Unit 9 Solution Chemistry, a Introduction Dissolution Of Ionic Crystals In Water Dissolution Of Molecular Substances In Water Electrical Conductivity Electrolytes van’t Hoff Factor Units of Concentration Diluting and Concentrating Solutions Solution Chemistry, b Works best when seen as a slide show. Click the ‘Slide Show’ button on the
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Slide 1
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Unit 9 Solution Chemistry, a Introduction Dissolution Of Ionic
Crystals In Water Dissolution Of Molecular Substances In Water
Electrical Conductivity Electrolytes vant Hoff Factor Units of
Concentration Diluting and Concentrating Solutions Solution
Chemistry, b Works best when seen as a slide show. Click the Slide
Show button on the lower right.
Slide 3
Matter Homogeneous Heterogeneous Compounds Elements Pure
substances Mixtures solutions
Slide 4
We will focus our attention primarily on solid-liquid
solutions. In this case the solid is the solute, or substance being
dissolved by the liquid, or solvent. The liquid is generally water.
An example would be dissolving salt in water. The composition of a
solution can vary and its components can be physically separated,
but each part is identical to every other part of the mixture (a
salt solution tastes salty throughout). Allowing a salt solution to
evaporate will leave the dry salt residue.
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A solution with a relatively high amount of solute is said to
be more than one that is more. One holding the maximum quantity of
solute is said to be. concentrated dilute saturated
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Dissolution Of Ionic Crystals In Water 1. First the water
molecules draw near the Na + and Cl - ions in the crystal lattice.
Then the negative part of the H 2 O dipole pulls on the sodium ion.
In a similar fashion the attraction of the positive end of the
water dipole pulls on the chloride ion. The ions actually separate
from the crystal lattice. Chemist6ry. Zumdahl. Copyright2000 by
Houghton Mifflin
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2. Then, the water molecules surround the freed ions. The
appropriate water dipole-ion attraction occurs (as diagrammed).
Click Chemistry. Zumdahl. Copyright2000 by Houghton Mifflin
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Sugars are molecular substances that, like ionic compounds,
dissociate from their crystal lattice. However, there is a big
difference. Sugar does not dissociate into ions. The individual
sugar molecule remains intact. Water surrounds each sugar molecule
after it is separated from the crystal. C 6 H 12 O 6 (s) C 6 H 12 O
6 (aq) H2OH2O C 12 H 22 O 11 Dissolution Of Molecular Substances In
Water C 12 H 22 O 11
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The passage of electricity occurs when a voltage is applied to
an electrical conductor such as metal. The electrical current
results from the movement of electrons through the conductor.
Electrical Conductivity Gold bar
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In addition, electrical current will result when a voltage is
applied through certain solutions. The NaCl solution will conduct a
current because it contains ions. The ions in the solution
environment are mobile and respond to a voltage. The sugar solution
will not conduct a current because there are no free ions only
neutral molecules (water and sucrose).
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For this reason NaCl is called an electrolyte, a substance that
conducts electricity when dissolved in water. Sugar is a
nonelectrolyte because it does not conduct a current when dissolved
in water. Pure water is a very low conductor due to some
ionization.
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No ions or free electrons in solution No No ions formed when
dissolved No No ions or free electrons in solution No No ions
formed when dissolved No No ions or free electrons in solution No
No ions formed when dissolved No Very few free ions in solution
Slight Dissociates into ions only slightly when dissolved Slight
Ions immobile in crystal lattice No Dissociates into ions when
dissolved Yes Mobile Ions in liquidYes Dissociates into ions when
dissolved Yes Mobile Ions in solution Yes Dissociates into ions
when dissolved Yes Electrons through metal conductor YesNot
solubleNo Electrons through metal conductor YesNot solubleNo Reason
Conductivity?Reason Electrolyte? Substance C 6 H 12 O 6 (l) C 6 H
12 O 6 (aq) C 6 H 12 O 6 (s) AgCl(aq) NaCl(s) NaCl(l) NaCl(aq)
Ag(l) Ag(s)
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The degree that substances dissociate when dissolved in water
is called the vant Hoff Factor, i. For sodium chloride it is 2
because one mole of sodium ion, Na +, and one mole of chloride ion,
Cl -, forms for every mole of NaCl dissolved. Q Specify the vant
Hoff Factor for each of the following: calcium bromide cesium oxide
ethanol CaBr 2 Ca 2+ (aq) + 2Br - (aq) = Cs 2 O 2Cs + (aq) + O 2-
(aq) = C 2 H 5 OH(l) C 2 H 5 OH (aq) = 3 3 1
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Units of Concentration When interested in experiments with a
salt such as CuSO 4 we weigh out the portion we need, usually in
grams.. www.sperlescales.com
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Many chemical reactions, however, are carried out by mixing
chemicals that are already dissolved in water. In that case, the
CuSO 4 is not weighed as crystals since it is immersed in water. To
solve problems when chemicals react in an aqueous environment we
use the idea of concentration instead of mass.
