Solutions Homogeneous mixture Solute Solvent

Solution Formation Rate Factors affecting it… –Temperature—think about dissolving sugar in tea –Agitation—again think about the tea –Particle size—granulated sugar vs. powdered sugar?

Henry’s Law The solubility (C) of a gas in a liquid is directly proportional to the pressure above the liquid C 1 = C 2 P 1 P 2 Think about carbonated beverages

Henry’s Law Problem At 20°C and 1.00atm, the solubility of oxygen gas in water is g/L. What will the solubility be if the pressure is increased to 1.75atm? Answer: g/L

Concentration Qualitative descriptions do not give amounts of solute in solution. –Concentrated—lots of solute –Dilute—not much solute Note: no quantities are provided

Concentration A few types… –Molarity –Mass percent –Mass/volume percent –Volume/volume percent –Mole fraction –Molality

Molarity Represented by M—always uppercase M = #moles of solute #liters of solution Units will be mol/L, or you may write it as M A 2M solution is described as “two molar”

Mass Percent Represented as m/m% m/m% = mass of solute x 100 mass of solution Mass units must be the same for the solute and the solution Unit-less

Mass/volume Percent Represented as m/v% m/v% = #g of solute x 100 #mL of solution Units are specific g/mL

Volume/volume Percent Represented by (v/v)% (v/v)% = volume of solute x 100 volume of solution Volume units must be the same for the solute and the solution Unit-less

Mole Fraction Represented by Greek letter, chi, χ χ solute = #mol of solute #mol of solution χ solvent = #mol of solvent #mol of solution Unit-less Sum will be one

Molality Represented by m—always lowercase m = #moles of solute #kg of solvent Units will be mol/kg, or you may write it as m A 2m solution is described as “two molal”

Dilutions I don’t usually stock every concentration of every acid that I need…so I start with the most concentrated form and make whatever molarity (the most common measurement of concentration) I need. (M 1 )(V 1 ) = (M 2 )(V 2 )

Dilutions The rule of thumb is if you’re making an acid dilution in a beaker, always add the acid to water…so if something splashes, it will be the water!

Practice #1 How many grams of sodium hydroxide are required to make 250mL of a 0.25M solution? Answer: 2.50g NaOH

Practice #2 How many grams of water are present in a 5.00%(m/m) solution containing 0.875g of calcium acetate? Answer: g H 2 O

Practice #3 How many grams of acetic acid are present in 4.50L of a 5.00%(m/v) solution? Answer: 225g HC 2 H 3 O 2

Practice #4 A 95% (v/v) solution of ethanol in double-distilled water is used to clean surfaces in a laboratory setting. If you have 500mL of the solution, how many of those mL are water? Answer: 25mL H 2 O

Practice #5 10.0g of calcium acetate are added to 100g of water. What are the mole fractions of both the solute and the solvent? Answer: χ calcium acetate = χ water = 0.989

Practice #6 10.0g of calcium acetate is added to 100g of water. What is the molality of this solution? Answer: 0.632mol/kg or m

Practice #7 If I need to make 250mL of 6M sulfuric acid, and all I have in my cabinet is full strength, 18M sulfuric acid, describe how I make the dilution. Answer: measure 83.3mL of 18M H 2 SO 4 and pour into the 250-mL volumetric flask. Fill the flask to the etched line with water, cork it, and agitate it.

A Doozie Problem A solution is prepared by mixing 1.00g of ethanol, CH 3 CH 2 OH, with 100.0g of water to give a final volume of 101.0mL of solution. Calculate the molarity, mass percent, mass/volume percent, molality, mole fraction of the ethanol, and mole fraction of the water.

A Doozie’s Answers… molarity = 0.215M or 0.215mol/L mass percent = 0.990% mass/volume percent = 0.990(g/mL)% molality = 0.217m or 0.217mol/kg mole fraction ethanol = mole fraction water = 0.996