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SOLUTIONS
Definitions
A solution is a system in which one or more substances are
homogeneously mixed or dissolved in another substance
• homogeneous mixture
-- uniform appearance
-- similar properties throughout mixture
• The solvent is the dissolving agent
-- i.e., the most abundant component of the solution
• The solute is the component that is dissolved
-- i.e., the least abundant component of the solution
Types of solutions: solid-liquid
Solute: Potassium permanganate
(KMnO4)
Other types of solutions
Gas-gas
Example: Air
Solvent: Nitrogen gas (N2)
Solutes: O2, Ar, CO2, water
vapor, etc.
Gas-liquid
Solvent:
Water
solution
(solid-liquid)
Example: Carbonated drinks
(soda, sparkling water, etc.)
Solvent: Water
Solute: CO2
Other types of solutions
Other types of solutions
Liquid-liquid
Example: Antifreeze
Solvent: Water
Solute: Ethylene glycol
Solid-solid
H2
Gas-solid
Example: Brass
Solvent: Copper
Solute: Zinc
Example: H2 / Pt catalyst
Solvent: Platinum metal (Pt)
Solute: Hydrogen gas (H2 )
Pt
Concentration of solutions
concentration -- the amount of solute dissolved in a given
quantity of solvent or solution
Concentration of solutions
concentration -- the amount of solute dissolved in a given
quantity of solvent or solution
There are many different types of concentration units:
molarity
we will focus on this
mass %
volume %
mass/volume %
parts per million (ppm)
These units are covered in
the supplemental notes
(posted on course website)
parts per billion (ppb)
mole fraction
molality (not to be confused with molarity)
molarity -- number of moles of solute per
liter of solution
Molarity (M ) =
moles of solute
liters solution
Preparation of a 1 molar solution
What is the molarity of a solution made by dissolving 2.00 g of
potassium chlorate in enough water to make 150. ml of solution?
Step 1: Start with the definition of molarity:
Molarity (M ) =
moles of solute
liters solution
Step 2: Determine the number of moles of solute
Molar mass of KClO3 = 39.10 + 35.45 + 3(16.00) = 122.6 g / mol
2.00 g KClO3
1 mole KClO3
122.6 g KClO3
What is the molarity of a solution made by dissolving 2.00 g of
potassium chlorate in enough water to make 150. ml of solution?
Molarity (M ) =
moles of solute
liters solution
Step 3: Determine the number of liters of solution
150. ml
1000 ml
What is the molarity of a solution made by dissolving 2.00 g of
potassium chlorate in enough water to make 150. ml of solution?
Molarity (M ) =
=
0.150 L
moles of solute
liters solution
Step 4: Plug values into molarity equation
Molarity (M ) =
1 liter
= 0.0163 moles KClO3
0.0163 moles KClO3
0.150 L
Molarity (M ) = 0.109 moles KClO3 / L = 0.109 M KClO3
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
Step 1: Start with the definition of molarity:
Molarity (M ) =
moles of solute
Molarity (M ) =
liters solution
moles of solute
liters solution
Step 3: Plug known values into molarity equation and solve for unknown
(moles of solute)
Step 2: Determine the number of liters of solution
0.450 M KOH =
600. ml
1 liter
=
0.450 moles KOH
1 L solution
=
x moles of KOH
0.600 L
0.600 L
1000 ml
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
moles of solute
Molarity (M ) =
Molarity (M ) =
liters solution
Step 3: Plug known values into molarity equation and solve for unknown
(moles of solute)
(0.600 L)
0.450 moles KOH
1L
=
x moles of KOH
0.600 L
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
moles of solute
liters solution
Step 4: Convert moles KOH to grams KOH
Molar mass of KOH = 39.10 + 16.00 + 1.008 = 56.11 g / mol
