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. 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.

10. 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

11. 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

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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. . 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

20. 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