1. UNIT 5 Aqueous Reactions and Solution Stoichiometry Molarity

2. Concentration of an Aqueous Solution - Molarity An important property of a solution is its concentration : the amount of solute dissolved in a given quantity of solvent. There are many ways to express concentration. We will learn one: molarity. Molarity (symbol M) = moles of solute volume of solution in liters Molarity can be treated as a conversion factor (like density) to go from volume of a solution to moles of solute and vice versa.

3. Two Ways Calculate Moles You now know two ways to calculate the number of moles of a substance. 1. If you have a known mass of the substance, you divide by the molar mass to get moles. mass of A x 1 mol A = mol A molar mass of A 2. If you have a known volume of a solution of known molarity, you multiply the volume of the solution in liters by the molarity to get moles. liters of solution of A x molarity of A = mol A

4. Calculations Involving Molarity How to prepare 1.00 liter of a 1.00 M solution of copper(II) sulfate: 1.00 L x 1.00 mole CuSO 4 = 1.00 mole CuSO 4 1 L solution 1.00 mole CuSO 4 x 159.61 g CuSO 4 = 160. g CuSO 4 1 mole CuSO 4 Put 160. g CuSO 4 in a 1L volumetric flask. Fill about ¼ full with deionized (DI) water and swirl to dissolve. Fill to the mark with DI water and up end the flask to mix.

6. Calculations Involving Molarity How to prepare 250.0 mL of a 1.00 M solution of copper(II) sulfate: 250.0 mL x 1 L x 1.00 mole CuSO 4 = 0.250 mole CuSO 4 1000 mL 1 L solution 0.250 mole CuSO 4 x 159.611 g CuSO 4 = 39.9 g CuSO 4 1 mole CuSO 4 Put 39.9 g CuSO 4 in a 250 mL volumetric flask and follow previous procedure. Both flasks contain 1.00M CuSO 4.

7. Calculations Involving Molarity What is the molarity of a solution made by dissolving 25.0 g of NaCl in enough water to make 125 mL? molarity = moles NaCl = 25.0 g NaCl x 1 mole NaCl vol soln in L 58.44 g NaCl 125 mL x __1L___ 1000 mL = 0.428 moles NaCl 0.125 L = 3.42 M

8. Molarity of Electrolytes Molarity is just a unit that describes concentration. It can describe the concentration of the entire compound or any part of that compound. A 0.500 M solution of potassium carbonate has what concentration of potassium ions? Carbonate ions? 0.500 M K 2 CO 3 = 0.500 mol K 2 CO 3 x 2 mol K + = 1.00 M K + 1 L soln 1 mol K 2 CO 3 0.500 M K 2 CO 3 = 0.500 mol K 2 CO 3 x 1 mol CO 3 2- = 0.500 M CO 3 2- 1 L soln 1 mol K 2 CO 3

9. Using Molarity to Convert Between Moles and Volume We have learned how to use the stoichiometry of a reaction to predict the yield of a product in grams or moles. We can now extend this knowledge to include reactions involving solutions as reactants.

10. How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? The steps are the same as with the stoichiometry we learned in Chapter 3. 1. Write and balance the equation for the reaction: HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) Using Molarity to Convert Between Moles and Volume

11. How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? 2. Write what you know and what you need to know: 15.0 mL 6.0M ? mL 1.50M HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 3. Convert “what you know” to moles. 15.0 mL 1 L 6.0 mol HNO 3 1000 mL 1 L solution Using Molarity to Convert Between Moles and Volume

12. How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? 15.0 mL 6.0M ? mL 1.50M HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 4. Convert to “moles of what you need to know” using the stoichiometry of the reaction: 15.0 mL 1 L 6.0 mol HNO 3 1 mol KOH 1000 mL 1 L solution 1 mol HNO 3 Using Molarity to Convert Between Moles and Volume

13. How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? 15.0 mL 6.0M ? mL 1.50M HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 5. Convert your answer from moles to the required units. We have been asked for volume. We have moles. Molarity lets us convert between the two: 15.0 mL 1 L 6.0 mol HNO 3 1 mol KOH L 1000 mL = 60. mL 1000 mL 1 L 1 mol HNO 3 1.50 mol KOH 1 L 60. mL of 1.50M KOH are required. Using Molarity to Convert Between Moles and Volume

