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Section 6.4—Solubility & Precipitation How can we make sure everything that’s added to the sports drink will dissolve? A Review of DoubleReplacement Reactions Double Replacement Reactions The cations from two compounds replace each other. NaCl + AgNO3 AgCl + NaNO3 Two ionic compounds switch ions Double Replacement Reactions General format of a double replacement reaction: A X B Z A Z B X Products of a Double Replacement 1 CaCl2 Combine the cation of the first reactant with the anion of the second reactant + AgNO3 Products of a Double Replacement 2 CaCl2 Combine the cation of the second reactant with the anion of the first reactant + AgNO3 Products of a Double Replacement 3 Remember to write cations first … & balance charges with subscripts when writing formulas Only leave subscripts that are in the original compound there if they are a part of a polyatomic ion! CaCl2 CaCl2 + AgNO3 + AgNO3 Ca(NO3)2 + AgCl Precipitation Reactions Precipitation Reactions A precipitation reaction is when 2 soluble substances are mixed together and they form an insoluble substance. This is called a precipitate. Reactants 2 soluble chemicals: NaOH Cu(NO ) NaOH and Cu(NO3)2 3 2 Precipitation Reactions(DR Rxns) Na+1 OH-1 Cu+2 NO3 -1 Cu(OH)2(S) Products: 1 soluble chemical: NaNO3 1 insoluble chemical (the precipitate): Cu(OH)2 Na+1 NO3 -1 Solubility Rules Solubility Rules Table Use the table on the reference sheet! Insoluble = Precipitate Let’s Practice #1 NaNO3 Example: Decide whether each is soluble or not Fe(C2H3O2)2 CaBr2 Ba(OH)2 Cu(OH)2 Let’s Practice #1 Example: Decide whether each is soluble or not NaNO3 Soluble Fe(C2H3O2)2 Soluble CaBr2 Soluble Ba(OH)2 Soluble Cu(OH)2 Not Soluble Let’s Practice #2 Example: Write the products for this reaction & predict the precipitate Remember to indicate compounds that dissolve with “aq” for “aqueous” and compounds that don’t dissolve with “s” for “solid” AgNO3 (aq) + NaCl (aq) Let’s Practice #2 Example: Write the products for this reaction AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) precipitate Let’s Practice #3 Example: Write the products for this reaction & identify the precipitate Remember to indicate compounds that dissolve with “aq” for “aqueous” and compounds that don’t dissolve with “s” for “solid” BaCl2 (aq) + K2CO3 (aq) Let’s Practice #3 Example: Write the products for this reaction BaCl2 (aq) + K2CO3 (aq) KCl (aq) + BaCO3 (s) precipitate Net Ionic Reactions Shows the details of aqueous reactions that involve ions in aqueous solution Molecular Equation: the typical equation you are use to writing keeping all molecules together Complete Ionic Equation: shows all the particles in a solution as they really exist, as IONS or MOLECULES. Anything aqueous needs to be split apart into the cation and anion Anything solid stays intact Coefficients need to be multiplied by subscripts to determine the exact amount of each cation and anion. Spectator ions: ions that do not participate in a reaction; they are identical on both sides of the equation & are crossed out! Net Ionic Equation: the final equation showing the major players. All spectator ions have been removed. NET IONIC REACTIONS for Precipitation Reactions Molecular equation: KI(aq) + AgNO3(aq) AgI(s) + KNO3(aq) Complete Ionic equation: K+1 + I-1 + Ag+1+ NO3-1 AgI + K+1 + NO3-1 Spectator ions: ions that do not participate in a reaction; they are identical on both sides of the equation & are crossed out! Net Ionic equation: I-1 + Ag+1 AgI NET IONIC REACTIONS for Precipitation Reactions Molecular equation: 2 NaOH(aq) + CuCl2(aq) 2 NaCl(aq) + Cu(OH)2(s) Complete Ionic equation: 2 Na+1 + 2 OH-1 + Cu+2 + 2 Cl-1 2 Na+1 + 2 Cl-1 + Cu(OH)2 Net Ionic equation: 2 OH-1 + Cu+2 Cu(OH)2 Take Home Practice: Predict products and balance Iron (III) chloride reacts with sodium hydroxide Molecular equation: 1 FeCl3(aq) + 3 NaOH(aq) 1 Fe(OH)3(s) +3 NaCl(aq) Complete Ionic equation: 3 Na+1 + 3 OH-1 + Fe+3 + 3 Cl-1 3 Na+1 + 3 Cl-1 + Fe(OH)3 Net Ionic equation: 3 OH-1 + Fe+3 Fe(OH)3 Section 6.5—Stoichiometry How can we determine in a lab the concentration of electrolytes? What do those coefficients really mean? The coefficient of the balanced chemical equation tells how many moles of each substance is used in the reaction. For every 2 moles of H2… 2 2 and 2 moles of H2O are produced 2 H2 + O2 2 H2O No coefficient = 1 1 mole of O2 is need to react… Mole Ratio Is a conversion factor that relates 2 substances in moles; must use a balanced chemical equation to create it 2 H2 + O2 2 H2O Examples of Mole Ratio’s 2mol H2 1 mol O2 1 mol O2 2 mol H2O 2 mol H2O 2 mol H2 What is stoichiometry? Stoichiometry – Calculations using the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the equation. Example: What is the mole ratio of chlorine to sodium? 