Transcript Chapter Two Atoms & The Periodic Table


A solution is a homogeneous mixture
 Gas example: air
 Liquid liquid: salt water
 Solid example: brass

Solute: substance being dissolved
 Typically lesser in quantity

Solvent: substance doing the dissolving
 Typically greater in quantity

Electrolyte: substance that when dissolved in
water conducts electricity
 Sodium Chloride (or table salt)
 Has ions in solution (dissociation)

Nonelectrolyte: substance that when dissolved
in water does NOT conduct electricity
 Sucrose (or sugar)
 Does NOT have ions in solution, but molecules

All water-soluble ionic compounds will
dissociate completely
 Therefore, they are strong electrolytes (i.e.
substances that completely dissociate)
 There are only 7 molecular compounds that are
also considered strong electrolytes
▪ HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4

Most molecular compounds are weak
electrolytes OR nonelectrolytes
 Weak electrolytes produce some ions upon
dissolving but exist mostly of molecules that aren’t
ionized

Acids are electrolytes (they produce H+ ions)
 HCl(g)  H+(aq) + Cl-(aq)

Bases are electrolytes (they produce OH- ions)
 NH3(g)  NH4+(aq) + OH-(aq)

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For acids/bases that are WEAK, the reaction
goes in both directions simultaneously
HC2H3O2(l)  H+(aq) + C2H3O2-(aq)
“” reaction occurs in both
directions

 Dynamic Chemical Equilibrium
 A + B2  AB2
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Sucrose (C12H22O11)

Fructose (C6H12O6)
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Sodium Citrate (Na3C6H5O7)
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Potassium Citrate (K3C6H5O7)

Ascorbic Acid (H2C6H6O6)
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Reaction where a “precipitate” forms

Maximum
amount of
solute that will
dissolve in a
given quantity of
solvent at a
specific
temperature

Pb(NO3)2(aq) + NaI(aq) 

Ionic Equation: Shows equation with ions
dissociated

Net Ionic Equation: Shows only what’s involved in
the reaction
 Removes “Spectator Ions”

For the following reaction, correctly predict
the products to write the balanced molecular
equation. Then write the ionic equation and
the net ionic equation.

Aqueous solutions of Lead Acetate and
Calcium Chloride

Arrhenius Model:
 Acids produce H+ ions
 Bases produce OH- ions

Bronsted Model:
 Acids are H+ donors (or proton donors)
 Bases are H+ acceptors (or proton acceptors)

Reaction between an acid and base
 Produce water (most of the time) and a salt (ionic
compound)

A.K.A. “Redox” Reactions

Chemical Reaction where electrons are being
transferred from one reactant to another.

Consider Zn(s) + CuCl2(aq)  ZnCl2(aq) + Cu(s)

Oxidation is loss of electrons
Reduction is gain of electrons

“OIL RIG”

Oxidizing Agent: species that causes oxidation

 Takes the electrons

Reducing Agent: species that causes reduction
 Gives the electrons

A.K.A. Oxidation State (or charge)

Help us determine what elements were
oxidized and reduced

In order to determine an element’s oxidation
number, you must follow the guidelines on
the next two slides:

What is the oxidation number of each atom in
the following:

SO2
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NaH
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CO32-
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H2SO4
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2Fe + 6HBr  3H2 + 2FeBr3
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N2 + 3H2  2NH3
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2KClO3  2KCl + 3O2
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What is the oxidation number for chlorine in
the compound HClO4?
What species is the reducing agent in the
following equation?
 Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

Does the following equation represent a
redox reaction? Why?
 2Mg(s) + O2(g)  2MgO(s)
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Measure of amount of solute dissolved in a
certain amount of solvent or solution

More solute:
 Concentrated

Less solute:
 Diluted
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Molarity = moles of solute/ L of solution
 A.K.A. molar concentration
 Represented by “M” ex: 1.5 M
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If you have exactly 1 L of 1.5 M glucose, it
contains 1.5 moles of glucose
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Suppose you wanted to make a 0.150 M
solution of KMnO4 using a 25o.00 mL
volumetric flask. How would you do this?
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You need to make 500. mL of a 0.650 M
solution of Sodium Hydroxide (NaOH). What
mass of NaOH do you need to use?
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What is the molar concentration (M) of a
solution prepared by dissolving 58.50 g of
Copper Chloride (CuCl2) in water to yield a
1.50 L solution?
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Preparing less concentrated solutions
 Typically done by adding water to concentrated
solution
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Dilution formula: McVc = MdVd
 C = concentrated
 D = diluted
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What volume in mL of a 1.20 M HCl solution
must be diluted in order to prepare 1.00 L of
0.0150 M HCl?
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How much water was added?
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Recall: Soluble Ionic Compounds dissociate
completely (all ionize)
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If you have 0.500 M of KMnO4, then there is 0.500
M of K+ and 0.500 M of MnO4- (1:1 ratio between
ions)
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[ ] are usually used to show concentration
 [KMnO4] = 0.500 M, [K+] = 0.500 M, [MnO4-] = 0.500M
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If you have soluble ionic compounds with ratios
other than 1:1 for ions, use subscripts to determine
ion concentration
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Ex: Na2SO4
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[Na2SO4] = 0.35 M,
[Na+] = 0.70 M,
[SO42-] = 0.35 M
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Suppose you had a
1.55 L solution of this
ionic compound. How
many moles of each ion
do you have? How
many individual ions do
you have?
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Analytical technique based on mass
 Uses percent composition

Ex: A 0.8633-g sample of an ionic compound
containing chloride ions and unknown metal
cations is dissolved in water and treated with
excess AgNO3. If 1.5615 g of AgCl
precipitate, what is the percent by mass of Cl
in the original compound?

Process where
 Solution of known concentration (standard
solution) is added gradually to
 Another solution of unknown concentration till
 The reaction is complete
▪ Equivalence point: # of moles of H+ ions equals # of
moles of OH- ions
▪ End point: Color change in solution (visually indicates
the equivalence point)

What volume of a 0.203 M NaOH solution is
needed to neutralize 25.0 mL of a 0.188 M
H2SO4 solution?

If it takes 26.79 mL of 0.560 M HCl solution to
neutralize 85.70 mL of Ba(OH)2, what is the
molarity of the base?
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What is the molar mass of a diprotic acid if
30.5 mL of 0.1112 M NaOH is required to
neutralize a 0.1365-g sample?
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How many milliliters of a 1.89 M H2SO4
solution are needed to neutralize 91.9 mL of a
0.336 M KOH solution?
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Explain the difference between an endpoint
and an equivalence point.