1. LECTURE Fourteen CHM 151 ©slg Topics: 1. Titration Calculations 2. Dilution Problems

2. How many mL of solution A,.250 M Pb(NO 3 ) 2, would be required to react with 30.00 mL of solution B,.150 M KI? Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M ? mL 30.00 mL Pathway: mL “B” soltn  mol “B”  mol “A”  mL “A” soltn 30.00 mL B soltn = ? mL A soltn

3. Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M ? mL 30.00 mL 30.00 mL soltn.150 mol B 1 mol A 1000 mL soltn 1000 mL soltn 2 mol B.250 mol A = 9.00 mL soltn A Molarity equation Molarity

4. TITRATION CALCULATIONS Titration: Procedure in which measured increments of one reactant are added to a known amount of a second reactant until some indicator signals that the reaction is complete. This point in the reaction is called “the equivalence point.” Indicators include many acid/base dyes, potentiometers, color change in one reagent...

5. Titrations are run with buret and Erlenmeyer flasks as seen on the CD ROM or demo just observed... Two types of problems are typically encountered: Standardizing an acid or a base solution (determining the exact molar concentration of an acid or base solution) Determining amount of acidic or basic material in a sample ( or other substances detectable by some indicator)

6. Ex 5.13 type: Standardizing an acid solution: Suppose a pure, dry sample of Na 2 CO 3 weighing 0.379 g is dissolved in water and titrated to the equivalence point with 35.65 mL of HCl solution. What is the molarity of the HCl solution? Na 2 CO 3(aq) + 2HCl (aq) -----> 2 NaCl (aq) + H 2 O + CO 2(g) 105.99 g/mol 35.65 mL soln 0.379 g M= ? = #mol HCl / L soltn 2 Na = 2 X 22.99 = 45.98 1 C = 1 X 12.01 = 12.01 3 O = 3 X 16.00 = 48.00 105.99 g/mol

7. Calculations for Standardization 1) Calculate moles of solute from balanced equation, using information about known reagent 2) Calculate M, using volume of solution required for titration of known to equivalence point

8. Na 2 CO 3(aq) + 2HCl (aq) -----> 2 NaCl (aq) + H 2 O + CO 2(g) 105.99 g/mol 35.65 mL soln 0.379 g M=? = Mol HCl / L soltn Step One: Calculate Moles of HCl from equation.379 g Na 2 CO 3 = ? moles HCl

9. M, HCl soln = ? = # mol HCl = L soln.00715 mol HCl 1000 mL =.201 mol HCl =.201 M HCl 35.65 mL soltn 1 L L soltn Na 2 CO 3(aq) + 2HCl (aq) -----> 2 NaCl (aq) + H 2 O + CO 2(g) 105.99 g/mol 35.65 mL soln 0.379 g.00715 mol M=? Step Two: Calculate molarity: Calculated in Step One

10. Group Work: Suppose a pure, dry sample of Na 2 CO 3 weighing 0.437 g is dissolved in water and titrated to the equivalence point with 39.85 mL of HCl solution. What is the molarity of the HCl solution? Na 2 CO 3(aq) + 2HCl (aq) -----> 2 NaCl (aq) + H 2 O + CO 2(g) 105.99 g/mol 39.85 mL soln 0.437 g M= ? = #mol HCl / L soltn #1: Find moles of HCl from g Na 2 CO 3 #2: Find Molarity (moles HCl / volume soltn)

11. Na 2 CO 3(aq) + 2HCl (aq) -----> 2 NaCl (aq) + H 2 O + CO 2(g) 105.99 g/mol 39.85 mL soln 0.437 g M= ? = #mol HCl / L soltn #1: Find Moles of HCl.437 g Na 2 CO 3 1 mol Na 2 CO 3 2 mol HCl =.008246 mol HCl 105.99 g Na 2 CO 3 1 mol Na 2 CO 3 #2: Calculate Molarity.008246 mol HCl 1000 mL =.2069 mol HCl =.207 M HCl 39.85 mL soltn 1 L L soltn Group Work Solution:

12. Standardization of a Base: Let’s use the 0.201 M HCl which we standardized in the first problem to determine the exact molar concentration of a sodium hydroxide solution (“standardize the base”!): Problem: If titrating 25.00 mL of NaOH to a phenolphthalein endpoint required 31.25 mL of.201 M HCl, what is the molar concentration of the base?

