Chapter 14 - Mixtures and Solutions What is a heterogeneous mixture? Substances do not mix evenly (on the molecular level)

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1 Chapter 14 - Mixtures and Solutions What is a heterogeneous mixture? What is a suspension? What is a solution? Notes Substances do not mix evenly (on the molecular level) mixture with particles that will settle out and separate into layers Homogeneous mixture- evenly mixed at molecular level, no separation What are the two parts of a solution and their definitions? Solvent: Dissolving medium, substance that tears apart other component (substance in the greater amount) Solute: Substance that GETS dissolved, usually in LESSER quantity What phases of matter can combine to form solutions? What two terms qualitatively describe how much solute is in a given amount of solvent? Give their definitions. What is solvation? What is our rule for which things dissolve in which? Write down the examples. solid + liquid = sugar in water gas + liquid = ammonia in water liquid + liquid = ethyl alcohol in water antifreeze in water gas + gas = oxygen and nitrogen solid + solid = alloys (bronze, brass) DILUTE: Qualitative way of saying weak, little solute compared to solvent CONCENTRATED: Qualitative way of saying strong, LOT of solute compared to solvent Attractive forces between solute and solvent particles is greater than forces between solute particles and the Solute is pulled out and surrounded by solvent Like dissolves like (polar will dissolve in polar and nonpolar will dissolve in nonpolar but not polar and nonpolar) Salts (NaCl) - polar water molecules attracted to ionic charges Molecular (sugar, alcohol) - water H-bonds with -OH s Oil - no charges or -OH s so it doesn t dissolve in water 3 Things that affect How much will dissolve Nature of solute Temperature (except for gases usually increasing Temp will increase solubility) Pressure only affects gases. Increasing P will increase the amount that dissolves

2 3 Things that affect the Rate (how fast) of salvation Temperature - higher temperature = faster particle movement and faster dissolving Surface Area - greater surface area allows more solvent-solute contact and faster dissolving Grinding/crushing = greater surface area Agitation -stirring moves dissolved solute away and allows faster dissolving To dissolve fast: heat it up, crush it up, stir it up What is the difference between the solubility of gases in water versus solids? see above What are the definitions of: Saturated Unsaturated Supersaturated Solution is holding MAXIMUM amount of solute at constant T & P. Solution is holding LESS than maximum. Solution is holding MORE than saturated amount solution can become supersaturated by allowing a warm, saturated solution to cool carefully so no crystals form What is Molarity? Quantitative way of expressing concentration It s a number of particles dissolved in a given volume of solvent M = Moles of solute Liters of solution Important to remember: moles=grams given molar mass Also remember 1000.mL= 1.000L First type of molarity problem: Ex. 1- Find the Molarity of saline solution if there is 2.23 g of NaCl in 225 ml of solution Second type of molarity problem: Ex. 2 - Find the moles of solute in L of a 0.333M H 2 SO 4 solution g NaCl x 1 mol NaCl = mol NaCl g NaCl 225 ml =.225 L M = mol =.170 M NaCl.225 L 333 M = x mol x = (.333 mol/l)(.678l).678 L x =.226 mol H 2 SO 4

3 Third type of molarity problem: Ex. 3- Find the grams of solute in a 2.20M NaHCO 3 solution if there is 5.51 liters of it. 1st find moles of NaHCO M = x mol 5.51 L x = 12.1 mol NaHCO 3 Then find grams 12.1 mol NaHCO 3 x g NaHCO 3 = 1 mol NaHCO g NaHCO 3 What is the formula we use for DILUTION problems? Dilution: lower the ratio of solute to solvent by the addition of water. Make less concentrated Addition of water WILL NOT change Moles of Solute M 1 V 1 = M 2 V 2 Original = Conditions New, Diluted Conditions Example dilution problem: How many ml of 18.1 M KMnO 4 is needed to produce 125mL of 0.100M KMnO 4 solution? M 1 = 18.1 M M 2 = M V 1 =?? V 2 = 125 ml (18.1 M) V 1 = (0.100 M) (125 ml) V 1 = (0.100 M) (125 ml) =.691 ml (18.1 M) What is the definition of COLLIGATIVE PROPERTIES? Show how you determine the number of particles a solute will make when it dissolves: Properties of a solution that are determined by how many particles are present and not what kind of particle NaCl Na + + Cl - = 2 particles CaCl 2 Ca Cl - = 3 particles AlCl 3 Al Cl - = 4 particles What 3 things are affected by colligative properties? vapor pressure,boiling point, freezing point Vapor Pressure is the Force per unit area exerted by molecules as they escape the surface of a liquid & turn to gas or vapor. If we take a Non-Volatile Solute (One that DOES not evaporate, NaCl does not evaporate

