Download - Let’s study solutions
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Let’s study solutionsSolutions• homogeneous mixtures of two or more
substances• solvent & one or more solutes
Solutes• spread evenly throughout• cannot separate
• filtration• Separated
• evaporation. • not visible
• unless colored
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Most common solvent “water”
• polar molecule.• forms hydrogen bonds
• hydrogen atom in one molecule - oxygen atom of different water molecule
Na+ and Cl- ions• surface of a NaCl crystal attracted
• polar water molecules.
• hydrated solution • many H2O molecules surrounds
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Like dissolves like
Two substances form solution • an attraction between
• particles of solute and solvent.• polar solvent dissolves polar solutes
• water and sugar• water ionic solutes (NaCl)
• non-polar solvent hexane (C6H14)• dissolves nonpolar solutes
• (oil or grease)Substances in water are:• strong electrolytes produce ions, conduct an electric current
• NaCl(s) + H2O Na+(aq) + Cl− (aq) • weak electrolytes produce few ions
• HF(g) + H2O(l) H3O+(aq) + F- (aq) • non-electrolytes do not produce ions
Water and polar solute
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Solubility• maximum amount of solute
• dissolves in specific amount of solvent• expressed grams of solute in 100 grams of
solvent water g of solute100 g water
Unsaturated solutions • less than maximum amount of solute• dissolve more soluteSaturated solutions • contain maximum amount of solute than
dissolves • undissolved solute at bottom of container
Dissolved solute
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Effect of Temperature on SolubilitySolubility• Depends on temperature.• Of most solids increases as temperature
increases.• Of gases decreases as temperature
increases.
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How to calculate percent concentration?Solubility
amount of soluteamount of solution
• mass or volume of solute in a solution expressed in “100”
percent = amt.solute x 100 amt. solute + amt. solvent
• a. mass percent (m/m) = g of solute x 100 100 g of solution
b. volume % (v/v) = mL of solute x 100
100 mL of solutionc. mass/volume % (m/v) = g of solute x 100
100 mL of solution
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Example calculating percentages!
calculation of mass percent• grams of solute (g KCl) and• grams of solution (g KCl solution).
g of KCl = 8.00 gg of solvent (water) = 42.00 gg of KCl solution = 50.00 g
8.00 g KCl (solute) x 100 = 16.0% 50.00 g KCl solution
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Using percent concentration as conversion factors!
How many grams of NaCl are needed to prepare
225 g of a 10.0% (m/m) NaCl solution?STEP 1 Given: 225 g solution; 10.0% (m/m)
NaCl Need: g of NaCl
STEP 2 g solution g NaClSTEP 3 Write the 10.0% (m/m) as conversion
factors.10.0 g NaCl and 100 g solution
100 g solution 10.0 g NaCl
STEP 4 Set up using the factor that cancels g solution.225 g solution x 10.0 g NaCl = 22.5 g NaCl 100 g solution
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Learning Check
A solution is prepared by mixing 15.0 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution.
A) 15.0% (m/m) Na2CO3
B) 6.38% (m/m) Na2CO3
C) 6.00% (m/m) Na2CO3
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Solution
C) 6.00% (m/m) Na2CO3
STEP 1 mass solute = 15.0 g Na2CO3
mass solution = 15.0 g + 235 g = 250. g
STEP 2 Use g solute/ g solution ratio
STEP 3 mass %(m/m) = g solute x 100 g solution
STEP 4 Set up problem mass %(m/m) = 15.0 g Na2CO3 x 100 = 6.00%
Na2CO3 250. g solution
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What is Molarity (M)?Molarity (M)
• concentration term for solutions• gives moles of solute in 1 L solution• moles of solute
liter of solution M = m v
• or m = M x v or v = m M
1.00 M NaCl solution prepared• weigh 58.5 g NaCl (1.00 mole) and• add water to make 1.00 liter of solution
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Steps for calculation of molarityWhat is the molarity of 0.500 L NaOH solution if
itcontains 6.00 g NaOH?STEP 1 Given 6.00 g NaOH in 0.500 L solution
Need molarity (mole/L)STEP 2 Plan g NaOH mole NaOH
molaritySTEP 3 Conversion factors 1 mole NaOH =
40.0 g1 mole NaOH and 40.0 g NaOH40.0 g NaOH 1 mole NaOH
STEP 4 Calculate molarity.6.00 g NaOH x 1 mole NaOH = 0.150
mole 40.0 g NaOH
0.150 mole = 0.300 mole = 0.300 M NaOH
0.500 L 1 L
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Learning Check
What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?
