Chapter 04: Stoichiometry 1. Insight 2. Fundamentals of Stoichiometry 3. Limiting Reactants 4. Theoretical and Percentage Yields 5. Solution Stoichiometry CONCENTRATION Molarity (M) %v/v %w/w Number of moles/ volume %w/v Composition of Solutions: qualitative • Mixtures have variable composition? • Need a way of indicating the amounts in the composition • Relatively speaking, solutions are often described as dilute or concentrated • Dilute solutions: low amount of solute compared to solvent • Concentrated solutions: large amount of solute compared to solvent 2 Tro: Chemistry: A Molecular Approach, 2/e Solution concentration • Molarity: moles of solute per liter of solution • Describes #molecules per liter of solution Molarity (M) = moles of solute volume of solution in liters Total volume not just the volume of solution used Preparing 1 L of a 1.00 M NaCl Solution DI water (Distilled, deionized water) 4 Tro: Chemistry: A Molecular Approach, 2/e Practice — Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution moles solute M= Total Volume ( L ) Step 1: g KBr mol KBr M Step 2: 5 Tro: Chemistry: A Molecular Approach, 2/e L sol’n Practice — Determine the mass of CaCl2 (MM = 110.98) in 1.75 L of 1.50 M solution L sol’n 6 Tro: Chemistry: A Molecular Approach, 2/e mol CaCl2 g CaCl2 Molarity Calculator • http://www.tocris.com/molarityCalculator.php 7 Given a concentrated stock solution, prepare a less concentrated (more dilute) working solution. Glacial acetic acid (99+%) m.p. 16.7 °C Flash point 40 °C The Dilution processes of concentrated acid and base are exothermic. Dilution • Given a concentrated stock solution, prepare a less concentrated (more dilute) working solution # moles of solute remains constant Concentration (M) X volume (V) = # moles 9 Dilution • # moles of solute remains constant McVc = MdVd where “c” stands for the concentrated solution & “d” for the diluted solution. (In the following exams) 10 Dilution Exam ple Add 28.7 ml of 17.4M concentrated stock A student added 28.7 ml of 17.4M concentrated stock solution to a 500.0 ml volumetric flask, after adding D.I. water to the curve, after well mixing, what is the final molar concentration (molarity) of this solution? Fill to the mark with water M= ( 28.7ml ) (17.4 M ) = 1.0M 500.ml 1.00M 500 ml volumetric flask Figure 4-12 p153 Chapter 04: Stoichiometry 1. Insight 2. Fundamentals of Stoichiometry 3. Limiting Reactants 4. Theoretical and Percentage Yields 5. Solution Stoichiometry 6. Insight STOICHIOMETRY CONCENTRATION MOLARITY (M) Solution Stoichiometry • Molarity • A function of the # moles of solute #molecules • Can be used to convert between amount of reactants and/or products in a chemical reaction 13 Tro: Chemistry: A Molecular Approach, 2/e Solution vs. Solid Stoichiometry 14 Tro: Chemistry: A Molecular Approach, 2/e Example: What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)? Method 2 Given: 0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2 Find: L KCl Conceptual Plan: L Pb(NO3)2 mol Pb(NO3)2 mol KCl L KCl Relationships: 1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl Solution: Check: because you need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x the volume of Pb(NO3)2 16 Tro: Chemistry: A Molecular Approach, 2/e Acids & bases Electrolytes & nonelectrolytes Strong Acids • completely dissociate (100%) • HCl(aq) H+(aq) + Cl−(aq) • H2SO4(aq) 2 H+(aq) + SO42−(aq) Form ula N am e HCl Hy drochloric acid HBr Hy drob ro mic acid HI Hy droio dic acid HNO3 N itric acid H2 SO4 Su lfu ric acid HClO4 Perch loric acid notice that the polyvalent ion stays together Form ula N am e LiOH Lith iu m h yd ro xide NaOH So diu m hy dro xide KOH Po tass iu m h yd ro xide Ba( OH) 2 Barium hy droxide Weak Acids • only partially dissociate • HF(aq) H+(aq) + F−(aq) • Acetic acid: only 4 out every 1000 molecules (0.4%) are converted to acetate ions CH3 COOH(aq) + H 2 O( l) Acetic acid - CH3 COO ( aq) + H 3 O+ ( aq) A cetate ion Common Acids 21 Tro: Chemistry: A Molecular Approach, 2/e Common Bases 22 Tro: Chemistry: A Molecular Approach, 2/e Practice – Write the equation for the process that occurs when the following strong electrolytes dissolve in water CaCl2 CaCl2(aq) Ca2+(aq) + 2 Cl−(aq) HNO3 HNO3(aq) H+(aq) + NO3−(aq) (NH4)2CO3 (NH4)2CO3(aq) 2 NH4+(aq) + CO32−(aq) 23 Tro: Chemistry: A Molecular Approach, 2/e Acid-Base Systems ① An acid MUST react with a base ② A base always reacts with an acid They always come in pairs Acid – Base Definitions • Acid: proton (H+) donor • Base: hydroxide (OH-) donor 𝐻 + 𝑂𝐻 − → 𝐻2 𝑂 If you put the Base in water…….. NH3(aq) + H+(aq) NH4+(aq) proton acceptor H+(aq) + OH-(aq) + H2O NH3(aq)+ H2O NH4+(aq) + OH-(aq) OH- donor 26 Acids & Bases • Acid: proton donor HCl(aq) H+(aq) + Cl-(aq) • Base: proton acceptor: NH3(aq)+ H+(aq) NH4+(aq) + HCl(aq) + NH3(aq) NH4+(aq) + Cl-(aq) Bases are anti-acids (antacids) • What happens when you add a base to an acid? HCl(g) H+(aq) + Cl-(aq) NaOH(s) Na+(aq) + OH-(aq) HCl(g) + NaOH(s) H+(aq) + OH-(aq)+ Cl-(aq) + Na+(aq) HCl(g) + NaOH(s) H2O(l) + Cl-(aq) + Na+(aq) You get water! 