1. 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

2. 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

3. 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

4. – Example 1: What volume of 0.500 M HCl (aq) is required to react completely with 0.100 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 : 0.100 mol Pb(NO 3 ) 2 x 2 mol HCl = 0.200 mol HCl 1mol Pb(NO 3 ) 2 –convert mol HCl to vol. HCl solution: 0.200 mol HCl x 1 L sol’n = 0.400 L or 0.500 mol HCl 400 mL HCl 4 Ppt16b

5. Example 2: What is the molarity of an NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of 0.144 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) 0.144 M ? M 0.0350 L 0.0480 L –determine mol of H 2 SO 4(aq) : 0.0350 L H 2 SO 4 x 0.144 mol H 2 SO 4 = 0.00504 mol 1 L sol’n H 2 SO 4 5 Ppt16b

6. 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

7. Example 3: What is the molarity of an NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of 0.144 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) : 0.0350 L H 2 SO 4 x 0.144 mol H 2 SO 4 = 0.00504 mol 1 L sol’n H 2 SO 4 0.144 M ? M 0.0350 L 0.0480 L 7 Ppt16b

8. –determine mol of NaOH: 0.00504 mol H 2 SO 4 x 2 mol NaOH = 0.0101 mol 1 mol H 2 SO 4 NaOH –determine Molarity of NaOH: 0.0101 mol NaOH = 0.210 M NaOH 0.0480 L sol’n H 2 SO 4(aq) + 2NaOH (aq  2H 2 O + Na 2 SO 4(aq) 8 Ppt16b

9. 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

10. Example 4: What mass of chloride ion is present in a sample of water if 15.7 mL of 0.108 M AgNO 3 is required to titrate the sample? –AgNO 3(aq) + Cl - (aq)  AgCl (s) + NO 3 - (aq) 0.0157 L ? g 0.108 M 10 Ppt16b

11. 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

12. – Example 5: What mass of chloride ion is present in a sample of water if 15.7 mL of 0.108 M AgNO 3 is required to titrate the sample? AgNO 3(aq) + Cl - (aq)  AgCl (s) + NO 3 - (aq) 0.0157 L ? g 0.108 M 0.0157 L x 0.108 mol AgNO 3 = 0.00170 mol AgNO 3 1 L sol’n 0.00170 mol AgNO 3 x 1mol Cl - = 0.00170 mol Cl - 1 mol AgNO 3 0.00170 mol Cl - x 35.5 g Cl- = 0.0602 g Cl - 1 mol 12 Ppt16b