1 SOLUTION STOICHIOMETRY - Previously, we dealt with comparing components of a chemical reaction by converting mass of substances to moles. - Now we will consider solutions where we will need to convert volumes to moles to make comparisons. Scheme: Mass of reactant (g) Mass of product (g) M Molar mass M Molar mass Moles of reactant (mol) Balanced equation Moles of product (mol) M Molarity M Molarity Volume of reactant (L) Volume of product (L) GRAVIMETRIC ANALYSIS Measure the concentration of solution by measuring the mass of a precipitate formed. - Use grams of precipitate to find moles of solute. - Divide by volume of solution to find concentration. Example: 25.00 mL of Pb(NO3)2 solution with an unknown concentration reacts with excess aqueous Rb3AsO4. After filtering and drying, 0.0814 g of precipitate is found. What is the concentration of the lead(II) nitrate solution? M(Pb3(AsO4)2) = 692.2 g/mol 3 Pb(NO3)2 (aq) + 2 Rb3AsO4 (aq) → Pb3(AsO4)2 (s) + 6 RbNO3 (aq) 0.0814 g Pb3 (AsO4 )2 ⋅ 1mol Pb3 (AsO4 )2 ⋅ 3 mol Pb( NO3 )2 692.2 g Pb3 (AsO4 )2 1molPb3 (AsO4 )2 ⋅ 0.0141 molPb( NO3 )2 1 = 0.02500 L L 2 VOLUMETRIC ANALYSIS Measure the concentration of solution by measuring the volume of solution need to find stoichiometric equivalence between reactants using a titration. - Use volume of titrant to find moles of analyte. - Divide moles of analyte by volume of analyte to find concentration. - Acid/base reactions are commonly used but not always. Example: What is the concentration of a HBr solution when 50.00 mL is titrated with 41.88 mL of 0.176 M KOH? KOH (aq) + HBr (aq) → H2O (l) + KBr (aq) 0.04188 L ⋅ 0.1762 molKOH 1mol HBr 0.1476 molHBr 1 ⋅ ⋅ = = 0.1476 M HBr L 1molKOH 0.05000 L 1L LIMITING REAGENT PROBLEMS Recall the following about limiting reagent problems. - To find limiting reactant, calculate number of moles of product formed from each number of moles of reactant. - Limiting reactant will yield lowest number of moles produced. Example: How many grams of Zn(OH)2 are produced when 350 mL of 0.152 M of ZnSO4 is mixed with 250 mL of 0.275 M of LiOH? First write balanced equation ZnSO4 (aq) + 2 LiOH (aq) → Zn(OH)2 (s) + Li2SO4 (aq) Now find number of moles of each reactant. ZnSO 4 : 0.350 L ⋅ 0.152 mol = 0.0532 mol ZnSO 4 L LiOH: 0.250 L ⋅ 0.275 mol = 0.0688 mol LiOH L Now calculate possible amount of product that each reactant can produce. ZnSO 4 :0.0532 mol ZnSO 4 ⋅ LiOH :0.0688mol LiOH ⋅ 1mol Zn ( OH ) 2 1mol ZnSO 4 1mol Zn ( OH ) 2 2mol LiOH Therefore, LiOH is the limiting reagent. 0.0688mol LiOH ⋅ = 0.0532mol Zn ( OH ) 2 = 0.0344mol Zn ( OH ) 2 1mol Zn ( OH ) 2 99.39g Zn ( OH ) 2 ⋅ = 3.42g Zn ( OH ) 2 2mol LiOH mol Zn ( OH ) 2 3 Example: When 732 L of 1.81 M of Ag2SO4 is mixed with 1148 L of 2.07 M of KBr, a) kilograms of AgBr (s) formed b) concentration of all metal ions remaining a) Ag2SO4 (aq) + 2 KBr (aq) → 2 AgBr (s) + K2SO4 (aq) Ag 2SO 4 : 732 L ⋅ KBr : 1148 L ⋅ 1.81 mol Ag 2SO4 L ⋅ 2 mol AgBr 1mol Ag 2SO4 = 2660 molAgBr 2.07 mol KBr 1 mol AgBr ⋅ = 2380 mol AgBr L 1mol KBr KBr is the limiting reactant. KBr : 2380 molAgBr ⋅ 187.772 g AgBr mol AgBr = 447000 g AgBr = 447 kg AgBr In photographic film, AgBr decomposes on exposure to light which darkens the negative. 2 AgBr(s) + hν → 2 Ag(s) + Br2 (l) b) Find concentration of metal ions remaining. The metal ions we might have in solution are Ag+ and K+. i) Consider the concentration of K+ first. - K+ is a spectator ion. - It hasn’t participated in any chemical change. Number of moles of K+ ion is same as number of moles of KBr initially. Volume of solution is 732 L + 1148 L = 1880 L. c K+ = 2380 mol K + 1880 L = 1.26 M Note that although KBr is limiting reagent, none of the K+ is used. This means that we could be more precise by saying that Br- is the limiting reagent. 4 ii) Consider the concentration of Ag+. Once reaction is complete, most of the Ag+ ion is now part of AgBr solid. But since Ag+ ion is not limiting reagent some of it remains. How much remains? First of all how much Ag2SO4 went into AgBr solid. 2380 mol KBr ⋅ 1 mol Ag2SO4 2 mol KBr = 1190 mol Ag2SO4 Therefore subtract Ag2SO4 used from Ag2SO4 total. total - used remaining 1330 mol -1190 mol 140 mol We need to be careful about dissociation of Ag2SO4. 140 mol Ag 2SO 4 ⋅ 2 mol Ag + 1 mol Ag 2SO 4 = 280 mol Ag + Thus concentration of Ag+ is c Ag + = 280 mol Ag + 1880 L = 0.15 M