1. Solutions Chapter 15

2. Mixtures Heterogeneous mixture- unevenly mixed substance (separation can be seen) Homogeneous mixture- evenly mixed substance (no separation can be seen)

3. Suspensions ~Small but visible particles suspended or floating in a gas or liquid (heterogeneous mixture) Like a snow globe or dust or “shake before using” the particles are too big to float forever without being stirred If a suspension sits, the particles will settle Can be filtered out

4. Colloids or Colloidal Suspension ~mixture that appears uniform unless under a high powered microscope. Particles are a little larger than the wavelength of light Extremely light particles float almost indefinitely. Milk, blood, smoke These can be separated in a centrifuge

5. Tyndall Effect ~Scattering of light by a colloid or suspension Both a colloid and a suspension have particles larger than the wavelength of light, so when light shines through it should be deflected every which way. This will make the beam of light visible.

6. Solutions Particles are smaller than the wavelength of light. Therefore, it will not scatter light. With solutions, no separation can be seen even under a high powered microscope. Cannot be separated by any filter or by a centrifuge. Can be separated by boiling/ melting points. salt water, metal alloys, air

7. Tyndall Effect Colloid/suspension solution

8. Parts of a solution Solvent- what the substance is dissolved in Solute- what is being dissolved Water is called the “universal solvent” because it dissolves a lot of substances and is very common. Water solutions are called aqueous.

9. Mass and volume In a solution, mass is conserved, however, volume is not. That is to say, the mass of a solution = mass of the solute + solvent. The volume of a solution may not equal the volume of the solute +solvent.

10. Example It is easy to think of sand and water (not a solution, but it works for the general concept) If you mix a liter of sand and a liter of water you get… A mixture that is more than one liter but less than 2 liters. Now this applies to solutions, if you mix 1 L of water with.5 liter of Na 2 CO 3 the resultant solution is more than 1 L but less than 1.5 L

11. Density of solutions Increasing the mass of the solution and not increasing the volume comparatively will increase the density. Dissolving solids into water almost always increases the density. How much the density increases, depends on how much is dissolved.

12. Solution misconceptions Solutions don’t have to be a solid in a liquid. carbonated water is CO 2 dissolved in water, streams have dissolved O 2 in them. The solvent doesn’t have to be water or even a liquid. Alloys (two or more metals) are a solution as is air. Several things dissolve in oils.

13. Gases Gases dissolved in water tend to decrease the density of the solution. Again the volume of the solution does NOT increase anywhere near the volume of the gas + water, but it does increase at a greater rate than the mass.

14. Liquids Liquids may increase or decrease the density of the solution dependent on whether they are more or less dense than the solvent. Rubbing alcohol will decrease the density of a water solution, where acetic acid will increase the density of a water solution.

15. Coke v. Diet Coke Coke cans sink in water, diet coke floats. That means a coke can is more dense than water, diet coke is less dense. Aluminum is more dense than water, but there is head space, a little air pocket, at the top of the can. Diet Coke (and all diet beverages) use artificial sweeteners like Nutrasweet. Nutrasweet is 200x sweeter than sugar, so you need to dissolve less in the solution, making it less dense

16. Concentration ~How much solute is present in a solution compared to the solvent. Molarity (M)- moles of solute per liter of solution. M = mol/L 2.1 M AgNO 3 means 2.1 mol of AgNO 3 for every one liter of solution

17. Other measures of concentration Name Abbrev. What it is molalitymmol solute/kg solvent parts per millionppmg solute/g solvent x 10 6 parts per billionppbg solute/g solvent x 10 9 mole fractionxmol solute/mol solution percent by mass%g solute/g solution x 100 percent by volume%mL solute/mL solvent x100

18. Molarity Problems Molarity = mol/L Molarity = moles of solute / Liters of solution

19. Molarity problems How many moles of HCl are in 125 mL of 2.5 M HCl? 2.5 mol HCl 1 L of soln..125 L of soln =.31 mol HCl

20. Here we go What concentration solution would be prepared if 39 g of Ba(OH) 2 were mixed in a 450 mL solution? 39 g Ba(OH) 2 1 mol Ba(OH) 2 171.316 g Ba(OH) 2 =.2276 mol Ba(OH) 2.2276 mol Ba(OH) 2.45 L of solution M = mol/L =.51 M Ba(OH) 2

21. More For a lab in this chapter, I need to make.60 L of 3.0 M NaOH, what mass of NaOH did I need?.6 L x 3.0 M NaOH = 1.8 mol NaOH 1.8 mol NaOH x 39.998 g/mol = 72 g NaOH

22. Molarity Problems A 0.24 M solution of Na 2 SO 4 contains 0.36 moles of Na 2 SO 4. How many liters were required to make this solution? 0.36 mol Na 2 SO 4 1 L soln 0.24 mol = 1.5 L Na 2 SO 4

23. Getting tougher AgNO 3 + BaCl 2  AgCl + Ba(NO 3 ) 2 Balance the equation. If 1.2 L of.50 M AgNO 3 is reacted completely, what molarity solution of Ba(NO 3 ) 2 will be created if the volume increased to 1.5 L? 22 AgNO 3 =.6 mol AgNO 3 1.2 L x.5 M AgNO 3 =.6 mol AgNO 3. 6 mol AgNO 3 1 mol Ba(NO 3 ) 2 2 mol Ag NO 3 NO 3 ) 2 =.3 mol Ba( NO 3 ) 2 1.5 L NO 3 ) 2 =.20 M Ba(NO 3 ) 2

24. HNO 3 + Zn  H 2 + Zn(NO 3 ) 2 If you have.65 L of 1.2 M HNO 3 and you react it completely what volume of H 2 gas will you produce at STP? 2 HNO 3 x.65 L =.78 mol HNO 3 1.2 M HNO 3 x.65 L =.78 mol HNO 3. 78 mol HNO 3 1 mol H 2 2 mol HNO 3 =.39 mol H 2.39 mol H 2 22.4 L at STP 1 mol H 2 = 8.7 L at STP

25. . 78mol HNO 3 1 mol Zn(NO 3 ) 2 2 mol HNO 3 Zn(NO 3 ) 2 =.39 mol Zn(NO 3 ) 2.75 L Zn(NO 3 ) 2 =.52 M Zn(NO 3 ) 2 HNO 3 + Zn  H 2 + Zn(NO 3 ) 2 If you have.65 L of 1.2 M HNO 3 and you react it completely, what conc. of Zn(NO 3 ) 2 will be left if the volume increases to.75 L? HNO 3 x.65 L =.78 mol HNO 3 1.2 M HNO 3 x.65 L =.78 mol HNO 3 2

26. Fe + H 2 SO 4  Fe 2 (SO 4 ) 3 + H 2 If 350 mL of 2.3 M H 2 SO 4 is completely reacted, what is the volume of hydrogen gas produced at 24 o C and 114 kPa? 2 3 3 H 2 SO 4.35 L x 2.3 M =.805 mol H 2 SO 4. 805 mol H 2 SO 4 1 mol H 2 1 mol H 2 SO 4 =.805 mol H 2 PV = nRT 114 kPa V =.805 mol (8.31) 297 K =17 L H 2