Transcript Solutions

Solutions
Chapter 14
Common Solutions
Chemical solutions encountered in everyday life:
air
coffee
tap water
gasoline
shampoo
cough syrup
orange soda
Gatorade
Solutions
. . . the components of a mixture
are uniformly intermingled
(the mixture is homogeneous).
A Solute
-
dissolves in water (or other “solvent”)
-
changes phase (if different from the
solvent)
-
is present in lesser amount (if the same
phase as the solvent)
A Solvent
-
retains its phase (if different from the
solute)
-
is present in greater amount (if the same
phase as the solute)
Figure 14.1: When
solid sodium
chloride dissolves,
the ions are
dispersed randomly
throughout the
solution
Aqueous Solutions
Aqueous solutions are solutions in which
water is the solvent.
Aqueous solutions are the most common type
of solution.
Some Properties of Water
Water is able to dissolve so many substances
because:
-
Water is “bent” or V-shaped.
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The O-H bonds are covalent.
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Water is a polar molecule.
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Hydration occurs when salts dissolve in
water.


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H

2
105
O
H


Water is a polar molecule because it is a bent
molecule. The hydrogen end is + while the oxygen
end is -, Delta () is a partial charge--less than 1.
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+



 +
+ 
 +
+ 
H
O H
+
Cation




+ 
 +
+ 
+

+
H
H O



+

+


Anion

Polar water molecules interact with the positive
and negative ions of a salt, assisting in the
dissolving process. This process is called hydration.
Solubility
The general rule for solubility is:
“Like dissolves like.”
Polar water molecules can dissolve other polar
molecules such as alcohol and, also, ionic
substances such as NaCl.
Nonpolar molecules can dissolve other
nonpolar molecules but not polar or ionic
substances. Gasoline can dissolve grease.
Miscibility
Miscible -- two substances that will mix
together in any proportion to make a
solution. Alcohol and water are miscible
because they are both polar and form
hydrogen bonds.
Immiscible -- two substances that will not
dissolve in each other. Oil and vinegar are
immiscible because oil is nonpolar and
vinegar is polar.
Figure 14.6: An oil layer floating on water
Structure & Solubility
Like dissolves like.
Hydrophobic --water-fearing. Fat soluble
vitamins such as A, D, E, & K.
Hydrophilic --water-loving. Water soluble
vitamins such as B & C.
Hypervitaminosis--excessive buildup of
vitamins A, D, E, & K in the body.
Solubility
How does the rule “Like dissolves like.”
apply to cleaning paint brushes used for
latex paint as opposed to those used with
oil-based paint?
Common Terms of Solution
Concentration
Saturated - when a solution contains as much
solute as will dissolve at that temperature.
Unsaturated - when a solution has not reached
its limit of solubility.
Supersaturated - when a solution has more of
the solute dissolved than it normally would -very unstable.
Common Terms of Solution
Concentration
Stock - routinely used solutions prepared in
concentrated form.
Concentrated - relatively large ratio of solute
to solvent. (5.0 M NaCl)
Dilute - relatively small ratio of solute to
solvent. (0.01 M NaCl)
Solution Composition
Qualitative -- general Quantitative -- mathematically
defined.
and relative.
dilute
Mass % -- biology
concentrated
Molarity -- most solutions in
chemistry
Normality -- chemical titrations
Molality -- molar mass, freezing
point depressions, & boiling
point elevations
Mass %
mass of solute
Mass (weight) percent =
(100%)
mass of solution
Mass % Calculations
If 1.00 g of ethanol is added to 100.0 g of
water, what is the mass % of the ethanol?
 massethanol 
%C 2 H 5OH  
100% 
 masssolution 
 1.00 g 
100% 
 
 101.0 g 
 0.990%C 2 H 5OH
Mass % Calculations
Cow’s milk typically contains 4.5 % by mass of the
sugar lactose, C12H22O11. Calculate the mass of
lactose present in 175 g of milk.
mass of solute
(100%)
Mass % =
mass of solution
mass of solute = (mass %)(mass of solution)/(100%)
mass of solute = (4.5 % lactose)(175 g)/(100%)
mass of solute = 7.9 g lactose
Molarity
Molarity (M) = moles of solute per volume of
solution in liters:
moles of solute
M  molarity 
liters of solution
6 moles of HCl
3 M HCl 
2 liters of solution
Molarity Calculations
If 1.00 g of ethanol is dissolved in enough
water to make 101 mL of solution, what is
the molarity of the solution?
 1.00 gethanol  1000mL  1mol 
  0.215M