www.comeau.ca/beaker.jpg Units of Concentration
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Chemists measure the concentration of solutions several ways.
In chemistry we will frequently use the unit called molarity, which
is the ratio of moles of solute per liter of solution: molarity, M=
In dilute solutions, the concentration of solute is relatively low
compared to a concentrated solution. Units of Concentration
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The diagrams and descriptions following show how to prepare one
liter of a 1 molar (1.000 M), solution of NaCl. This is the recipe:
1. Weigh out 1.000 mole of NaCl, or 58.44 g. 2. Carefully pour it
into a specially marked 1.000-liter volumetric flask. 3. Add about
200 mL distilled water. 4. Shake until dissolved. 5. Add distilled
water to the 1.000-liter mark. Note: this step is done after the
initial mixing. Adding a solute to water changes the volume in
different ways. 6. Mix thoroughly. Units of Concentration
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1. Weigh out 1.000 mole of NaCl, or 58.44 g. 2. Carefully pour
it into a specially marked 0.500-liter volumetric flask. 3. Add
about 200 mL distilled water. 4. Shake until dissolved. 5. Add
distilled water to the 0.50-liter mark. 6. Mix thoroughly. Q
Suppose you needed to prepare 500 mL of a 2.000 M NaCl solution.
How would you carry out this procedure? What would you need? Units
of Concentration
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Solving problems involving molarity will at times require
simple plug-in-the-formula procedures, at other times lengthier
dimensional analysis sentences. Q What is the molarity of a 500-mL
solution containing 1.75 mole of sodium hydroxide? Molarity = moles
solute/liter of solution = mol NaOH0.300 = 1.75 mol/0.500 L = 3.50
mol/L Q How many moles of sodium hydroxide are needed to make 750.
mL of a 0.400 M solution? Units of Concentration
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Q What is the molarity of a 250.-mL solution containing 56.0 g
of sodium hydroxide? Q What volume solution will be prepared if
42.0 g of sodium hydroxide are used to make a 0.42 M solution? =
mol/L NaOH5.60 = L NaOH2.50 Units of Concentration
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More Practice 1.Calculate the molarity of a solution that
contains 5.78 grams of calcium chloride dissolved to get a 500.0-mL
solution. (ans. 0.104 M) 2.How many grams of solute are in 250.0 mL
of a 1.500 M hydrochloric acid solution? (ans. 13.7 g) 3.How many
liters of solution can be prepared if 35.8 grams of sodium chloride
are used to make a 0.500 M solution? (ans. 1.22 L)
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Diluting and Concentrating Solutions
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suegregg.com Step 1 Preparing a special solution Diluting and
Concentrating Solutions
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Step 2 Diluting and Concentrating Solutions
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Step 3 Serves eight Diluting and Concentrating Solutions
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Dilution is a process where water is added to an existing
solution. The solutes concentration thereby decreases. The problems
associated with this topic use the molarity formula. Keep in mind,
though, that the total number of moles of solute is the same after
the water is added. Diluting and Concentrating Solutions
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Dilution diminishes the concentration because the particles are
separated. Add water Add 50 mL CuSO 4 Diluting and Concentrating
Solutions
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Dilution reduces the concentration of the solute
proportionally. 0.40 L water Diluting and Concentrating
Solutions
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Diluting solutions follows a numerical format moles 1 = moles 2
moles 1 = M 1 L 1 moles 2 = M 2 L 2 Then M 1 L 1 = M 2 L 2 Diluting
and Concentrating Solutions
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Q Water is added to a beaker containing 200. mL of 0.750 M HCl.
