Transcript Chapter 7 Solutions - Oklahoma City Community

Concentration of a Solution
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Quantity of solute:
grams or moles
Quantity of solution:
grams or volume
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amount of solute
Concentration of a solution =
amount of solution
amount of solute in: grams , volume, moles
amount of solution in : grams or volume
• Mass percent (m/m) concentration
• Volume percent (v/v) concentration
• Mass/volume percent (m/v) concentration
• Molarity (mol/L)
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Mass/Volume Percent (%m/v)
Concentration
percent mass (g) of solute divided by volume
(mL) of solution.
Calculate:
mass/volume % (m/v) =
g of solute x 100%
mL of solution
ex: 5.0 g of KI in 250 ml solution
%(m/v) concentration =
4
% m/v concentration
• percent mass (g) of solute to volume (mL) of solution.
mass/volume % (m/v) =
g of solute x 100
mL of solution
• solute (g) in 100 mL of solution.
mass/volume % (m/v) =
g of solute
100 mL of solution
ex:
a 1.0 % (m/v) solution of a compound
contains 1.0 g of compound in 100 ml solution
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Sample Problems
1) Calculate the %(m/v) concentration for the solute in the following solutions:
a) 2.50 g of LiCl in 40.0 ml of LiCl solution (2.50 g/ 40.0 ml)x100%= 6.25 %(m/v)
b) 7.5 g of casein in 120 ml of low-fat milk
(7.5 g/ 120 ml)x100%= 6.3 %(m/v)
2) Calculate the grams of solute needed to prepare the following solution:
450 ml of a 2.0 %(m/v) KOH solution
2.0 %(m/v) = 2.0 g/100 ml, so 450 ml x(2.0 g/100 ml) = 9.0 g KOH needed
3) A patient received 2.0 g of NaCl in 8 hours. How many milliliters of a 0.9 %(m/v)
NaCl (saline) solution were delivered?
9.0 %(m/v) = 9.0 g /100 ml, so 2.0 g/ (0.9 g/100 ml) = 220 ml solution
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Molarity Concentration
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Molarity (M)
• a concentration term for solutions.
• gives the moles of solute in 1 L solution.
• moles of solute
1 liter of solution
Written as molar solution,
eg: 6.0 M = 6.0 moles/1.0 L = 6.0 mol/L
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Preparing a 1.0 Molar Solution
A 1.00 M NaCl solution is prepared
• by weighing out 58.5 g NaCl (1.00 mole) and
• adding water to make 1.00 liter of solution.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Use concentration as conversion
factors
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Calculation of Molarity
What is the molarity of 0.500 L NaOH solution if it
contains 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
molarity
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Calculation of Molarity (cont.)
STEP 3 Molar mass 1 mole NaOH = 40.0 g
1 mole NaOH
and 40.0 g NaOH
40.0 g NaOH
1 mole NaOH
STEP 4 Calculate moles, then divide by solution volume.
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
1L
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Learning Check
What is the molarity of 325 mL of a solution containing 46.8 g of
NaHCO3?
1)
0.557 M
2)
1.44 M
3)
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?
34.8 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 L
In 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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Dilutions
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Dilution
In a dilution
• water is added.
• volume increases.
• concentration decreases.
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Dilution Equation: C1xV1 = C2xV2
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Dilution example problem
Calculate the concentration of the following diluted solutions:
a) 1.0 L of a 4.0 M HNO3 solution is added to water so that the final volume is 8.0 L.
C1 = 4.0 M, V1 = 1.0 L, V2 = 8.0 L
C2 = 0.50 M
b) A 5.0 ml sample of a 50.0 %(m/v) acetic acid solution is added to water to give a
final volume of 25 ml.
C1 = 50.0 M, V1 = 5.0 mL, V2 = 25 L
C2 = 10. M
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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?
Prepare a table:
C1= 9.00 %(m/v)
V1 = 10.0 mL
C 2= ?
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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Learning Check
What volume (mL) of a 1.80 M KOH solution is needed to prepare
90.0 ml of a 0.300 M KOH solution?
Prepare a table:
C1= 1.80 M
V1 = ?
C2= 0.300 M
V2 = 90.0 mL
Solve dilution equation for V1 and enter values:
C1V1 = C2V2
V1
= C2V2
C1
= (0.300 M)(90.0 mL) = 15.0 mL
1.80 M
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Properties of Solutions
Solutions and colloids are
homogenous, suspension
is heterogeneous.
Colloids and suspensions
contain large particles.
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Osmosis
In osmosis,
• water (solvent) flows from
the lower solute
concentration into the
higher solute
concentration.
• the level of the solution
with the higher
concentration rises.
• the concentrations of the
two solutions become
equal with time.
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Osmotic Pressure of the Blood
Cell membranes in biological systems
• have cell walls that are semi-permeable membranes,
osmosis occurs all the time. So solutes in body
solutions like blood, tissue fluids, lymph and plasma
excert osmotic pressure.
Red blood cells
• maintain an osmotic pressure that cannot change or
damage occurs.
• must maintain an equal flow of water between the red
blood cell and its surrounding environment.
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IV solutions are
Isotonic Solutions
• exert the same osmotic
pressure as red blood cells.
• are known as a
“physiological solution”.
• 5.0% glucose or 0.90%
NaCl is used medically
because each has a solute
concentration equal to the
osmotic pressure of red
blood cells.
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Hypotonic Solution
• has a lower osmotic
pressure than red blood
cells.
• has a lower concentration
than physiological
solutions.
• causes water to flow into
red blood cells.
• causes hemolysis: RBCs
swell and may burst.
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Hypertonic Solution
• has a higher osmotic
pressure than RBCs.
• has a higher
concentration than
physiological solutions.
• causes water to flow out
of RBCs.
• cause crenation: RBCs
shrinks in size.
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Dialysis
• solvent and small solute
particles pass through an
artificial membrane.
• large particles are
retained inside.
• waste particles such as
urea from blood are
removed using
hemodialysis (artificial
kidney).
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waste particles such as urea from blood are
removed using hemodialysis (artificial kidney).
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