Student Ch 15 Solutions

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Solutions Why does a raw egg swell or shrink when placed in different solutions? Chemistry I – Chapters 15 & 16 Chemistry I HD – Chapter 15 ICP – Chapter 22 SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

Some Definitions: 

Some Definitions A solution is a _______________ mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

Parts of a Solution: 

Parts of a Solution SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) Solute + Solvent = Solution


Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

Example: Saturated and Unsaturated Fats: 

Example: Saturated and Unsaturated Fats Unsaturated fats have at least one double bond between carbon atoms; monounsaturated means there is one double bond, polysaturated means there are more than one double bond. Thus, there are some bonds that can be broken, chemically changed, and used for a variety of purposes. These are REQUIRED to carry out many functions in the body. Fish oils (fats) are usually unsaturated. Game animals (chicken, deer) are usually less saturated, but not as much as fish. Olive and canola oil are monounsaturated. Saturated fats are called saturated because all of the bonds between the carbon atoms in a fat are single bonds. Thus, all the bonds on the carbon are occupied or “saturated” with hydrogen. These are stable and hard to decompose. The body can only use these for energy, and so the excess is stored. Thus, these should be avoided in diets. These are usually obtained from sheep and cattle fats. Butter and coconut oil are mostly saturated fats.


Definitions SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: Warm the solvent so that it will dissolve more, then cool the solution Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

Supersaturated Sodium Acetate: 

Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”

IONIC COMPOUNDS Compounds in Aqueous Solution: 

IONIC COMPOUNDS Compounds in Aqueous Solution Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO4 in water

Aqueous Solutions: 

How do we know ions are present in aqueous solutions? The solutions _________________________ They are called ELECTROLYTES HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions

Aqueous Solutions: 

Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugar ethanol ethylene glycol

It’s Time to Play Everyone’s Favorite Game Show… Electrolyte or Nonelectrolyte!: 

It’s Time to Play Everyone’s Favorite Game Show… Electrolyte or Nonelectrolyte!

Electrolytes in the Body: 

Electrolytes in the Body Carry messages to and from the brain as electrical signals Maintain cellular function with the correct concentrations electrolytes

Concentration of Solute: 

Concentration of Solute The amount of solute in a solution is given by its concentration.


1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity.: 

PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate Molarity [NiCl2•6 H2O ] = 0.0841 M


Step 1: Change mL to L. 250 mL * 1L/1000mL = 0.250 L Step 2: Calculate. Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles Step 3: Convert moles to grams. (0.0125 mol)(90.00 g/mol) = 1.13 g USING MOLARITY moles = M•V What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution?

Learning Check: 

Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 1) 12 g 2) 48 g 3) 300 g

Concentration Units: 

An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this! Concentration Units

Two Other Concentration Units: 

Two Other Concentration Units grams solute grams solution MOLALITY, m % by mass = % by mass

Calculating Concentrations: 

Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate molality and % by mass of ethylene glycol.

Calculating Concentrations: 

Calculating Concentrations Calculate molality Calculate weight %

Learning Check: 

Learning Check A solution contains 15 g Na2CO3 and 235 g of H2O? What is the mass % of the solution? 1) 15% Na2CO3 2) 6.4% Na2CO3 3) 6.0% Na2CO3

Using mass % : 

Using mass % How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?

Try this molality problem: 

Try this molality problem 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution. m = mol solute / kg solvent 25 g NaCl 1 mol NaCl 58.5 g NaCl = 0.427 mol NaCl Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl 5 kg water = 0.0854 m salt water

Colligative Properties: 

Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified. Vapor pressure decreases Melting point decreases Boiling point increases Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

Change in Freezing Point : 

Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent Pure water Ethylene glycol/water solution

Change in Freezing Point : 

Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals

Change in Freezing Point : 

Common Applications of Freezing Point Depression Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? sand, SiO2 Rock salt, NaCl Ice Melt, CaCl2 Change in Freezing Point

Change in Boiling Point : 

Change in Boiling Point Common Applications of Boiling Point Elevation

Boiling Point Elevation and Freezing Point Depression: 

Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -) Compound Theoretical Value of i glycol 1 NaCl 2 CaCl2 3 Ca3(PO4)2 5

Boiling Point Elevation and Freezing Point Depression: 

Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i m = molality K = molal freezing point/boiling point constant

Change in Boiling Point : 

Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution? Kb = 0.52 oC/molal for water (see Kb table). Solution ∆TBP = Kb • m • i 1. Calculate solution molality = 4.00 m 2. ∆TBP = Kb • m • i ∆TBP = 0.52 oC/molal (4.00 molal) (1) ∆TBP = 2.08 oC BP = 100 + 2.08 = 102.08 oC (water normally boils at 100)

Freezing Point Depression: 

Calculate the Freezing Point of a 4.00 molal glycol/water solution. Kf = 1.86 oC/molal (See Kf table) Solution ∆TFP = Kf • m • i = (1.86 oC/molal)(4.00 m)(1) ∆TFP = 7.44 FP = 0 – 7.44 = -7.44 oC (because water normally freezes at 0) Freezing Point Depression

Freezing Point Depression: 

At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1 oC FP = 0 – 20.1 = -20.1 oC Freezing Point Depression

Preparing Solutions: 

Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.


ACID-BASE REACTIONS Titrations H2C2O4(aq) + 2 NaOH(aq) ---> acid base Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION.


Setup for titrating an acid with a base


Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION.

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