The new concentration is 0.500 M. a) What is the volume of the
solution after dilution? M l L 1 = M 2 L 2 200. mL 0.750 M HCl
0.500 M HCl L 2 0.750 x 0.200 = 0.500 x L 2 0.300 L = L 2
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b) How much water had to be added to the original solution to
dilute it to 0.500 M? 200. mL 0.750 M HCl 0.500 M HCl L 2 L 2 L 1
0.300 L 0.200 L = 0.100 L
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1. Add your concentrated solution to the volumetric flask 2.
Dilute with water 3. Invert several times How chemists dilute
solution
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In some cases, more of a solute is added to a solution to make
it more concentrated. Q A person adds 0.50 mole of sugar crystals
to a 1.00-liter solution containing already 1.00 mole of sugar. The
new sugar completely dissolves and makes the solution volume
increase to 1.05 liter. What is the concentration of the new
solution? New molarity = 1.00-liter 1.05-liter 1.00 mole sugar add
0.50 mole of sugar moles 1 + moles added Total volume 1.00 mol +
0.50 mol 1.05 L 1.43 mol/L
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www.oxbowhay.com/images/ Medicinal Casey Recommended Daily
Intake: 50cc/kg Body weight (divided between 3-5 feedings per day/)
Example: 2 lb. (0.9 kg) guinea pig would be fed a total 45 cc per
day when mixed at a 1:1 ratio, adjust feeding amounts accordingly
if more water is added, to account for dilution effect
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Chemists use fancy equipment to do dilutions
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The process of dilution can be harmful, such as the dilution of
harsh reagents in our ground water.
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Microbiologists go to great lengths to make serial dilutions to
study enzyme activity and other metabolic phenomena
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More Practice 1.250. mL of a 0.300 M glucose (C 6 H 12 O 6 )
solution is diluted with water to a concentration of 0.200 M. a)
What is the volume of the new solution? (ans. 375 mL) b) How much
water had to be added to the original solution? (ans. 125 mL)
2.Water was added to 1,250 mL of a 0.682 M potassium sulfate
solution. The final volume was 1,860 mL. Determine the final
concentration of the solution. (ans. 0.458 M)
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Other Concentration Units
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Besides molarity we will present two other units of solution
concentration. One is called weight fraction or percentage with the
symbol w/w: Q A sample of 0.892 g of potassium chloride is
dissolved in 54.600 g of water. a) What is the fraction (w/w) of
KCl in this solution? b) What are the units for this number? Mass
of solution Mass of solute 0.892 g + 54.600 g 0.892 g 55.492 g
0.892 g = 0.0161 Generally unitless but in this case it is g KCl/g
solution ==
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Other Concentration Units Density is a measure of the
compactness of substances. A relative measure of density is given
in this table: SubstanceDensity (g/mL) Air0.001 Balsa wood0.16
Water1.00 Table salt2.16 Iron7.9 Gold19.32
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Q A solution is prepared by dissolving 3.76 g of calcium
nitrate in 37.60 g of water. The total volume of the solution is
39.4 mL. a) Calculate the density of the solution. b) Calculate the
molarity of the solution. Volume of solution Mass of solution 39.4
mL 37.60+3.76 g Ca(NO 3 ) 2 = 1.05 g/mL = mol/L Ca(NO 3 ) 2 0.582
=
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More Practice 1.Calculate the fraction by mass (w/w) of 3.76
grams of KNO 3 in 95.0 grams of water. (ans. 0.0381) 2.If a chemist
dissolves 6.88 g of glucose, C 6 H 12 O 6, in 100.0 g of water the
total volume is 102.0 mL. Calculate the density of this solution.
(ans. 1.05 g/mL)