13. If titrating 25.00 mL of NaOH to a phenolphthalein endpoint required 31.25 mL of.201 M HCl, what is the molar concentration of the base? NaOH(aq) + HCl(aq) ---> H 2 O(l) + NaCl(aq) 25.00 mL 31.25 mL M=?.201 M Steps: 1) calculate moles NaOH reacted 2) calculate M, moles NaOH/ volume soln

14. NaOH(aq) + HCl(aq) ---> H 2 O(l) + NaCl(aq) 25.00 mL 31.25 mL M=?.201 M Step 1) 31.25 mL HCl soln = ? mol NaOH Step 2) Calculate molarity:

15. A 30.0 mL sample of vinegar requires 39.35 mL of.843 M NaOH solution for titration to the equivalence point. What is the molar concentration of the acetic acid in vinegar? CH 3 CO 2 H(aq) + NaOH(aq) ----> H 2 O + NaCH 3 CO 2 (aq) 30.0 mL 39.35mL.843 M M=?mol CH 3 CO 2 H /L Step 1) solve for moles, CH 3 CO 2 H Step 2) solve for M CH 3 CO 2 H Group Work

16. CH 3 CO 2 H(aq) + NaOH(aq) ----> H 2 O + NaCH 3 CO 2 (aq) 30.0 mL 39.35mL M=?.843 M #1: #2: 39.35 mL soltn.843 mol NaOH 1 mol CH 3 CO 2 H =.03317 mol CH 3 CO 2 H 1000 mL soltn 1 mol NaOH.03317 mol CH 3 CO 2 H 1000 mL = 1.1056 mol CH 3 CO 2 H = 1.11 M 30.00 mL soltn 1 L L soltn

17. Let’s do another base solution, standardizing it with Potassium acid phthalate, KHC 8 H 4 O 4, a popular solid for this purpose. (Structure, next slide!) It reacts with strong bases according to the following net ionic equation: KHC 8 H 4 O 4 (aq) + NaOH (aq) ------> H 2 O (l) + KNaC 8 H 4 O 4 (aq) If a.896 g sample of this compound is dissolved in water and titrated to the equivalence point with 16.95 mL of NaOH solution, what is the molarity of the NaOH soln?

20. KHC 8 H 4 O 4 (aq) + NaOH (aq) ---> H 2 O (l) + K NaC 8 H 4 O 4 (aq) 204.22 g/mol.896 g 16.95 mL M=? mol NaOH /L 1K = 1 X 39.10 = 39.10 8C = 8 X 12.01 = 96.08 5H = 5 X 1.008 = 5.04 4O = 4 X 16.00 = 64.00 204.22 g/mol If a.896 g sample of potassium acid phthalate is dissolved in water and titrated to the equivalence point with 16.95 mL of NaOH solution, what is the molarity of the NaOH soln?

21. KHC 8 H 4 O 4 (aq) + NaOH (aq) ---> H 2 O (l) + K NaC 8 H 4 O 4 (aq) 204.22 g/mol 16.95 mL.896 g M=? mol NaOH / L #1: Find moles of NaOH #2: Calculate Molarity:.896 g KHC 8 H 4 O 4 1 mol KHC 8 H 4 O 4 1 mol NaOH =.00439 mol NaOH 204.22 g KHC 8 H 4 O 4 1 mol KHC 8 H 4 O 4

22. “Redox” titration You wish to determine the weight percent of copper in a copper containing alloy. After dissolving a sample of an alloy in acid, an excess of KI is added, and the Cu 2+ and I - ions undergo the reaction: 2Cu 2+ (aq) + 5 I - (aq) -----> 2 CuI(s) + I 3 - (aq) The I 3 - which is produced in this reaction is titrated with sodium thiosulfate according to the equation: I 3 - (aq) + 2 S 2 O 3 2- (aq) -----> S 4 O 6 2- (aq) + 3 I - (aq) If 26.32 mL of 0.101 M Na 2 S 2 O 3 is required for titration to the equivalence point, what is the wt % of Cu in.251 g alloy?