(0.600 L)
0.270 moles KOH
56.11 g KOH
1 mole KOH
0.270 moles KOH = x
= 15.1 g KOH
Calculate the number of moles of nitric acid in 325 ml of 16
M HNO3
Step 1: Start with the definition of molarity:
Molarity (M ) =
Step 1: Start with the definition of molarity:
moles of solute
Molarity (M ) =
liters solution
Step 2: Plug known values into molarity equation and solve for unknown
(moles of solute)
16 M HNO3 =
Calculate the number of moles of nitric acid in 325 ml of 16
M HNO3
16 moles HNO3
1L
=
liters solution
Step 2: Plug known values into molarity equation and solve for unknown
(moles of solute)
x moles of HNO3
0.325 L
moles of solute
(0.325 L)
16 moles HNO3
1L
=
x moles of HNO3
0.325 L
5.2 moles HNO3 = x
Dilutions
Dilution: Reducing the concentration of a solution by adding more
solvent to the solution
•
More solvent is added:
-- volume of the solution increases
•
No additional solute is added
-- number of moles of solute stays the same
NO3Na+
Moles = 1.0
Volume = 1.0 L
Molarity = 1.0 M
Net result: The molarity of the solution decreases
Molarity (M) =
moles of solute (unchanged)
liters solution
NaNO3 solution
(0.325 L)
Calculate the molarity of a solution prepared by diluting 125
ml of 0.400 M HCl with 875 ml of water
Step 1: Start with the definition of molarity:
NO3Na+
moles of solute
Molarity (M ) =
Moles = 1.0
liters solution
Step 2: Plug in known values and solve for the unknown -- i.e., the number
of moles of solute contained in the initial volume of undiluted solution
Volume = 2.0 L
Molarity = 0.50 M
(0.125 L) 0.400
moles HCl
L
• Solution volume is doubled
• Moles of solute remain the same
• Solution concentration is halved
Calculate the molarity of a solution prepared by diluting 125
ml of 0.400 M HCl with 875 ml of water
Molarity (M ) =
liters solution
Step 3: Calculate new molarity of solution based on final volume of
solution after dilution
Initial
moles HCl: 0.050 moles
Final
moles HCl: 0.050 moles
liters solution: 0.125 L
liters solution: 0.125 L + 0.875 L
= 1.000 L
Molarity =
0.125 L
Molarity = 0.400 M
Molarity =
x moles of HCl
0.125 L
(0.125 L)
0.050 moles HCl = x
Calculate the molarity of a solution prepared by diluting 125
ml of 0.400 M HCl with 875 ml of water
ALTERNATIVE METHOD (SHORTCUT): Use dilution factor based on
initial solution volume and final solution volume
moles of solute
0.050 moles
=
Initial solution volume: 0.125 L
Final solution volume: 0.125 L + 0.875 L = 1.000 L
initial volume
Dilution factor =
final volume
=
0.125 L
1.000 L
Final concentration = (dilution factor) x (initial concentration)
0.050 moles
1.000 L
Molarity = 0.050 M
=
0.125 L
1.000 L
0.400 M HCl
= 0.050 M HCl
Acids and Bases
Properties of acids
1. Sour taste
2. Changes the color of litmus from blue to red
3. Reacts with:
• metals (e.g., Zn and Mg) to produce hydrogen gas
• hydroxide bases (e.g., NaOH) to produce water and
an ionic compound (salt)
• carbonates (e.g., CaCO3) to produce CO2
Examples of acids:
• lemon juice
• Vitamin C (ascorbic acid)
• vinegar (acetic acid) • hydrochloric acid
Properties of bases
1. Bitter taste
2. Slippery, soapy feel
3. Changes the color of litmus from red to blue
4. Reacts with acids
Examples of bases:
• ammonia -- NH3
• Milk of Magnesia -- Mg(OH)2
• sodium hydroxide -- NaOH
Classical definition of acids and
bases (Arrhenius)
acid -- a substance that produces hydrogen ions (H+)
in aqueous solutions
HCl
H+ (aq)
+
Cl- (aq)
hydrochloric
acid
base -- a substance that produces hydroxide ions (OH-)
in aqueous solutions
NaOH
sodium
hydroxide
Na+ (aq)
+
OH- (aq)