A) 0.557 M B) 1.44 MC) 1.71 M
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Solution
C) 1.71 M 46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557
mole NaHCO3
84.0 g NaHCO3
0.557 mole NaHCO3 = 1.71 M NaHCO3
0.325 L
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Learning Check
What is the molarity of 225 mL of a KNO3 solution containing 34.8 g KNO3?
A) 0.344 MB) 1.53 MC) 15.5 M
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Solution
B) 1.53 M34.8 g KNO3 x 1 mole KNO3 = 0.344 mole KNO3
101.1 g KNO3
M = mole = 0.344 mole KNO3 = 1.53 M
L 0.225 LIn one setup:
34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M 101.1 g KNO3 0.225 L
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Molarity Conversion Factors
The units of molarity are used as conversion factors in calculations with solutions.
TABLE 7.8
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Calculations Using Molarity
How many grams of KCl are needed to prepare 125mL
of a 0.720 M KCl solution?
STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl
Need Grams of KCl
STEP 2 Plan L KCl moles KCl g KCl
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Learning Check
How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
A) 20.0 g AlCl3
B) 16.7g AlCl3
C) 2.50 g AlCl3
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Solution
C) 2.50 g AlCl3
0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3
1 L 1 mole
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Learning Check
How many milliliters of 2.00 M HNO3 contain 24.0 g
HNO3?
A) 12.0 mLB) 83.3 mLC) 190. mL
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Solution
24.0 g HNO3 x 1 mole HNO3 x 1000 mL =
63.0 g HNO3 2.00 mole HNO3
Molarity factor inverted
= 190. mL HNO3
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Dilution
In a dilution• water is added.• volume increases.• concentration decreases.
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Comparing Initial and Diluted Solutions
In the initial and diluted solution,• the moles of solute are the same.• the concentrations and volumes are related
by the following equations:For percent concentration:C1V1 = C2V2
initial diluted
For molarity:M1V1 = M2V2
initial diluted
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Dilution Calculations with Percent
What volume of a 2.00% (m/v) HCl solution can be
prepared by diluting 25.0 mL of 14.0% (m/v) HCl
solution?Prepare a table:
C1= 14.0% (m/v) V1 = 25.0 mL
C2= 2.00% (m/v) V2 = ?
Solve dilution equation for unknown and enter values:C1V1 = C2V2
V2 = V1C1 = (25.0 mL)(14.0%) = 175 mL
C2 2.00%
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Learning Check
What is the percent (% m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
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Solution
What is the percent (%m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
Prepare a table:C1= 9.00 %(m/v) V1 = 10.0 mL
C2= ? V2 = 60.0 mL
Solve dilution equation for unknown and enter values:C1V1 = C2V2
C2 = C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v)
V2 60.0 mL
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Dilution Calculations with Molarity
What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO3 to 0.540 L?
Prepare a table:M1= 0.600 M V1 = 0.180 L
M2= ? V2 = 0.540 L
Solve dilution equation for unknown and enter values:M1V1 = M2V2
M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M
V2 0.540 L
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Learning Check
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
A) 27.0 mLB) 60.0 mL C) 90.0 mL
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Solution
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?Prepare a table:
M1= 1.80 M V1 = 15.0 mL
M2= 0.300 M V2 = ? Solve dilution equation for V2 and enter
values:M1V1 = M2V2
V2 = M1V1 = (1.80 M)(15.0 mL) = 90.0 mL
M2 0.300 M
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How colloids and suspension different than solutions?Solutions • contain small particles (ions or molecules).• are transparent.• do not separate.• cannot be filtered.• do not scatter light.
Colloids
• have medium size particles.
• cannot be filtered.
• can be separated by semipermeable membranes.
• scatter light (Tyndall effect).
Suspensions• have very large particles.• settle out. • can be filtered. • must be stirred to stay suspended, blood platelets, muddy water
Examples
• Fog
• Whipped cream
• Milk
• Cheese
• Blood plasma
• Pearls
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What is osmosis?• water (solvent) flows from lower solute concentration into
higher solute concentration. • level of solution with higher concentration rises• concentrations of two solutions become equal with timeOsmotic pressure • produced by solute particles
• dissolved in solution.• equal to pressure that
• prevent flow of additional water • into more concentrated solution.
• greater as number of dissolved particles in solution increases.isotonic solution• same osmotic pressure hypotonic solution • has a lower osmotic pressurehypertonic solution• has a higher osmotic pressure