Quantitative Aspects of Reactions in Solution Sections 4.5-4.7

1. Quantitative Aspects of Reactions in SolutionSections 4.5-4.7

2. Terminology In solution we need to define the • SOLVENT the component whose physical state is preserved when solution forms; usually the component in the largest proportion • SOLUTE the other solution component

3. Concentration of Solute The amount of solute in a solution is given by its concentration. Concentration (M) = [ …]

4. The Nature of a CuCl2 Solution:Ion Concentrations CuCl2(aq) --> Cu2+(aq) + 2 Cl-(aq) If [CuCl2] = 0.30 M, then [Cu2+] = 0.30 M [Cl-] = 2 x 0.30 M

5. Preparing a Solution 2. Dissolve it in a little water in a volumetric flask 1. Weigh out mass of solute. 3. Add ENOUGH water to fill the solution up to the line on the flask

6. Do NOT add 1.0 L of water!!! 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 206

7. PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate molarity. Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate molarity [NiCl2•6 H2O] = 0.0841 M

8. USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g moles = M•V

9. Preparing Solutions • Weigh out a solid solute and dissolve to a given quantity of solution. • OR • Dilute a concentrated solution to give one that is less concentrated.

10. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution! But how much water do we add?

11. moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? How much water is added? The important point is that --->

12. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M • V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 mL

13. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add enough water to the 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

14. Preparing Solutions by Dilution A shortcut Cinitial • Vinitial = Cfinal • Vfinal

15. pH, a Concentration Scale pH: a way to express acidity -- the concentration of H+ in solution. Low pH: high [H+] High pH: low [H+] Acidic solution pH < 7 Neutral pH = 7 Basic solution pH > 7

16. The pH Scale pH = - log [H+] In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC pH = - log [H+] = -log (1.00 x 10-7) = - [0 + (-7)] = 7

17. [H+] and pH If the [H+] of soda is 1.6 x 10-3 M, the pH is ____? Because pH = - log [H+] then pH= - log (1.6 x 10-3) pH = -{log (1.6) + log (10-3)} pH = -{0.20 - 3.00) pH = 2.80

18. pH and [H+] If the pH of Coke is 3.12, its [H+] is ____________. Because pH = - log [H+] then log [H+] = - pH Take antilog and get [H+] = 10-pH [H+] = 10-3.12 = 7.6 x10-4M

19. SOLUTION STOICHIOMETRY • Zinc reacts with acids to produce H2 gas. • Have 10.0 g of Zn • What volume of 2.50 M HCl is needed to convert the Zn completely?

20. Convert something to Moles Coefficient Ratio Convert to what you’re looking for GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS

21. Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g) Step 1: Calculate amount of Zn Step 2: Use the stoichiometric factor

22. Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 2: Use the stoichiometric factor Step 3: Calculate volume of HCl req’d

23. Oxalic acid, H2C2O4 ACID-BASE REACTIONSTitrations H2C2O4(aq) + 2 NaOH(aq) ---> acid base Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION.

24. Setup for titrating an acid with a base Active Figure 5.23

25. Titration • 1. Rinse buret with titrant. Remove bubbles. Add solution from the buret to flask. 2. Flask contains analyte and indicator. Base reacts with acid in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H+ + OH- --> H2O 5. At equivalence point (end point) moles H+ = moles OH-

26. LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration. 10.65 mL of 1.2 M H2C2O4(oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

27. 10.65 mL of 1.2M H2C2O4(oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H2C2O4 Step 2: Calculate amount of NaOH req’d

28. 10.65 mL of 1.2M H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amountof H2C2O4 = 0.01278 mol acid Step 2: Calculate amount of NaOHreq’d = 0.02556 molNaOH Step 3: Calculate concentration of NaOH [NaOH] = 0.718 M