Brønsted-Lowry acids and bases
A Brønsted-Lowry acid is any substance that is able to give
hydrogen ions (H+) to another molecule or ion
Brønsted-Lowry acids and bases
A Brønsted-Lowry acid is a proton (H+) donor :
A Brønsted-Lowry base is a proton (H+) acceptor :
A Brønsted-Lowry base is any substance that accepts
hydrogen ions (H+) from another molecule or ion
conjugate acid-base pair
HCl (aq) +
OH–
acid
base
(aq)
H2O (l) +
Cl- (aq)
H–O–H
NH3 (l) + H2O (l)
NH4+ (aq)
base
acid
acid
+ OH- (aq)
base
conjugate acid-base pair
• a Brønsted-Lowry acid is a proton (H+) donor
• a Brønsted-Lowry base is a proton (H+) acceptor
Naming acids
The hydronium ion
A hydrogen ion (H+) does not exist by itself in an aqueous
solution
• In water, H+ combines with a polar H2O molecule to form
a hydrated hydrogen ion (H3O+) called a hydronium ion
H+
+
HO
H
HOH
H
hydronium
ion
+
Binary acids are formed from hydrogen and one other
nonmetallic element
Note: Not all binary compounds containing hydrogen are
binary acids
HCl
CH4
hydrochloric acid
methane
not an acid
acid
If the binary compound is an acid, the H will appear first in the
chemical formula
Naming polyatomic acids
Naming binary acids
Binary compound
• both elements are nonmetals
• the compound is an acid ( H appears first in formula )
Rule: The compound name is based on the stem of the nonhydrogen element
Add hydro- before the stem and -ic after the stem, followed by the
word acid
Example:
H+
Formula:
Cl -
Polyatomic acids are made up of hydrogen and a polyatomic anion
-- the chemical formula of a polyatomic acid begins with H
-- the second part of the formula is a polyatomic anion containing
oxygen (an oxy-anion)
Rule: The name of a polyatomic acid is derived from its anion
-- the ending of the anion name is modified
-ate
-ic
-ite
-ous
-- the modified anion name is followed by the word acid
HCl
Elements:
hydrogen (nonmetal)
Compound name:
chlorine (nonmetal)
hydrochloric acid
anion:
anion name:
modified anion name:
polyatomic acid name:
Example: HNO3
Naming polyatomic acids
nitric
Naming bases
Rule: The name of a polyatomic acid is derived from its anion
Common bases are formed from group 1A and 2A metals and
the polyatomic hydroxide ion (OH–)
-- the ending of the anion name is modified
-ate
-ic
-ite
-ous
These bases therefore follow the naming rules for polyatomic
ions
-- the modified anion name is followed by the word acid
Rules:
-- identify the ions
Acid
Anion
Anion name
Acid name
H2SO4
SO4
sulfate
sulfuric acid
-- name the cations in the order given
H2SO3
SO32-
sulfite
sulfurous acid
-- follow them with the name of the anion
2-
NO3–
nitrate
nitrate
nitric acid
Compound
Cation(s)
Anion
Compound name
KOH
K+ (potassium)
OH– (hydroxide)
potassium hydroxide
Ca(OH)2
Ca2+
OH– (hydroxide)
calcium hydroxide
(sodium)
Reactions of acids and bases
Neutralization reaction between an acid and a base
• in aqueous solutions, the products of a neutralization
reaction are water and a salt
acid + base
water + salt
Reactions of acids and bases
Reactions between acids and carbonates
• similar to acid-base neutralization reaction
• the products are carbon dioxide, water, and a salt
acid + carbonate
carbon dioxide + water + salt
H2SO4 (aq) + MgCO3 (s)
HBr (aq) + NaOH (aq)
H2O (l) + NaBr (aq)
H2CO3 (aq) + MgSO4 (aq)
Carbonic acid spontaneously decomposes into carbon dioxide and water
H2CO3 (aq)
CO2 (g) + H2O (l)
Net reaction:
H2SO4 (aq) + MgCO3 (s)
Homework
Chapter 9 Problems:
9.38, 9.42, 9.48, 9.52, 9.62, 9.63, 9.64, 9.69, 9.73
Chapter 10 Problems:
10.47, 10.51, 10.52, 10.53
CO2 (g) + H2O (l) + MgSO4 (aq)