Lecture 6: Lec4a Chemical Reactions in solutions

Transcription

1 Lecture 6: Lec4a Chemical Reactions in solutions Zumdahl 6 th Ed, Chapter 4 Sections Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solutions 4.4 Types of Chemical Reactions precipitation, acid-base, redox reactions 4.5 Precipitation Reactions 4.6 Describing Reactions in Solution Use Molarity to keep track of number of moles; serial dilution. Problems: , , , , , Discussion: 4.1, 4.2, 4.4, 4.5, 4.6

2 Water: The best known solvent

3 Because of hydrogen bond formation, water boils at a much higher temperature than CH 4 (90 K), which has the same molecular mass (to three sig figs).

4 Polar and Nonpolar Molecules Nonpolar: Electrons Equally Shared Polar: Electrons Unequally Shared

5 Which of the following molecules is non-polar? 1. HCl 2. CHCl 3 3. CO 4. BrCl 5. NH 3 6. None of the above

6 Dissolve a salt in water

7 Polar water molecules interact with the positive and negative ions of a salt, making the dissolving process possible.

8 The Role of Water as a Solvent: The solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong Electrolytes. ( ) ( ) + ( ) NaCl s + H O Na aq + Cl ( aq) 2 When Sodium Chloride dissolves into water the ions become solvated, and are surrounded by water molecules. These ions are called aqueous and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

9 Electrical Conductivity of Ionic Solutions

10 Completely Ionize Strong Acid (HCl) and Strong Base (NaOH) + ( ) ( ) + ( ) ( ) ( ) ( ) HCl aq H aq Cl aq H aq H O aq H O H n + ( ) ( ) + ( ) NaOH aq Na aq OH aq

11 Partially Ionize in Solution Weak Acid (HAc) and a Weak Base (NH 3 ) + + ( ) ( ) + ( ) NH3( aq) NH 4 ( aq) + OH ( aq) ( ) ( ) HAc aq H aq Ac aq Ac aq CH COO aq 3

12 Strong Electrolytes Produce ions in aqueous solution and conduct electricity well. Strong electrolytes are soluble salts, strong acids and strong bases. Strong acids produce H + ions when they dissolve in water. HCl, HNO 3 and H 2 SO 4 are strong acids HNO 3 (aq) H + (aq) + NO 3- (aq) NaOH and KOH are strong bases: NaOH(s) Na + (aq) + OH - (aq) (Produce OH- ions) All of the above species are ionized nearly 100% in water.

13 Weak Electrolytes Produce relatively few ions in aqueous solution The most common weak electrolytes are weak acids and weak bases. Acetic acid is a typical weak acid: HCH COO aq H aq + C H O aq ( ) + ( ) ( ) Ammonia is a common weak base: NH aq + H O NH aq + OH aq ( ) ( ) + ( ) ( ) Weak bases like NH 3 and weak acids like acetic acid don t conduct electricity well; they partially ionize. These two (acetic acid and ammonia) ionize only ~1%

14 Nonelectrolytes Dissolve in water but produce no ions in solution. Nonelectrolytes do not conduct electricity because they dissolve as whole molecules, and produce no ions. Common nonelectrolytes include ethanol and table sugar (sucrose, C 12 H 22 O 11 ) Non electrolytes contain polar parts; can hydrgeon bond with water. Nonpolar covalent compounds can t form hydrogen bonds and have little or no interactions with water molecules. Examples are the hydrocarbons in gasoline and oil, which don t mix with water.

15 Interaction of Water and Ethanol A water soluble non-electrolyte, has polar components (COH) that hydrogen-bonds; fits in with water network.

16 Carbohydrates Molecules that contain carbon and water! C x H 2y O y or C x (H 2 O) y ardemo.htm H C HO H CH 2 OH C H OHH C C OH O C Sucrose CH 2 OH H C H O C OH O OH C H C H CH 2 OH Sucrose C 12 H 22 O 11, C 12 (H 2 O) 11 a disaccharide Glucose C 6 H 12 O 6, (CH 2 O) 6 a monosaccharide

17 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds Problem: How many moles of each ion type are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water Other examples later: Both mass and moles are conserved in solutions, even if chemical forms change. H 2 O a) Na 2 CO 3 (s) 2 Na + (aq) + CO 3-2 (aq) moles of Na + = 4.0 moles Na 2 CO 3 x 2 mol Na + 1 mol Na 2 CO 3 = 8.0 moles Na + and 4.0 moles of CO -2 3 are present

18 Dissolve a salt in water How do you describe it? Problem: Take 28 grams of NaCl and put it in 250 mls of solution. What do you write on the bottle? Possibilities?

19 Figure 4.9: Steps involved in the preparation of a standard solution.

20 Molarity (Concentration of Solutions) = M Moles of Solute mol M = = Liters of Solution L The solute is material dissolved into the solvent In sea water, water is the solvent, and NaCl, MgCl 2, etc are the solutes. In brass, copper is the solvent (90%), and zinc is the solute (10%). Compare with density which is the mass of the solute in the solution. mass moles M moles = = = = = Volume Volume Volume W d MW Molarity MW M MW Does it matter whether the salt dissociates or not?

