and Ionic Equations are the same in this case Mg(C2H3O2)2(aq) + 2H2O
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Learning Check HNO3(aq) + Ca(OH)2(aq) H+(aq) + OH–(aq) H2O
What is the net ionic equation for the reaction between the following reactants: HNO3(aq) + Ca(OH)2(aq) H+(aq) + OH–(aq) H2O 2. N2H4(aq) + HI(aq) N2H4(aq) + H+(aq) N2H4+(aq) 3. CH3NH2(aq) + HC4H7O2(aq) CH3NH2(aq) + HC4H7O2(aq) CH3NH3+(aq) C4H7O2–(aq) HCHO2 is a weak acid N2H4 is a weak base
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Your Turn Which is the net ionic equation for the reaction:
NaOH(aq) + HF(aq) ? Na+(aq) + OH–(aq) + H+(aq) + F–(aq) H2O NaF(aq) OH–(aq) + H+(aq) H2O Na+(aq) + OH–(aq) + HF(aq) H2O + NaF(aq) OH–(aq) + HF(aq) H2O + F–(aq) No reaction
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Metathesis and Gas Formation
Metathesis reactions involving certain ions lead to formation of a gas. Low solubility of gas in solvent (water) leads to escape of gas. Once escaped, can’t go back in, drives reaction to completion Many anions that give rise to gases are insoluble Dissolving in acid followed by gas formation drives reaction to completion Dissolves insoluble salt
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Metathesis and Gas Formation
Gases formed by metathesis H2S, HCN Unstable compounds—decompose and form gas H2CO3 H2O and CO2(g) H2SO3 H2O and SO2(g) NH4OH H2O and NH3(g) The reactions are driven to completion because the gas escapes and is unavailable for back reaction
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Reactions that Release CO2
b) Acid with Bicarbonate (HCO3–) NaHCO3(aq) + HI(aq) NaI(aq) + H2O + CO2(g) Acid with Carbonate (CO32–) CaCO3(aq) + 2HCl(aq) CaCl2(aq) + H2O + CO2(g) Fig 5.22 and 5.23
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Reactions that Release Gases
Acid with Sulfites (SO32–) or Bisulfites (HSO32–) K2SO3(aq) + 2HClO4(aq) SO2(g) + 2KClO4(aq) + H2O LiHSO3(aq) + HClO3(aq) SO2(g) + H2O + LiClO3(aq) Acid with Sulfides 2HCl(aq) + Na2S(aq) 2NaCl(aq) + H2S(g) Acid with Cyanides HNO3(aq) + CsCN(aq) HCN(g) + CsNO3(aq) Bases with Ammonium NaOH(aq) + NH4Cl(aq) NH3(g) + H2O + NaCl(aq)
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Learning Check Write the molecular, ionic and net ionic equations for the reaction of Li2SO3 with formic acid, HCHO2 Molecular Equation: Li2SO3(aq) + 2HCHO2(aq) 2LiCHO2(aq) + SO2(g) + H2O Ionic Equation: 2Li+(aq) + SO32–(aq) + 2HCHO2(aq) 2CHO2–(aq) + 2Li+(aq) + SO2(g) + H2O Net Ionic Equation: SO32 –(aq) + 2HCHO2(aq) 2CHO2–(aq) + SO2(g) + H2O
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Your Turn What is the net ionic equation for the reaction of HCl with KHCO3? HCl(aq) + KHCO3(aq) KCl(aq) + H2CO3(aq) H+(aq) + HCO3–(aq) H2CO3(aq) HCl(aq) + KHCO3(aq) KCl(aq) + CO2(g) + H2O H+(aq) + Cl–(aq) + K+(aq) + HCO3–(aq) K+(aq) + Cl–(aq) + CO2(g) + H2O H+(aq) + HCO3–(aq) CO2(g) + H2O
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Metathesis Overview Precipitation: Neutralization: Gas-forming:
2 solutions form solid product Neutralization: Acid + metal hydroxide or oxide forms water and salt Gas-forming: Metathesis reaction forms one of these products: HCN, H2S, H2CO3(aq) , H2SO3(aq) , NH4OH(aq) Formation of Weak Electrolyte: Salt of weak acid reacts with base to form molecule
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Predicting Reactions and Writing Their Equations
What reaction, if any, occurs between potassium nitrate and ammonium chloride? Need to know whether net ionic equation exists. Determine formulas of reactants KNO3 + NH4Cl ? Write molecular equation KNO3 + NH4Cl KCl + NH4NO3 Check Solubilities All are soluble
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Predicting Reactions and Writing Their Equations
Predicted Molecular Equation KNO3(aq) + NH4Cl(aq) KCl(aq) + NH4NO3(aq) Write Ionic Equation K+(aq) + NO3–(aq) + NH4+(aq) + Cl–(aq) K+(aq) + Cl–(aq) + NH4+(aq) + NO3–(aq) Same on both sides All ions cancel out No gases, solids, water, or weak electrolytes formed
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Learning Check Determine the net ionic equation for the following reactions. Co(OH)2 + HNO2 Co(OH)2(s) + 2H+(aq) Co2+(aq) + 2H2O KCHO2 + HCl CHO2–(aq) + H+(aq) HCHO2(aq) CuCO3 + HC2H3O2 CuCO3(s) + 2HC2H3O2(aq) Cu2+(aq) + CO2(g) C2H3O2–(aq) + H2O