28 Acid – Base reactions: look to form water and exothermic • For the acid base reaction: HNO3(aq) + Ca(OH)2(aq) • Dissociate the acid and the base into ions (inverse criss-cross rule): (H+ + NO3−) + (Ca2+ + 2OH−) • Rearrange: H+ combines with OH− to make H2O(l) H+ + 2OH− + Ca2+ + NO3− • – notice there are 2OH- ions need 2 H+ ions 2H+ + 2OH− + Ca2+ + 2NO3− • Now form water 2H2O + Ca2+ + 2NO3− • write the formula of the salt Ca2+ + NO3− Ca(NO3)2 • Combine and make sure it’s balanced HNO3(aq) + Ca(OH)2(aq) 2H2O + Ca(NO3)2 Practice(Diacid/binary acid/base) – Predict the products and balance the equation HCl(aq) + Ba(OH)2(aq) (H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−) HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2 2 HCl(aq) + Ba(OH)2(aq) 2 H2O(l) + BaCl2(aq) H2SO4(aq) + Sr(OH)2(aq) (H+ + SO42−) + (Sr2+ + OH−) → (H+ + OH−) + (Sr2+ + SO42−) H2SO4(aq) + Sr(OH)2(aq) → H2O(l) + SrSO4 H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4 H2SO4(aq) + Sr(OH)2(aq) 2 H2O(l) + SrSO4(s) 30 Tro: Chemistry: A Molecular Approach, 2/e Acid-Base Titrations • Neutralization reaction • [acid] or [base] known • Other is unknown • Stoichiometry of neutralization known • Volumes of acid (VH) & base (VOH) needed to reach neutralization known 31 Acid-Base Titration Reactions (Exp 8 inFor Chem301) Titration exp 8 Analytical procedure involving acids & bases in Chem301 1) Goal: determine [unknown reactant] 2) Materials: acid solution, base solution, indicator a) [1 reactant] known b) [2nd reactant] unknown (to be determined) 3) Stoichiometry of reaction known 4) Procedure: Slowly add 1 reactant into another until neutrality is reached (molesH = molesOH) a)Result: volumes of acid & base used 5) Analysis: use MHVH=MOHVOH or solution stoichiometry to determine [unknown reactant] Conclusion: [unknown reactant] 32 Titration Here the titrant is the base solution in the burette. As the titrant is added to the flask, the H+ reacts with the OH– to form water. Once the acid is neutralized (endpoint), the phenolphthalein will show a stable light pink color for 30 secs 33 Tro: Chemistry: A Molecular Approach, 2/e Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? ① Start with a balanced eqn ② Convert given moles HCl + NaOH H2O + NaCl æ 0.100moles_ NaOH ö -3 0.01254L ç ÷ =1.254x10 moles_ NaOH è ø L ③ Convert moles given desired (balanced eqn) æ 1HCl ö -3 1.254x10 moles_ NaOH ç ÷ =1.254x10 moles_ HCl è 1NaOH ø -3 ④ Convert desired moles Molarity using V & definition of M 1.254x10-3 moles_ HCl = 0.1254M _ HCl -2 1.000x10 L Let’s go to the Lab. Please do not forget to 1. Sign-in 2. Keep your PPEs on. 3. Checkout with my signature on your datasheet before you leave Practice − What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4? ① Start with a balanced eqn ② Convert given moles H2SO4 + 2NaOH 2H2O + Na2SO4 æ 0.1015moles_ H 2 SO4 ö -3 0.050L ç ÷ = 5.075x10 moles_ H 2 SO4 è ø L ③ Convert moles given desired (balanced eqn) æ 2NaOH ö -2 5.075x10 moles_ H 2 SO4 ç ÷ =1.015x10 moles_ NaOH è 1H 2 SO4 ø -3 ④ Convert desired moles Molarity using V & definition of M 1.015x10-2 moles_ NaOH = 0.369M _ NaOH -2 2.75x10 L Practice — Solution stoichiometry • 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. The unbalanced equation is HCl + Ba(OH)2 BaCl2 + H2O What is the molarity of the base? 37 Tro: Chemistry: A Molecular Approach, 2/e Practice 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. The unbalanced equation is HCl + Ba(OH)2 BaCl2 + H2O What is the molarity of the base? Given: Find: 0.0438 L of 0.107 M HCl, 0.0376 L Ba(OH)2 M Ba(OH)2 Conceptual Plan: Balance the equation L HCl mol HCl mol Ba(OH)2 M Ba(OH)2 Relationships: L Ba(OH)2 1 mL= 0.001 L, 1 L HCl = 0.107 mol, 1 mol Ba(OH)2 : 2 mol HCl Solution: 2HCl + Ba(OH)2 BaCl2 + 2H2O