Unit 13: Solutions

Solvation - Definitions Solution - homogeneous mixture Solute - substance being dissolved Solvent – substance that dissolves the solute

A. Definitions Solute - KMnO4 Solvent - H2O

B. Solvation First... Then... Solvation – the process of dissolving solute particles are surrounded by solvent particles First... solute particles are separated and pulled into solution Then...

B. Solvation Non- Electrolyte Weak Electrolyte Strong Electrolyte + sugar - + acetic acid - + salt Non- Electrolyte Weak Electrolyte Strong Electrolyte solute exists as molecules only solute exists as ions and molecules solute exists as ions only DISSOCIATION IONIZATION View animation online.

B. Solvation “Like Dissolves Like” NONPOLAR POLAR

B. Solvation Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

Solubility Unsaturated When the solvent holds less solute than it normally can at a given temperature.

Solubility Saturated When the solvent holds as much of a solute as it normally can at a given temperature.

Solubility Supersaturated When the solvent holds more dissolved solute than it normally can at that temperature.

C. Solubility UNSATURATED SOLUTION more solute dissolves no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

C. Solubility Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln

C. Solubility Solubility Curve shows the dependence of solubility on temperature

C. Solubility Solids are more soluble at... high temperatures. Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

Solubility Curve

Things to remember The graph is set in 100 mL or 100 grams of water. The line represents the saturation point at that temperature. Anything above the line (at that temp) is supersaturated Anything below the line (at that temp) is unsaturated.

Rules to solubility problems 1. Look up the amount of water given in the problem. (highlight it) If it is 100 mL or 100 grams, look the answer up on the graph.

Example What amount of NaCl would make a saturated solution in 100 mL of water at 800C? Just look it up on the graph. Answer : 40 grams GRAPH

What if it is not 100 mL or 100 grams of water? Set up a proportion: solid solid = liquid liquid This will be x From graph From problem Always 100 mL

Example What amount of KNO3 would make a saturated solution in 177mL of water at 500C? 60 g x = 100mL 177mL = 106.2 g GRAPH

What if there is no amount of water given? Set up a proportion: solid solid = liquid liquid From graph From problem Always 100 mL This will be x

Example What amount of water at 200C would make a saturated solution with 63g of KNO3? 20 g 63 g = 100mL X = 315 mL GRAPH

Stuff from yesterday a. Rate of Solubility depends upon 1.the size of the solute crystals 2.the vigor and duration of stirring 3.the temperature of the solvent b. Substances may be soluble or insoluble 1. refers to substance than be dissolved into water 2. insoluble refers to substances which are only very, very slightly soluble in water (ex. Chalk is insoluble in water)

Precipitation Reactions: What are they? Type of double replacement reaction. Two solutions of ionic compounds are mixed One of the products of the reaction is an insoluble salt called a precipitate (chunky milk) Use solubility rules to determine the precipitate.

Solubility Rules Solubility (in water) Compounds Exceptions Soluble Alkali metal (Group 1) salts none Ammonium salts (NH4+) Acetates (C2H3O2 -) Nitrates (NO3 -) Chlorides (Cl -), Bromides (Br -), Iodides (I -) Compounds of Ag, Hg, and Pb Sulfates (SO4 2-) Compounds of Sr, Ba, Hg, and Pb ********** *************************** Insoluble Carbonates (CO3 2-), Phosphates (PO4 3-), Sulfites (SO3 2-), Chromates (CrO42-) Compounds of Alkali metal (Group 1) and NH4+ Sulfides (S 2-), Hydroxides (OH -) Compounds of Alkali metal (Group 1) and NH4+, Ca, Sr, Ba

Example using chart Insoluble Pb(OH)2 NH4Cl K2SO4 Soluble Ba3(PO4)2 RULES

How to use it. Write the products for the double replacement reactions and balance. Then, using the solubility rules identify the precipitate by a subscript (s) for solid (insoluble in water). Identify the soluble compounds by marking them with a subscript (aq) for aqueous (soluble in water).

Example Ag(NO3)2 + LiCl  Ag(NO3)2 + 2 LiCl  AgCl2 + 2 LiNO3 AgCl2 +1 -1 +2 -1 AgCl2 + LiNO3 aq aq s aq RULES

Practice Problems 1) Na2CO3 + Ca(NO3)2  2) Pb(C2H3O2)2 + Na2S  3) K3PO4 + MgCl2  4) (NH4)2SO4 + BaI2 

Practice #1 Na2CO3 + Ca(NO3)2  Na2CO3 + Ca(NO3)2 2NaNO3 +CaCO3 NaNO3 +2 -2 -1 +1 NaNO3 + CaCO3 aq aq s aq RULES

Practice #2 Pb(C2H3O2)2 + Na2S  PbS NaC2H3O2 -1 +1 +2 -2 PbS + NaC2H3O2 s aq aq aq RULES

Practice #3 K3PO4 + MgCl2  2K3PO4 + 3MgCl2 6 KCl + Mg3(PO4)2 KCl +2 -1 -3 +1 KCl + Mg3(PO4)2 aq aq aq s RULES

Practice #4 (NH4)2SO4 + BaI2  (NH4)2SO4 + BaI2  2 NH4I + BaSO4 NH4I +2 -1 NH4I + BaSO4 +1 -2 aq aq s aq RULES

Practice Problems 1) Na2CO3 (aq) + Ca(NO3)2(aq) 2 NaNO3 (S)+ CaCO3(aq) 2) Pb(C2H3O2)2 + Na2S PbS + 2NaC2H3O2 3) 2K3PO4 + 3MgCl2 6 KCl + Mg3(PO4)2 4) (NH4)2SO4 + BaI2  2 NH4I + BaSO4

A. Concentration The amount of solute in a solution. Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

Percent Concentration Percent concentration is solid (solid + liquid) (solid + liquid) x 1000 = ppt (solid + liquid) x 1000000 = ppm x 100 = pph

Example What is the percent concentration (pph) of NaCl when 42 g are dissolved in 379g of water? 42g (42g + 379g) x 100 = 421 9.98% 0r 9.98 pph

B. Molarity solvent only 1000mL = 1 L

B. Molarity Find the Molarity of a solution containing .5 mol of glucose in 2 L of water. GIVEN: M = ? Mol = .5 mol V = 2 L WORK: M= mol solute/ liters solvent M= .5 mol 2 L = .25 M

C. Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

C. Dilution What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

D. Preparing Solutions 1.54m NaCl in 0.500 kg of water 500 mL of 1.54M NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL water 45.0 g NaCl 500 mL mark volumetric flask