2 Na + Cl2 2 NaCl 2mol Na 1 mol Cl2 1 mol Cl2 2 mol NaCl 2 mol Na 2 mol NaCl Stoich (Mole-Mole) : 1 step problem using the mole ratio Example: If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed? 2 H2 + O2 2 H2O Stoichiometry with Moles Example: If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed? 2 H2 + O2 2 H2O 4.2 mole H2 1 mole O2 2 mole H2 From balanced equation: 2 mole H2 1 mole O2 = ________ 2.1 mole O2 Stoich (Mole-Mole) Example: If 0.67 moles of potassium nitrate reacts, how many moles of oxygen are produced? 2KNO3 2KNO2 + O2 0.67 mole KNO3 1 mole O2 2 Mole KNO3 From balanced equation: 2 mole KNO3 1 mole O2 = ________ 0.34 mole O2 But we can’t measure moles in lab! We can’t go to the lab and count or measure moles…so we need a way to work in measurable units, such as grams and liters! Molecular mass gives the grams = 1 mole of a compound! Stoich( Mole-Mass): 2 step problem use mole ratio & then molar mass conversion factors Example: How many grams of AgCl will be precipitated if 0.45 mole AgNO3 is reacted as follows: 2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2 Stoichiometry with Moles & Mass Example: How many grams of AgCl will be precipitated if 0.45 mole AgNO3 is reacted as follows: 2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2 From balanced equation: 2 mole AgNO3 2 mole AgCl Molar Mass of AgCl: 1 mole AgCl = 143.32 g 0.45 mole AgNO3 2 mole AgCl 2 mole AgNO3 143.32 g AgCl 1 mole AgCl = ________ 64 g AgCl Stoich( Mass- Mol): 2 step problem use molar mass & then mole ratio conversion factors Example: If 4.42 g of H2 reacts, how many moles of NH3 are produced? N2 + 3H2 2NH3 4.42 g H2 1 2.02 Mole H2 g H2 From balanced equation: 3 mole H2 2 mole N2 2 mole NH3 3 mole H2 =1.46 mole NH3 Stoich( Mass-Mass): 3 step problem use molar mass, then mole ratio & then molar mass conversion factors (Honors Only) Example: How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction: 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl Stoichiometry with Mass (Honors) Example: How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction: 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl Molar Mass of NaOH: 1 mole NaCl = 40.00 g From balanced equation: 2 mole NaOH 1 mole Ba(OH)2 Molar Mass of Ba(OH)2: 1 mole Ba(OH)2 = 171.35 g 14.5 g NaOH 1 mole NaOH 40.00 g NaOH 1 mole Ba(OH)2 171.35 g Ba(OH)2 2 mole NaOH 1 mole Ba(OH)2 31.1 = ________ g Ba(OH)2 Stoichiometry with Mass (Honors) Example: How many grams of HCl are needed to produce 65.0 g of magnesium chloride: __Mg + ____HCl ____MgCl2 + __H2 Balance the equation and fill in the missing information: 65 g MgCl2 1 mole MgCl2 g MgCl2 mole HCl mole MgCl2 g HCl mole HCl = g HCl What about the stoichiometry of gases? Recall Molar Volume of a Gas – at STP 1 mole of any gas = 22.4 liters Stoichiometry with Gases 1 mol= 22.4 L @STP Example: If you need react 1.5 g of zinc completely, what volume of hydrogen gas will be produced at STP? 2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g) Stoichiometry with Gases Example: If you need react 1.5 g of zinc completely, what volume of hydrogen gas will be produced at STP? 2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g) Molar volume of a gas: 1 mole H2 = 22.4 L From balanced equation: 1 mole Zn 1 mole H2 1.5 g Zn 1 Molar Mass of Zn: 1 mole Zn = 65.39 g mole Zn 1 mole H2 65.39 g Zn 1 mole Zn 22.4 1 L H2 mole H2 0.51 = ________ L H2 Stoichiometry with Gases Example: How many moles of water will be produced from the complete combustion of 7.3 L of oxygen gas? Assume STP C3H8 + 5O2 → 3CO2 + 4H2O Molar volume of a gas: 1 mole O2 = 22.4 L From balanced equation: 4 mole H2O 5 mole O2 7.3 L O2 Molar Mass of H: 1 mole H2O = 18 g 1 Mole O2 4 mole H2 