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Your Turn What is the net ionic reaction when aqueous solutions of NaOH and NiCl2 are mixed? Ni2+(aq) + 2OH–(aq) Ni(OH)2(s) NaOH(aq) + NiCl2(aq) NaCl(aq) + Ni(OH)2(s) 2NaOH(aq) + NiCl2(aq) 2NaCl(aq) + Ni(OH)2(s) 2Na+(aq) + 2OH–(aq) + Ni2+(aq) + 2Cl–(aq) Na+(aq) + 2Cl–(aq) + Ni(OH)2(s) No reaction.
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Your Turn Which of the following combinations will not react?
Na2CO3(aq) + HCl(aq) Na2SO3(aq) + CaCl2(aq) NaCl(aq) + HC2H3O2(aq) NH4Cl(aq) + HClO4(aq) KCN(aq) + H2SO4(aq) Yields a gas (CO2) Yields a gas (SO2) All ions are soluble, Acetic acid is a weak acid. Yields a gas (NH3) Yields a gas (HCN)
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Synthesize by Metathesis Reaction
Practical use of metathesis reactions Desired compound should be easily separated from reaction mixture. Two principal approaches Desired compound is insoluble in water Start with 2 soluble reactants Product isolated by filtration Desired compound is soluble in water Acid-base neutralization Reaction of metal carbonate and acid Either way, product isolated by evaporation of water Uncontaminated by Any reactant starting materials, or Any other reaction product Cation supplied by base. Anion supplied by acid Cation from metal carbonate. Anion from acid
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Learning Check What reaction might we use to synthesize nickel sulfate, NiSO4? Use solubility rules NiSO4 is soluble in water So, there are 2 possible methods Use acid + base H2SO4(aq) + Ni(OH)2(s) NiSO4(aq) + 2H2O Use acid + carbonate H2SO4(aq) + NiCO3(s) NiSO4(aq) + CO2(g) + 2H2O
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Molar Concentration Dissolve solutes. Make separate solutions.
Mix Solutions. Allow Reaction to occur. Need to know Quantitatively HOW MUCH? of each solute we used. Define 5.6 Molarity
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Molarity (M) Number of moles of solute per liter of solution.
Allows us to express relationship between moles of solute and volume of solution Hence, 0.100M solution of NaCl contains mole NaCl in 1.00 liter of solution Same concentration results if you dissolve mol of NaCl in liter of solution moles solute (n)=Volume (L)*Molarity (M)
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Molarity as Conversion Factor
Often have stoichiometry problems involving amount of chemical and volume of solution Solve the problem using molarity Molarity provides conversion factors between moles and volume M = mole per liter Molarity Moles of a Substance Volume of a solution of Substance
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Molarity as Conversion Factor
Gives equivalence relationship between “mol NaCl” and “L soln” Forms two conversion factors Three basic types of calculations: 0.100 mol NaCl 1.00 L soln
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Learning Check: Calculating Molarity (from grams and volume)
Calculate the molarity (M) of a solution prepared by dissolving 11.5 g NaOH (40.00 g/mol) solid in enough water to make 1.50 L of solution. g NaOH mol NaOH M NaOH
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Learning Check: Calculating Volume (from Molarity and moles)
How many mL of M NaCl solution are needed to obtain mol of NaCl? Use M definition Given molarity and moles, need volume = 400 mL of M NaCl solution
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Preparing Solution of Known Molarity
a) b) c) d) e) a b c d e Weigh solid and transfer to volumetric flask Add part of the water Dissolve solute completely Add water to reach etched line Stopper flask and invert to mix thoroughly Fig 5.24
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Learning Check: Preparing Solution of Known Molarity from Solid
How many grams of strontium nitrate are required to prepare mL of M Sr(NO3)2 solution? M × V mol × MM g Convert Molarity and Volume to mole = mol Sr(NO3)2 Convert mol to g Strontium nitrate, Sr(NO3)2, is used in fireworks to produce brilliant red colors. Weigh out 5.29 g Sr(NO3)2 and dilute to mark. = 5.29 g Sr(NO3)2
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Your Turn How many grams of KMnO4 must you weigh out if you want to make 250.mL of a M KMnO4 solution? 7900 g 50.0 g 0.316 g 7.90 g 198 g M mol g Can do all conversions in series of steps. Weigh out 7.90 g KMnO4 & dilute to mark.
