Ppt16b, Stoichiometry when Reactions are in Aqueous Solution / Titrations 1.Recall: Stoichiometry involves using a “mole to mole” ratio from the balanced equation. 2.Like recent labs (24 and 30): Use molarity and volume to get moles (or variant). 3.Reactions may be exchange reactions (ppt, acid- base, etc.) or redox. Kind of reaction is not relevant to the calculations—it’s just stoichiometry! 1 Ppt16b

Solution Stoichiometry Chemical Analysis of Solutions All discussions of stoichiometry apply to solutions as well as solid reactants and products use the same format for stoichiometric problems as in chapter 3 determine the moles of reactant, convert to moles of product can use solution volume and concentration to give you solute moles (rather than using grams and molar mass, as before) can have a combination of both now! 2 Ppt16b

General Sequence of Conversion: vol or M of A moles of A M or vol. of A moles of B mole ratio molarity of B volume of B note: the central theme is conversion of moles react to moles prod 3 Ppt16b

– Example 1: What volume of M HCl (aq) is required to react completely with mol of Pb(NO 3 ) 2(aq), forming a precipitate of PbCl 2(s) ? –write a correct equation: 2HCl (aq) + Pb(NO 3 ) 2(aq)  PbCl 2(s) + 2HNO 3(aq) –determine amt. HCl to react w/ Pb(NO 3 ) 2 : mol Pb(NO 3 ) 2 x 2 mol HCl = mol HCl 1mol Pb(NO 3 ) 2 –convert mol HCl to vol. HCl solution: mol HCl x 1 L sol’n = L or mol HCl 400 mL HCl 4 Ppt16b

Example 2: What is the molarity of an NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of M H 2 SO 4 ? –write the equation for the reaction: H 2 SO 4(aq) + 2NaOH (aq)  H 2 O + Na 2 SO 4(aq) M ? M L L –determine mol of H 2 SO 4(aq) : L H 2 SO 4 x mol H 2 SO 4 = mol 1 L sol’n H 2 SO 4 5 Ppt16b

General Sequence of Conversion: vol or M of A moles of A M or vol. of A moles of B mole ratio molarity of B volume of B 6 Ppt16b

Example 3: What is the molarity of an NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of M H 2 SO 4 ? H –write the equation for the reaction: H 2 SO 4(aq) + 2NaOH (aq)  2H 2 O + Na 2 SO 4(aq) –determine mol of H 2 SO 4(aq) : L H 2 SO 4 x mol H 2 SO 4 = mol 1 L sol’n H 2 SO M ? M L L 7 Ppt16b

–determine mol of NaOH: mol H 2 SO 4 x 2 mol NaOH = mol 1 mol H 2 SO 4 NaOH –determine Molarity of NaOH: mol NaOH = M NaOH L sol’n H 2 SO 4(aq) + 2NaOH (aq  2H 2 O + Na 2 SO 4(aq) 8 Ppt16b

Titrations Acid-base, precipitation, or oxidation- reduction reactions commonly used the first reactant is titrated with the second reactant until stoichiometric equivalence is reached (i.e., until all of the 1 st reactant is just used up and there is essentially none of the 2 nd left over either) –the second reactant is added slowly, in small aliquots from a buret (for high accuracy and precision of volume added) this is used to determine: –the moles of the first reactant (and related qtys) or –the molarity of the second reactant 9 Ppt16b

Example 4: What mass of chloride ion is present in a sample of water if 15.7 mL of M AgNO 3 is required to titrate the sample? –AgNO 3(aq) + Cl - (aq)  AgCl (s) + NO 3 - (aq) L ? g M 10 Ppt16b

General Sequence of Conversion: vol or M of A moles of A M or vol. of A moles of B mole ratio amount of B volume of B 11 Ppt16b

– Example 5: What mass of chloride ion is present in a sample of water if 15.7 mL of M AgNO 3 is required to titrate the sample? AgNO 3(aq) + Cl - (aq)  AgCl (s) + NO 3 - (aq) L ? g M L x mol AgNO 3 = mol AgNO 3 1 L sol’n mol AgNO 3 x 1mol Cl - = mol Cl - 1 mol AgNO mol Cl - x 35.5 g Cl- = g Cl - 1 mol 12 Ppt16b