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Transcript of Solutions Chemistry 5th & 8th Hall ©Hall2010. Solutions Solution –Homogenous mixture containing 2...
Solutions
Chemistry 5th & 8th
Hall©Hall2010
Solutions
• Solution– Homogenous mixture containing 2 or more
substances called solute and solvent
• Solute– Substance being dissolved
• Solvent– Dissolving medium
• Example– Salt into Water
Solutions Vocabulary
• Soluble– Substance that
dissloves in a solvent– E.g. salt into water,
sugar into water
• Insoluble– Substance that does
not dissolve into water
– E.g. sand into water
• Immiscible– 2 liquids that are not
soluble with each other– E.g. oil/water, oil/vinegar
• Miscible– 2 liquids that do dissolve
into each other– E.g. antifreeze,
water/vinegar
Solvation
• To form a solution, solute particles must separate.
• Solvation is the process of surrounding solute particles with solvent particles to form a solution
Ionic Compounds
• Water molecules are polar and in constant motion
• When an ionic compound is placed in water, the water molecules collide with the ionic crystal until entire crystal dissolves.
• Table salt into water
Molecular Compounds
• Like ionic, but the molecules are much bigger
• Sugar C12H22O11 into water
Factors that Affect Rate of Solvation
Increase in collisions
1. Agitation
2. Increase in surface area of solute
3. Increase Temperature of solvent
Heat of Solution
• During solvation, solute must separate into particles
• Solvent particles also must move apart• Energy is required to break attractive forces
with in solvent and solute• When solute and solvent particles mix, they
are attracted to each other and energy is released (exothermic)
• Heat of Solution is the overall change in energy
Solubility
• Maximum amount of solute that will dissolve in a given amount of solvent at a specific Temperature and Pressure.– Usually in
• Example: 10g of NaCl in 100g of H2O
€
gramsofsolute
100gofsolvent
Saturation
• Saturated– Max amount of dissolved solute for a given
amount of solvent at a specific T & P
• Unsaturated– Contains less dissolved solute for a given
T & P than a saturated solution
Factors that affect Saturation
• Temperature– Increase in temperature=more soluble– Not true for gases though
• Supersaturation– Contains more dissolved solute than a saturated
solution at the same Temperature• To make, saturate a solution at high temperature and
allow to cool slowly• If a tiny amount of solute is added or the solution is
agitated, a precipitate will form
Pressure and Solubility
• Affects gases
• Solubility Pressure
• Henry’s Law– At a given temperature, solubility (S) of a
gas in a liquid is directly proportional to the pressure (P) of gas above the liquid
€
S1P1=S2P2
€
S1P2 = S2P1
Henry’s Law Example
• If 0.85g of a gas at 4.0 atm of pressure dissolves in 1.0L of water at 25C, how much will dissolve in 1.0L of water at 1.0 atm of pressure and the same temperature
S1= 0.85 g/L
P1= 4.0 atm
S2= ? g/L
P2= 1.0 atm
Henry’s Law Example
S1= 0.85 g/L
P1= 4.0 atm
S2= ? g/L
P2= 1.0 atm
Work problem 2 on pg. 461
€
S1P2 = S2P1
€
S2 =S1P2P1
€
S2 =(0.85g /L)(1.0atm)
4.0atm=0.85g /L
4.0= 0.21g /L
Concentration
€
Massofsolute
massofsolution•100
Percent by Mass
Whether it is mass or volume, the solution is the solvent + the solute Example:
Mass of Solute= 3.6 g
Mass of Solvent = 100.0 g
Find mass of solution first
Then Percent by Mass
Now you work 8,9,10 and worksheet is homework
€
3.6g+100.0g =103.6g
€
3.6g
103.6g• 100 = 3.5%
Concentration
Percent by Volume
€
Volumeofsolute
Volumeofsolution• 100
Example:
Volume of solute= 35 mL
Volume of solvent = 115 mL
Now you do problems 11 and 12 on page 464
Find volume of Solution
Next find the Percent by Volume
€
35mL +115mL =150mL
€
35mL
150mL• 100 = 23%
€
Volumeofsolute
Volumeofsolution• 100
Molarity
Number of Moles of solute dissolved per liter of solution
Know your UNIT CONVERSIONS and HOW TO GO FROM GRAMS TO MOLES
€
Molarity(M) =Molesofsolute
litersofsolution
Molarity Example
A 100.5 mL intravenous solution contains 5.10 g glucose (C6H12O6). What is the molarity of this solution? Molar mass of glucose is 180.16 g/mol.
€
5.10gC6H12O6 ∗1molC6H12O6180.16gC6H12O6
= 0.0283molC6H12O6
€
100.5mL∗1L
1000mL= 0.1005L
€
M =0.0283molC6H12O6
0.1005L= 0.282M
Molarity Example 2
What is the molarity of a bleach solution containing 9.5 g of NaOCl per liter of bleach?
First change grams to moles
€
9.5gNaOCl∗1molNaOCl
74.352gNaOCl= 0.13molNaOCl
Next, find molarity. The volume of the solution is 1 L.
€
M =0.13molNaOCl
1Lsolution= 0.13M