14. Suppose the question had been “How many grams KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ?” Only step 2 and step 5 change! 15.0 mL 6.0M ? g KOH (step 2) HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 5. Convert your answer from moles to the required units. 15.0 mL 1 L 6.0 mol HNO 3 1 mol KOH 56.11 g KOH = 5.1 g KOH 1000 mL 1 L 1 mol HNO 3 1 mol KOH Using Molarity to Convert Between Moles and Volume

15. Dilutions liters of solution of A x molarity of A = mol A This equality is helpful in dealing with dilutions. Often we are called upon to make a solution of a known concentration not from a dry chemical, but from another solution. How do we make 200.0 mL of 2.50M HNO 3 from concentrated (12.0 M) HNO 3 ?

16. 200.0 mL x 1L x 2.50 mol HNO 3 = 0.500 mol HNO 3 1000 mL1L We must determine how many mLs of 12.0M HNO 3 contains this same number of moles. 0.500 mol HNO 3 x_____1L_____ = 0.0417 L 12.0 mol HNO 3 Answer: 41.7 mL Dilutions

17. How do we make 200.0 mL of 2.50M HNO 3 from concentrated (12.0 M) HNO 3 ? The answer is to take 41.7 mL of 12.0 M HNO 3 and add enough water to make 200.0 mL of solution…actually, you would add the acid to the water! Dilutions

18. Dilutions – A Quicker Way How do we make 200.0 mL of 2.50M HNO 3 from concentrated (12.0 M) HNO 3 ? If we set this up algebraically, we get 2.50M x 0.2000L = 12.0M x ?L. With the answer included, we have 2.50M x 0.2000L = 12.0M x 0.0417L.

19. Dilutions – A Quicker Way The general form for dilutions is M 1 V 1 = M 2 V 2 where V = volume (mL or L, just keep them the same) and M = molarity (NOT MOLES!!!)

20. Dilutions – Examples M 1 V 1 = M 2 V 2 The Dilution Equation How many mLs of 0.750M HCl are needed to make 150.0 mL of 0.125M HCl? How much water should be used? (0.125M)(150.0mL) = (0.750M) (x mL) x = 25.0 mL Water used = 150.0 – 25.0 = 125.0 mL

21. M 1 V 1 = M 2 V 2 What is the concentration of a solution made by diluting 250.0 mL of 18M sulfuric acid to 1500.0 mL? (18M)(250.0 mL) = (x M)(1500.0 mL) x = 3.0M Dilutions – Examples

22. Dilutions – Mixing Two Solutions Often reactions are carried out by mixing two solutions (one for each reactant) together. It then becomes necessary to calculate the concentration of the reacting species in the resulting solution. Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution? (NH 4 ) 3 PO 4 (aq) + Na 3 PO 4 (aq)  NR (no reaction…why?) This is just a dilution!

23. Dilutions – Mixing Two Solutions Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution? molarity = ________moles ion____________ volume of the final solution in liters volume of the final solution = 350.0 + 125.0 = 475.0 mL

24. Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution? final volume = 475.0 mL = 0.4750 L molarity = moles ion 0.4750 L Let’s use the dilution equation to find the concentration of ammonium phosphate (NH 4 ) 3 PO 4 in the resulting solution: 350.0 mL (0.820 M) = 475.0 mL M 2 M 2 = 0.6042 mol (NH 4 ) 3 PO 4 per liter of the final solution Dilutions – Mixing Two Solutions

25. Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution? final volume = 475.0 mL = 0.4750 L Let’s use the dilution equation to find the concentration of sodium phosphate Na 3 PO 4 in the resulting solution: 125.0 mL (0.750 M) = 475.0 mL M 2 M 2 = 0.1974 mol Na 3 PO 4 per liter of the final solution Dilutions – Mixing Two Solutions

26. Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution? Find the concentration of NH 4 + in the final solution: 0.6042 mol (NH 4 ) 3 PO 4 x 3 mol NH 4 + = 1.81 mol NH 4 + L of final soln 1 mol (NH 4 ) 3 PO 4 L 1.81 mol NH 4 + = 1.81 M NH 4 + L Dilutions – Mixing Two Solutions

27. Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution? Find the concentration of Na + in the final solution: 0.1974 mol Na 3 PO 4 x 3 mol Na + = 0.592 mol Na + L of final soln 1 mol Na 3 PO 4 L 0.592 mol Na + = 0.592 M Na + L Dilutions – Mixing Two Solutions

28. Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution? The phosphate comes from both solutions. To find its concentration in the final solution, add the concentrations of (NH 4 ) 3 PO 4 and Na 3 PO 4 in the final solution (why?): M of PO 4 3- in final solution = M of (NH 4 ) 3 PO 4 + M of Na 3 PO 4 = 0.6042 + 0.1974 = 0.802 M PO 4 3- in final solution Dilutions – Mixing Two Solutions Both molarities are for the final solution.

29. Titration Titration is a way to determine the concentration of a solution by reacting it with a known amount of a chemical (the standard) and monitoring the point of stoichiometric equivalence with an indicator. Titration can be performed on any solution that gives a product which can be monitored: acid-base titrations redox titrations precipitation titrations

30. Titration Titration is a way to determine the concentration of a solution by reacting it with a known amount of a chemical (the standard) and monitoring the point of stoichiometric equivalence with an indicator. You have already performed the calculations for an acid-base titration. This is just a stoichiometric problem with the concentration of one of the reactants known.

31. Titration – Example 1 KHP (potassium acid phthalate) is a primary standard with a molar mass of 204.2 g. NaOH is a secondary standard, the concentration of which is found by titration with a known amount of KHP, according to the following equation: 15.44 mL ? M 0.396 g NaOH (aq) + KHP (s)  KNaP (aq) + H 2 O(l) If 15.44 mL of NaOH solution are needed to reach a phenolphthalein endpoint in the titration of 0.396 g of KHP, what is the molarity of the NaOH solution?

32. Titration – Example 1 15.44 mL ? M 0.396 g NaOH (aq) + KHP (s)  KNaP (aq) + H 2 O(l) ? M = mol NaOH volume of NaOH solution in L 0.396 g KHP 1 mol KHP 1 mol NaOH _______ 1000 mL = 0.126 mol NaOH 204.2 g KHP 1 mol KHP 15.44 mL 1 L L of solution This is the same as 0.126 M NaOH.

33. Titration – Example 2 The NaOH solution we just titrated is a secondary standard and can be used to titrate acid samples such as vinegar, a solution of acetic acid in water. If 25.00 mL of the vinegar is neutralized by 26.63 mL of the NaOH solution, what is the molarity of the acetic acid in the vinegar? 26.63 mL 0.126 M 25.00 mL ? M NaOH (aq) + HC 2 H 3 O 2 (aq)  NaC 2 H 3 O 2 (aq) + H 2 O(l)

34. Titration – Example 2 26.63 mL 0.126 M 25.00 mL ? M NaOH (aq) + HC 2 H 3 O 2 (aq)  NaC 2 H 3 O 2 (aq) + H 2 O(l) The key to most titration problems is to keep track of which solution you need at what time. molarity of acetic acid = moles acetic acid volume of acetic acid (vinegar) in L 26.63 mL NaOH 1 L 0.126 mol NaOH 1 mol acetic acid ________ 1000 mL = 1000 mL 1L NaOH soln 1 mol NaOH 25.00 mL 1 L 0.134 mol acetic acid L This is the same as 0.134 M acetic acid.

35. Titration – Example 3: Polyprotic Acids If 25.00 mL of a solution of malonic acid (H 2 C 3 H 2 O 4 ) is neutralized by 26.63 mL of the NaOH solution, what is the molarity of the malonic acid? Note the position of the acid H’s in malonic acid. 26.63 mL 0.126 M 25.00 mL ? M 2NaOH (aq) + H 2 C 3 H 2 O 4 (aq)  Na 2 C 3 H 2 O 4 (aq) + 2H 2 O(l) 26.63 mL NaOH 1 L 0.126 mol NaOH 1 mol H 2 C 3 H 2 O 4 _________ 1000 mL = 1000 mL 1L NaOH soln 2 mol NaOH 25.00 mL 1 L 0.0671 mol H 2 C 3 H 2 O 4 L This is the same as 0.0671 M H 2 C 3 H 2 O 4.