 101mL  1L  46.07 g 
Molarity Calculations
Calculate the molarity of a solution prepared by
dissolving 11.5 g of solid NaOH in enough
water to make 1.50 L of solution.
(11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH)
= 0.192 M NaOH
Molarity Calculations
Calculate the molarity of a solution prepared by
dissolving 1.56 g of gaseous HCl into enough
water to make 26.8 mL of solution.
(1.56g HCl/26.8mL)(1mol/36.5g)(1000mL/1L) =
1.59 M HCl solution
Molarity Calculations
How many moles of nitrate ions are present in
25.00 mL of a 0.75 M Co(NO3)2 solution?
(25.00mL)(1L/1000mL)(0.75mol Co(NO3)2/1L)
(2 mol NO3-/1 mol Co(NO3)2) = 3.8 x 10-2 mol NO3-
Molarity Calculations
Typical blood serum is about 0.14M NaCl. What
volume of blood contains 1.0 mg of NaCl?
(1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol)
= 1.2 x 10-4 L blood serum
Molarity Calculations
A chemist needs 1.00 L of an aqueous 0.200 M
K2Cr2O7 solution. How much solid potassium
dichromate must be massed to make this
solution?
(1.00L)(0.200mol K2Cr2O7/L)(294.2g/1mol) =
58.8 g K2Cr2O7
Standard Solution
A standard solution is a solution whose
concentration is accurately known.
A volumetric flask is used to make a standard
solution.
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Volume marker
(calibration mark)
Wash Bottle
Weighed
amount
of solute
(a)
(b)
(c)
(d)
Steps involved in making a standard solution.
Dilution of Stock Solutions
When diluting stock solutions, the moles of solute
after dilution must equal the moles of solute
before dilution.
moles of solute before dilution = moles of solute after dilution
Stock solutions are diluted using either a
measuring or a delivery pipet and a volumetric
flask.
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Rubber bulb
500 mL
(a)
(b)
Steps to dilute a stock solution.
(c)
Dilution Calculations
What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of a 0.10 M H2SO4 solution?
(0.10mol H2SO4/1L)(1.5L)(1L/16mol)(1000mL/1L)
= 9.4 mL conc H2SO4
Dilution Calculations
What volume of 12 M HCl must be taken to
prepare 0.75 L of 0.25 M HCl?
(0.75L)(0.25mol HCl/L)(1L/12mol)(1000mL/1L)
= 16 mL conc. HCl
Steps For Solving Solution
Stoichiometry Problems
1. Write the balanced reaction.
2. Calculate moles of reactants.
3. Determine limiting reactant.
4. Calculate moles of required reactant/product.
5. Convert to grams or volume, as required.
Precipitation Calculations
When aqueous solutions of Na2SO4 & Pb(NO3)2 are mixed,
PbSO4 precipitates. Calculate the mass of PbSO4 formed
when 1.25 L of 0.0500 M Pb(NO3)2 & 2.00 L of 0.0250 M
Na2SO4 are mixed.
1. Pb(NO3)2(aq) + Na2SO4(aq) ----> PbSO4(s) + 2NaNO3(aq)
2. (1.25L)(0.0500mol Pb(NO3)2/L) = 0.0625 mol Pb(NO3)2
(2.00L)(0.0250mol Na2SO4/L) = 0.0500 mol Na2SO4
Precipitation Calculations
Continued
3. (0.0625mol Pb(NO3)2)(1mol Na2SO4/1mol Pb(NO3)2 =
0.0625 mol Na2SO4
Na2SO4 is the limiting reactant.
4. (0.0500mol Na2SO4)(1mol PbSO4/1mol Na2SO4)
(303.3g/1mol) = 15.2 g PbSO4
Neutralization Reactions
Acids and bases react to neutralize each other
and form a salt and water. This type of
reaction is called a neutralization reaction.
HCl(aq) + NaOH(aq) ----> NaCl(aq) + HOH(l)
Acid-Base Calculations
What volume of a 0.100 M HCl solution is needed to
neutralize 25.0 mL of 0.350 M NaOH?
HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)
(25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol
NaOH)(1L/0.100mol) = 87.5 mL HCl solution
Normality
 equivalentssolute 
Normality ( N )  

 Litersolut ion 
Acid-Base Equivalents = (moles) (total (+)
charge)
N = M ( total (+) charge)
Normality Calculations
.250 M H3PO4 =______N
N = M(total(+) charge)
N = (0.250)(3)
N = 0.750 N H3PO4
Normality Calculations
A solution of sulfuric acid contains 86 g of H2SO4
per liter of solution. What is its normality?
(86g H2SO4/1L)(1mol/98.0g) = 0.88 M H2SO4
N = M(total (+) charge)
N = (0.88M)(2)
N = 1.8 N H2SO4
Normality Stoichiometry
Calculations
What volume of a 0.075 N NaOH solution is
required to react exactly with 0.135 L of 0.45 N
H3PO4?
NaVa = NbVb
Na = 0.45 N
Vb = (NaVa)/Nb
Va = 0.135 L
Vb = (0.45N)(0.135L)/(0.075N)
Nb = 0.075 N
Vb = ?
Vb = 0.81 L NaOH