O 22.4 L L O2 5 Mole O2 = _0.26 mole H2O Keeping all these equalities straight! TO GO BETWEEN USE THE EQUALITY Grams & moles Molar mass (g)= 1 mole Particles & Moles 1 mol = 6.02 x 1023 particles Moles & liters of a gas at STP 1 mole = 22.4 L at STP 2 different chemicals in a reaction Coefficient ratio(MOLE RATIO) from balanced equation You Try! Given the UNBALANCED EQUATION: __MgCO3 __MgO + __CO2, how many liters of CO2 gas are produced from the reaction of 15 grams of MgCO3? Assume STP! Section 6.5b Percent Yield Percent Yield A “Yield” is a product Actual Yield(A): the actual amount of product you produce in the lab Theoretical Yield(T): the amount of product you should produce if nothing went wrong; use the balanced chemical equation to calculate this amount. Percent Yield: ratio of actual yield to theoretical yield Percent Yield %yield = A x 100 T Lets Practice in steps: 1a. If 4.20 moles H2 reacts completely with oxygen, how many grams of H2O are produced? 2 H2 + O2 2 H2O This is a mol-mass problem. Your answer is the theoretical yield of water? If 4.20 moles H2 react completely with oxygen how many grams of H2O are produced? 2 H2 + O2 2 H2O From balanced equation: 2 mole H2O 1 mole O2 4.2 mol H2 2 moleH2O 2 mol H2 Molar Mass of O2: 1 mole H2O = 18.02g 18.02 grams H2O 1 mole H2O 75.7 g = ______g H2O This is the theoretical yield. What is the percent yield if 60.0 grams of H2O are produced? A= 60.0 g T= 75.7 g %yield = A x 100 T %yield = 60.0x 100 75.7 79.3% yield You have precipitated 8.50 g of Ba(OH)2. If you start with 4.57 grams of NaOH, what is the % yield. 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl From balanced equation: 2 mole NaOH 1 mole Ba(OH)2 4.57 g NaOH 1 molNaOH 40.00g NaOH Molar Mass of Ba(OH)2: 173.25 g Molar Mass of NaOH: 40.00 g 1 molBa(OH)2 171.35g Ba(OH)2 2 mol NaOH 1mol Ba(OH)2 = 9.79 g Ba(OH)2 This is the theoretical yield. If 9.78 grams are obtained in the experiment, what is the percent yield? A= 8.50 g %yield = T= 9.79 g %yield = A x 100 T 8.50 x 100 9.79 86.8% yield Section 6.5c Titrations Titrations—Using Stoichiometry Titration – A technique where the addition of a known volume of a known concentration solution to a known volume of unknown concentration solution to determine the concentration. •Use a buret to titrate unknown concentration of solutions. Titrations—Using Stoichiometry The titrant is the known concentration in the buret and the analyte is the unknown concentration in the flask. Formula: nMaVa = nMbVb na= number of H+ in the acid nb= number of OH- in the base Ma= molarity of acid Mb= molarity of base V= volume End Point vs. Equivalence Point Equivalence Point (or Stoichiometric Point) – When there are no reactants left over—they have all been reacted and the solution contains only products -the point where the acid and the base are equal in equal moles moles acid = moles base Importance of Indicators Indicators – Paper or liquid that change color based on pH level. End Point: point at which the indicator in the solution changes color It signals the equivalence point and the stop of the titration Always select an indicator that has a pH value close to that of the pH of the equivalence point of the titration. Titration Process Titration Problem #1 How many liters of 0.10 M NaOH is needed to react with 0.125 L of 0.25 M HCl? NaOH + HCl H2O + NaCl Titration Problem #2 What is the molarity of a Ca(OH)2 solution if 30.0 ml of the solution is neutralized by 20.0 ml of a 0.50 M solution of HCl? Ca(OH) 2 + 2HCl 2H2O + CaCl2 Titration Problem #3 What volume of 2.0M solution of NH4OH is needed to neutralize 50.0 ml of a 0.50M solution of H2SO4? 2 NH4OH + H2SO4 2H2O + (NH4) 2SO4 Titration Curves Strong Base - Strong Acid Weak Base - Strong Acid Shows the changes of pH during a titration Identifies the pH of the equivalence point Strong Base - Weak Acid Weak Base - Weak Acid Titration curve for Titrating a strong acid with a strong base pH is always = 7 The titration curve graph shows the pH of the equivalence point. Take the vertical region and cut the length in half and then look to what pH value aligns to that point. Titration curve for Titrating a strong base with an strong acid pH is always = 7 Titration curve for Titrating a weak acid with an strong base pH is >7