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Solutions. “If you’re not part of the solution, you’re a part of the precipitate.” -Stephen Wright. Solutions, Colloids, and Suspensions. _______________ are homogeneous mixtures with particles large enough so they are visible to the naked eye. Examples: __________________.
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Solutions “If you’re not part of the solution, you’re a part of the precipitate.” -Stephen Wright
Solutions, Colloids, and Suspensions • _______________are homogeneous mixtures with particles large enough so they are visible to the naked eye. Examples: __________________. • _________ are “____" or “_____". They do not transmit light. • _________ separate on standing.
_______ • ___________look homogeneous to the naked eye. Colloids frequently appear “______" or “______". The particles are large enough to scatter light. (The _______ Effect). • Examples: __________.
____________ • _________are mixtures with particle sizes at the molecule or ion level. _______ can usually pass through the solution. If the solute is able to absorb visible light then the solution will have a color, but it will still be transparent.
________________ • Colloidal dispersion of liquid in liquid. • An agent needs to be added to make stable. • Examples: Mayonnaise, Shampoos, Cosmetics, lotions…
Some Definitions A solution is a _______________mixture of 2 or more substances in a single phase. One constituent is usually regarded as the ___________and the others as ____________.
Parts of a Solution • __________– the part of a solution that is being dissolved (usually the lesser amount) • __________– the part of a solution that dissolves the solute (usually the greater amount) • _______ + ________ = __________
Factors affecting rate of solvation • For you to discover in lab this week!
– – – – – – – – – + + + + + + + + + + + + + + + + + + Why oil and water don’t mix • The more similar the strength of their dipoles the more likely two compounds are to mix. Polar substances only dissolve other polar substances (________). Oil (non-polar) and water (polar) are ___________. • “_________________________”.
Solubility of Solids vs. Temperature ______________of several ionic solid as a function of temperature. MOST salts have ________ solubility in hot water. Why? A few salts have negative heat of solution, (exothermic process) and they become less soluble with increasing temperature.
Temperature & the Solubility of Gases • The solubility of gases __________at higher temperatures. Why? • Solubility of gas solute follows ____________which states that the amount of solute gas dissolved in solution is directly proportional to the amount of pressure above the solution.
Henry’s Law & Soft Drinks • Soft drinks contain “carbonated water” – water with dissolved carbon dioxide gas. • The drinks are bottled with a CO2 pressure greater than 1 atm. • When the bottle is opened, the pressure of CO2 decreases and the solubility of CO2 also decreases, according to Henry’s Law. • Therefore, bubbles of CO2 escape from solution.
Definitions Solutions can be classified as __________or ___________. A ___________solution contains the maximum quantity of solute that dissolves at that temperature. An _____________solution contains less than the maximum amount of solute that can dissolve at a particular temperature
Saturated Solution • ____________________________ • Rate of _______is equal to the rate of ______________
Example: Saturated and Unsaturated Fats 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. 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.
Solubility Curves • If your concentration is ON the line, your solution is ____________________. • If it is BELOW the line, your solution is ______________________. • If it is ABOVE the line, your solution is ______________________.
Definitions _________________ SOLUTIONS contain more solute than is possible to be dissolved _________________ solutions are unstable. The _____________ 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.
K+(aq) + MnO4-(aq) IONIC COMPOUNDS Many reactions involve ionic compounds, especially reactions in water — ____________ solutions. KMnO4 in water
Aqueous Solutions How do we know ions are present in aqueous solutions? The solutions _________ __________ They are called _________________ HCl, MgCl2, and NaCl are ____________They dissociate completely (or nearly so) into ions.
Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called ____________. Examples include:
___________ in the Body • Carry messages to and from the brain as electrical signals • Maintain cellular function with the correct concentrations electrolytes
( ) = _____________ of Solute The amount of solute in a solution is given by its _____________.
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. Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate Molarity
USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution?
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
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.
The Dilution Solution • Here's how to prepare a dilution from a stock solution. • As an example, say you need to prepare 50 ml of a 1.0 M solution from a 2.0 M stock solution. Your first step is to calculate the volume of stock solution that is required.
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!
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.
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 % How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?
Try this molality problem • 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution.
________ Properties On adding a solute to a solvent, the properties of the solvent are modified. • Vapor pressure _________ • Melting point _________ • Boiling point _________ These changes are called ______________ PROPERTIES. They depend only on the _________of solute particles relative to solvent particles, not on the ______of solute particles.
Change in Freezing Point Ethylene glycol/water solution Pure water The freezing point of a solution is _______than that of the pure solvent
Change in Freezing Point Common Applications of Freezing Point Depression Ethylene glycol – deadly to small animals Propylene glycol
Change in Boiling Point Common Applications of Boiling Point Elevation
Boiling Point Elevation and Freezing Point Depression ___________ 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
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?
Freezing Point Depression Calculate the Freezing Point of a 4.00 molal glycol/water solution. Kf = 1.86 oC/molal (See Kf table) Solution
DO NOW At what temperature will a 5.4 molal solution of NaCl freeze? Solution