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Preparing a Solution of Known Molarity by Dilution
When making solutions, don’t always begin with solute as pure solid, liquid or gas. Can take solution of high concentration and dilute to a lower concentration. Amount of MOLES does NOT change! Remains the same Small Volume Concentrated Solution Large Volume Dilute Solution Add lots of water
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Diluting Solutions Vdil Mdil = Vconc Mconc
Moles of solute do not change upon dilution Just changing volume # moles in dilute = # moles in concentrated Moles of solute in the dilute solution Moles of solute in the concentrated solution Vdil Mdil = Vconc Mconc
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Learning Check: Dilutions
What volume (in mL) of 16.0 M H2SO4 must be used to prepare 1.00 L of M H2SO4? Rearranging gives = 125 mL
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Preparing Solution of Known M
Using volumetric glassware ensures that the volumes are known precisely Use a volumetric pipette to transfer the stock solution Use a volumetric flask to receive the final solution Fig 5.27
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Your Turn What volume of 12.1 M HCl is needed to create mL of 3.2 M HCl? 66 mL 800 mL 3025 mL 945 mL 9680 mL Vconc = 66 mL
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Your Turn 25 mL of 6.0 M HCl are diluted to 500 mL with water. What is the molarity of the resulting solution? 150 M 3.0 M 0.120 M 120 M 0.30 M Mdil = 0.30 M
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Solution Stoichiometry
Often work with solutions when running reactions How do we determine amounts needed to completely react one compound? Like any other stoichiometry problem Now use Volume and Molarity to obtain moles of each substance. 5.7 Solution Stoichiometry
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Learning Check: Solution Stoichiometry
How many milliliters of M H3PO4 could be completely neutralized by 45.0 mL of M KOH? The balanced equation for the reaction is H3PO4(aq) + 3KOH(aq) K3PO4(aq) + 3H2O Strategy: Coefficients of Balanced equation mol KOH mol H3PO4 mol and M of H3PO4 soln Vol and M of KOH soln KOH solution H3PO4 soln
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Learning Check: Solution Stoichiometry
1. Calculate moles of KOH 2. Use coefficients to calculate the moles H3PO4 required 3. Calculate volume of H3PO4 needed = 4.50 × 10–3 mol KOH = 1.50 × 10–3 mol H3PO4 = 31.6 mL H3PO4
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Stoichiometry of Ionic Equations
Calculating Concentrations of Ions in Solutions of Electrolytes. When using solutions of electrolytes, need to know [ions] in solution. Ions don’t stay together, [ions] may not be same Can easily calculate from M electrolyte in molecular form. Concentration of particular ion equals concentration of salt multiplied by number of ions of that kind in one formula unit of salt. Note: [XX] = concentration of whatever species is in square bracket
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Learning Check: Ion Concentrations
If you have M Na2CO3 (aq), what is the concentration of each type of ion in solution? Means Na2CO3(aq) 2 Na+(aq) + CO32–(aq) Concentration of Na+ ions is: Concentration of CO32– ions is:
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Your Turn If the solution concentration of sulfate ion is M, what is the concentration of Al2(SO4)3, assuming that all of the sulfate ion comes aluminum sulfate? 0.750 M 2.25 M 0.250 M 1.50 M 0.500 M = M Al2(SO4)3
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Learning Check: Net Ionic Eqns in Solution Stoichiometry Calculations
What volume, in mL, of M KOH is needed to react completely with 60.0 mL of M FeCl2 to form Fe(OH)2 solid? 1. Write Balanced Net Ionic Equation Fe2+(aq) OH–(aq) Fe(OH)2(s) 2. Determine the game plan mol Fe2+ mol OH– M Fe2+ V OH– M FeCl2 V KOH
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Learning Check: Net Ionic Eqns in Solution Stoichiometry Calculations
3. Convert M FeCl2 M Fe2+ mol Fe2+ 4. Convert mol Fe2+ mol OH– 5. Convert mol OH– V OH– V KOH = mol Fe2+ = 60.0 mL KOH
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Learning Check: Solution Limiting Reagent Problem
How many grams of PbI2 (461.0) will form if mL of M FeI3 (436.5) is mixed with 50.0 mL of M Pb(NO3)2 (269.2)? 3Pb(NO3)2(aq) + 2FeI3(aq) 3PbI2(s) + 2Fe(NO3)3(aq) Net Ionic Eqn: Pb2+(aq) + 2I(aq) PbI2(s) Strategy Vol Pb(NO3)2 mol Pb(NO3)2 mol Pb2+ mol PbI2 g PbI2 Vol FeI3 mol FeI3 mol I mol PbI2 g PbI2 Whichever is less determines how much is formed and which reagent is limiting
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Limiting Reagent Problem
Starting with Pb(NO3)2 Starting with FeI3 = 6.92 g PbI2 = g PbI2 Pb(NO3)2 is limiting and only 6.92 g of PbI2 can be made.