23. 2Cu 2+ (aq) + 5 I - (aq) -----> 2 CuI(s) + I 3 - (aq) 63.55 g/mol produced 0.251 g alloy by reaction g Cu =? % Cu, alloy=? I 3 - (aq) + 2 S 2 O 3 2- (aq) -----> S 4 O 6 2- (aq) + 3 I - (aq) ? mol 26.32 mL.101 M Na 2 S 2 O 3

24. I 3 - (aq) + 2 S 2 O 3 2- (aq) -----> S 4 O 6 2- (aq) + 3 I - (aq) ? mol 26.32 mL.101 M Na 2 S 2 O 3

25. 2Cu 2+ (aq) + 5 I - (aq) -----> 2 CuI(s) + I 3 - (aq) 63.55 g/mol g=?.00133 mol % Cu, alloy=? 0.251 g alloy % Cu, alloy=? = 1.69 g Cu X 100 = 67.3% 0.251 g alloy

26. DILUTION PROBLEMS Suppose you would like to make up a more “dilute” solution (less moles/L) from a more “concentrated” one (more moles /L). This is how you might proceed: a) figure out how many total moles of solute you want in the desired volume of the dilute solution b) figure out what volume of the more concentrated solution will deliver this number of moles c) measure out the concentrated solution and add water to make up the desired volume of the dilute solution. (DEMO!)

27. Describe how you might makeup 750. mL of 1.00 M H 2 SO 4 from a concentrated sulfuric acid solution which is 12.0 M H 2 SO 4. a) Calculate desired number of moles in dilute soln: b) Calculate volume of concentrated solution which will contain this number of moles:

28. c) Measure out 62.5 mL of the concentrated soln and carefully add it to sufficient water to make up 750. mL of solution. NEVER ADD WATER TO CONCENTRATED ACIDS; ALWAYS ADD CON ACIDS TO WATER, SLOWLY...

29. Formula for Dilution Problems: Since we can calculate moles as follows: mL X # moles = moles of solute 1000 mL and since: moles, con soln = moles, dil soln We can say: mL, con X M, con = mL, dil X M, dil and do the last problem “the easy way”:

30. Describe how you might makeup 750. mL of 1.00 M H 2 SO 4 from a concentrated sulfuric acid solution which is 12.0 M H 2 SO 4. mL, con X M, con = mL, dil X M, dil ? mL X 12.0 M H 2 SO 4 = 750. mL X 1.00 M H 2 SO 4 mL con = 750 mL dil X 1.00 M H 2 SO 4 12.0 M H 2 SO 4 mL con = 62.5 mL Punch line the same: measure out 62.5 mL con acid and dilute to 750. mL.

31. Group Work: How many mL of 2.35 M AgNO 3 solution are required to makeup 2.00 L of.100 M AgNO 3 solution? Describe how you would makeup this new solution. mL con X M con = mL dil X M dil ? 2.35 M 2.00L.100 M

32. mL con X M con = mL dil X M dil ? 2.35 M 2.00L.100 M 2000 mL mL con = mL, dil X M, dil M, con ml, con = 2000 mL X.100 M 2.35 M mL, con = 85.1 mL Measure out 85.1 mL of con soltn, dilute to 2.00 L

33. Energy Units The “ calorie”: “The quantity of energy required to raise 1.00 g of water 1 o C”. Very small amount of energy, so Kcal are generally used: 1000 calories(cal) = 1 kilocalorie, kcal SI unit of energy is the “joule”, defined in terms of kinetic energy rather than heat energy. One joule is the amount of kinetic energy involved when a 2.0 kg object is moving with a velocity of 1.0 m/s. Again, a very small amount of energy, so kilojoules are generally used: 1000 joules (J) = 1 kilojoule kJ (kJ)

34. Relationship: 1 calorie = 4.184 joules Dietary or nutritional Calorie: 1 nutritional Calorie = 1 Cal = 1 kcal = 1000 cal Try the problems assigned in Chapter six! End, Lecture 14