4 with the water) the solute particles keep the solvent particles from reaching the surface and evaporating. Therefore the pressure is less. Adding a non volative solute will always lower the vapor pressure of the solvent boiling = Temperature at which vapor pressure of liquid is equal to atmospheric pressure. (When liquid has equivalent or HIGHER vapor pressure, liquid will boil.) Since the addition of a solute will lower the vapor pressure we must increase the temp to get the vapor pressure back to where it should be. addition of a non volatile solute will always raise the boiling point Freezing point= temp at which a liquid forms a solid solute particles interfere with the solvent particles trying to stick together addition of a non volatile solute always lowers the freezing point How do you approach solution stoichiometry (problems where we have a molarity of one and asked for g, l, mol, of another) How many grams of AgCl are produced if 10.0 ml of a.75 M Ag 2 SO 4 solution react with NaCl? Ag 2 SO 4 +2NaCl 2AgCl (s) + Na 2 SO 4 As usual, you will be give info on one substance and asked for another. In these one of the sets of info will be a volume and a molarity.(get to moles! see Molarity equation) Use your MOLE RATIO from the balanced equations Use conversion factors to solve for unknown This is stoich because they give us sodium sulfate and want AgCl Get to Moles! 75 M = x moles x =.0075 mol Ag 2 SO 4.010L Change to grams of AgCl.0075 mol Ag 2 SO 4 x 2mol AgCl x 143.4g AgCl 1 mol Ag 2 SO 4 1 mol AgCl = 2.2 g AgCl Math guys:.0075mol/1mol = xg/2(143g)

5 How many ml of 0.25 M Na 2 SO 4 solution is needed to produce 1.65 g NaNO 3? Na 2 SO 4 + Ba(NO 3 ) 2 BaSO NaNO 3 This is stoich because they give us sodium nitrate and want sodium sulfate Get to Moles! since they give us grams just do a normal massmol problem 1.65 g NaNO 3 x 1 mol NaNO 3 x 1 mol Na 2 SO g NaNO 3 2 mol NaNO 3 =.0097 mol Na 2 SO 4 now since we have M and moles just find volume.25 M =.0097 mol x L =.0097/.25 =.039 L 39 ml What ml volume of 0.25M Na 2 SO 4 solution is needed to precipitate all the barium as BaSO 4 from 12.5mL of 0.15M Ba(NO 3 ) 2 solution This is stoich because they give us barium nitrate and want sodium sulfate. Get an equation Na 2 SO 4 + Ba(NO 3 ) 2 BaSO 4 +2 NaNO 3 Get to moles because we have vol and M let s use our Molarity equation to get moles 0.15M x L = mol Ba(NO 3 ) mol Ba(NO 3 ) 2 x 2 mol Na 2 SO 4 1 mol Ba(NO 3 ) mol Na 2 SO 4. now since we havemoles and they give us M we can calc vol (L) 0.25M = mol/L L= mls = 15 AgNO 3 + NaCl AgCl (s) + NaNO 3(aq) Calculate the molarity concentration of 25.0 ml of NaCl solution required to precipitate27.2ml of 0.104M AgNO3 solution. same as above This is stoich because they give us silver nitrate and want NaCl. Get to moles! L x 0.104M = mol AgNO mol AgNO 3 x 1mol NaCl = mol 1mol AgNO mol NaCl/0.025L =.113 M NaCl