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Chemical Analysis Qualitative Analysis Quantitative Analysis
What substances are present in a sample Quantitative Analysis Measure the amounts of various substances in a sample Chemical reactions useful strategy Convert all of an element present in a sample into a substance of known formula Use the amount of this known to determine amount of element present in the orginial sample (unknown) 5.8 Chemical Analysis and Titration
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Learning Check: Chemical Analysis
Insecticide contains H, C, Cl. Carry out reactions on g sample to convert all chlorine to Cl– in H2O. Resulting solution treated with AgNO3 solution in excess. AgCl ppts and is collected, dried, and weighed. Mass AgCl = g. What is the percentage by mass of Cl in original sample? Strategy: g AgCl mol AgCl mol Cl g Cl
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How Much Cl in 2.022 g of AgCl? g AgCl mol AgCl mol Cl g Cl
% Cl in original Sample? = 50.02% Cl
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Titrations Widely used analytical technique
Used to determine concentration of solute Used daily to monitor: Water purity Quality control in food industry How it Works: Must know precise reaction that occurs Reaction must be rapid and complete Must know exact quantity of one reactant Use stoichiometry to find exact amount of any other substance in solution
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Titration Controlled addition of 1 reactant to known quantity of another until reaction is complete Acid-Base Titration Very common type of titration Ex. Analysis of citric acid in orange juice by neutralization with NaOH Know MNaOH and measure exact VNaOH needed to completely neutralize citric acid MNaOH × VNaOH = mol NaOH mol NAOH mol citric acid mol citric acid × MM citric acid = g citric acid
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Titration in practice:
Buret Volumetric measuring device with 0.10 mL markings Stopcock Permits flow of titrant to stop when reaction is complete Volume titrant used = Vf –Vi Figure 5.26
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Titration: Definitions
Titrant Solution in the buret Known concentration Can be either acid or base depending on nature of the analyte Analyte Solution being analyzed Solution in flask Solution of unknown concentration Equivalence Point Volume of titrant where moles of titrant and moles of analyte are exactly equal .
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Titration Indicator Endpoint:
Dye that is 1 color in acid and 2nd color in base Ex. phenolphthalein Colorless in acid and bright pink in base Color change signals end point of titration Endpoint: Volume of titrant required to complete reaction monitored by color change of indicator Choose indicator so endpoint and equivalence point are the same
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Learning Check Suppose that mL of a solution of oxalic acid, H2C2O4, extracted from rhubarb leaves, is titrated with M NaOH(aq) and that the stoichiometric point is reached when 37.5 mL of the solution of base is added. What is the molarity of the oxalic solution? Step 1: Write the balanced equation. H2C2O4(aq) + 2NaOH(aq) Na2C2O4(aq) + 2H2O
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Learning Check: Oxalic acid + NaOH
Step 2: Calculate moles of Base used Step 3: Calculate moles of Oxalic Acid Step 4: Calculate M H2C2O4 = M H2C2O4
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Summary of Stoichiometry Calculations
NA
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Your Turn 25.00 mL of HNO3 are titrated with mL of M Ca(OH)2. What is the concentration of HNO3 in the initial sample? 2HNO3(aq) + Ca(OH)2(aq) 2AgBr(s) + Ca(NO3)2(aq) 0.433 M 1.95 M 0.867 M 3.90 M 7.80 M = 7.80 M HNO3
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Your Turn A sample of metal ore is reacted according to the following reaction: Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g) If mL of 2.3 M HCl are used, what mass of Fe was in the ore? (Atomic mass of Fe is g/mol) 0.515 g 1.03 g 1.21 g 1.61 g 3.20 g = 1.61 g
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