21 Prepare a solution Problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to ml. What is the molarity, M, of the salt? What about (the M-ity of) each of the ion types? Strategy?

22 Prepare a solution Problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to ml. What is the molarity, M, of the salt? What about each of the ions? Density? Strategy: Think about what you would write on the bottle to describe the contents. The concentration is the quantity that will not change when you pour some out. Concentration can be expressed in moles per liter (Molarity) or grams per liter (Density). mass 3.95g d = = = 13.2 g Volume 0.30L l M ( ) W Na 3 PO 4 = g mol

23 Prepare a solution Problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to ml. What is the molarity, M, of the salt? What about each of the ions? Density? Strategy: Calculate the Concentration (as Molarity) of the salt. C n C ( ) Na PO 3 4 ( ) Na PO 3 4 ( ) Na PO g d = = l = MW g mol 3.95g = = mol M W M n mol = = = M V 0.30L

24 Concentration of Ions Strategy: Start with the Molarity of the salt, and the + 3 chemical reaction. Na PO s 3Na aq + PO aq Reaction shows: 3 moles Na + ions per mole of loss of salt. 1mole Reaction = 1 mole( loss) Na PO mole Reaction = 3moles Na ions + + ( + ) 3moles Na 3Na sc.. Na = = 1mole Reaction 1Reaction Assume the rxn goes to completion (no reactant left). ( 3 ) ( + ) sc.. Na ( + M Na ) = M ( Na3PO4) = M = 0.241M sc.. Na PO ( ) 3 4 ( 3 ) 4 sc.. PO M PO = M Na PO = M ( ) ( ) sc.. Na3PO4 ( ) ( ) ( ) 3 4 4

25 Dilution: Convert a Stock solution to a required concentration

26 Dilution Problem A medical team needs from pharmacy 500mls isotonic saline (0.14M) for a patient. You have 1 liter of 1.50 M saline (NaCl) solution. What do you (in pharmacy) do?

27 Dilute using Volumetric Pipets

28 Dilution Problem A medical team needs from pharmacy 500mls isotonic saline (0.14M) for a patient. You have 1 liter of 1.50M saline (NaCl) solution. What do you (in pharmacy) do? Need to find the initial volume of the 1.50M NaCl needed to make 500 mls of 0.14M NaCl. Use the fact that the number of moles Is what counts and what is the same (i.e. conserved). moles NaCl need = 0.14M 0.5L = 0.070moles ( ) moles NaCl available = 1.50M V ( ) M V0 = 0.14M 0.5L 0.14M 0.5L V0 = = L= 46.7mls 1.50M Take 46.7mls of stock solution and dilute to 500 mls and voila!

29 General Dilution Problem Need volume molarity M1 Have molarity M. What is? o Vo General Answer: Moles are conserved so: M o > M 1 moles = V M = V M o o 1 1 Rearrange the general expression for Vo. Can rearrange the general expression for any of the 4 quantities.

30 Two step serial Dilution Begin with a 3.5M (stock) solution of NaCl. Take 10 mls of this solution and dilute to 100 mls with water. Then tak 5 mls of the new solution and dilute to 300 mls with water (as a second serial dilution). What is the concentration of the salt after the second dilution?

31 Two step serial Dilution Begin with a 3.5M (stock) solution of NaCl. Take 10 mls of this solution and dilute to 100 mls with water. Then tak 5 mls of the new solution and dilute to 300 mls with water (as a second serial dilution). What is the concentration of the salt after the second dilution? Strategy: Note: final concentration means molarity, M mol n= M * V n= mls leave volume in mls, n is in millimoles l n M = : M1 = = 3.5 mol First Dilution factor V l 10 n= M * V: n2 = M1 5mls = 3.5 5mol moles taken from first dilution First Dilution Second Dilution n n M = 3.5 = mol l 2 = : M2 = = V V2 Serial Dilution as a product of dilution factors

32 Precipitation Reaction Pb(NO 3 ) 2 is in beaker (colorless and soluble in water) and NaI is in graduate (also colorless and soluble). The mixture produces a yellow ppt. which is PbI 2. All 4 ions are in solution. What pairings are possible? Hint: 6 is max. Many can be ruled out. 2+ Pb NO 3 Na I +

33 Table in Text Rules 1 and 2, for our purposes all are soluble (not most). Think Ammonium Nitrate the soluble salt. More info in chapter 8.

34 Molecular equation, Total ionic equation, and net ionic equation

35 Predicting if a precipitate forms, and which? Pb(NO 3 ) 2 (aq) + NaCl(aq) Pb +2 (aq) + 2 NO 3- (aq) + Na + (aq) + Cl - (aq) If any of the possible new species formed by combining anions with cations is insoluble, then that precipitate will form. Salts of sodium and nitrates do not ppt. USE TABLE 4.1 In this case, PbCl 2 is the only salt possibly insoluble, and a precipitate forms. Note: Like the demo with PbI 2 (s) above.

36 End of Lecture Extra Infromation available in the EXTRA file also on line.