Chapter 5 Chemical Reactions

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Chapter 5Chemical Reactions

• General Chemistry I
• T. ARA

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Chemical Nomenclature - Part IIPolyatomic Ions

• NH4 ammonium
• CO32 carbonate HCO3 hydrogen carbonate
• NO3 nitrate NO2 nitrite
• PO43 phosphate HPO42 hydrogen phosphate
• H2PO4 dihydrogen phosphate
• SO42 sulfate SO32 sulfite
• S2O32 thiosulfate
• CN cyanide OCN cyanate
• SCN thiocyanate
• Cr2O72 dichromate CrO42 chromate
• OH hydroxide MnO4 permanganate
• C2O42 oxalate CH3CO2 acetate

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Chemical Nomenclature - Part IISome Tricks
hypoite ClO- hypochlorite
ite ClO2- chlorite NO2- nitrite SO32- sulfite
ate ClO3- chlorate NO3- nitrate SO42- sulfate
perate ClO4- perchlorate
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Chemical Nomenclature - Part IINaming

• A compound consisting of polyatomic ions is named
  by listing the cationic atom or group of atoms
  first, followed by the anionic atom or group of
  atoms.
• Eg. KMnO4 is ionic
• (K MnO4)
• Potassium Permanganate

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Chemical Nomenclature - Part IIChemical Formulas

• Use the same method for predicting the chemical
  formula of a compound containing polyatomic ions
  as you did with those containing monatomic ions.
• Match the positive negative charges.
• Use parentheses when applying subscripts to
  polyatomic ions.
• Charges of polyatomic ions dont change from
  compound to compound!
• eg. Magnesium phosphate
• Mg2 PO43-
• Mg3(PO4)2

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Chemical Nomenclature - Part IIChemical Formulas

• What is the formula of ammomium sulfate?
• What is the formula of aluminum carbonate?

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Chemical Nomenclature - Part IIChemical Formulas

• What is the formula of ammomium sulfate?
• NH4 and SO42
• What is the formula of aluminum carbonate?

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Chemical Nomenclature - Part IIChemical Formulas

• What is the formula of ammomium sulfate?
• NH4 and SO42
• (NH4)2SO4
• What is the formula of aluminum carbonate?

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Chemical Nomenclature - Part IIChemical Formulas

• What is the formula of ammomium sulfate?
• NH4 and SO42
• (NH4)2SO4
• What is the formula of aluminum carbonate?
• Al3 and CO32

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Chemical Nomenclature - Part IIChemical Formulas

• What is the formula of ammomium sulfate?
• NH4 and SO42
• (NH4)2SO4
• What is the formula of aluminum carbonate?
• Al3 and CO32
• Al2(CO3)3

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Chemical Nomenclature - Part II

• Work through the problems on the worksheet for
  practice - do not turn in.
• You should be able to
• Name ionic compounds containing polyatomic ions.
• Determine the formulas of ionic compounds
  containing polyatomic ions.
• Use recognize ions on the handout.
• Formulas
• Charges
• Names

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A. Exchange Reactions

• Recall that exchange reactions have the following
  reaction pattern
• Many ionic compounds undergo aqueous exchange
  reactions in which the ionic partners are
  switched.
• The solubility of the reactants and products can
  determine whether a reaction will occur.

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1. Solubility of Ionic Compounds

• When ionic compounds dissolve in water, they
  dissociate into ions that are surrounded by water
  molecules - the ionic partners separate in
  solution.
• Molecular
• compounds
• generally do not
• ionize (separate)
• in solution.

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1. Solubility of Ionic Compounds

• Most, but not all, ionic compounds are soluble in
  water.
• General guidelines can be used to predict whether
  an ionic compound will be soluble.

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1. Solubility of Ionic Compounds

• Certain ions usually form insoluble ionic
  compounds.

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1. Solubility of Ionic Compounds
How does the solubility of an ionic compound
determine whether or not it will undergo an
exchange reaction?
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A. Exchange Reactions

• If both the reactants the potential products
  are soluble, then no overall reaction occurs.
• All of the ions remain dissociated in solution.

Ca(NO3)2, NaCl, CaCl2 NaNO3 are all soluble
NO EXCHANGE REACTION
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A. Exchange Reactions

• An exchange reaction will occur only if one of
  the potential products removes ions from the
  solution.
• This can occur in one of the following ways
• An insoluble ionic compound forms precipitates
  from the solution.
• A molecular compound forms (usually soluble) that
  will not dissociate back to its ionic components.
• A gaseous molecular compound forms escapes from
  the solution.

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2. Formation of a Precipitate

• An exchange reaction will occur if one of the
  products is insoluble - a precipitate forms and
  removes ions from the solution.
• eg. Precipitation of barium sulfate from a
  mixture of barium chloride sodium sulfate

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2. Formation of a Precipitate

• How can you predict whether an exchange reaction
  will occur?

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2. Formation of a Precipitate

• How can you predict whether an exchange reaction
  will occur?

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2. Formation of a Precipitate

• How can you predict whether an exchange reaction
  will occur?

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2. Formation of a Precipitate

• How can you predict whether an exchange reaction
  will occur?

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a) Net Ionic Equations

• A net ionic equation is a simplified way of
  writing an ionic exchange reaction that only
  includes the ions undergoing change during the
  reaction.
• The other ions that remain dissociated
  (dissolved) in solution are called spectator ions
  are not included in the net ionic equation.

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a) Net Ionic Equations

• Notice The sum of the charges and the
  individual atoms must be the same on each side
  of a net ionic equation. It must be balanced!

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a) Net Ionic Equations

• Follow these steps to convert an overall chemical
  equation into a net ionic equation
• Step 1 Write a balanced equation for the overall
  reaction
• -Make sure you use the correct formulas for the
  reactants products. Keep track of the proper
  charges of ions - charges do not change during an
  exchange reaction!

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a) Net Ionic Equations
Step 2 Use the general solubility guidelines to
determine the solubility of reactants products
- will a reaction occur?
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a) Net Ionic Equations

• Step 2 Use the general solubility guidelines to
  determine the solubility of reactants products
  - will a reaction occur?

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a) Net Ionic Equations
Step 3 Break all of the soluble ionic compounds
into their component ions.
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a) Net Ionic Equations

• Step 3 Break all of the soluble ionic compounds
  into their component ions.

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a) Net Ionic Equations
Step 4 Use the ions from step 3 to write a
complete ionic equation
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a) Net Ionic Equations

• Step 4 Use the ions from step 3 to write a
  complete ionic equation

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a) Net Ionic Equations
Step 5 Cancel out the spectator ions from the
two sides of the equation
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a) Net Ionic Equations
Step 5 Cancel out the spectator ions from the
two sides of the equation
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a) Net Ionic Equations
Step 5 Cancel out the spectator ions from the
two sides of the equation
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a) Net Ionic Equations

• Step 5 Cancel out the spectator ions from the
  two sides of the equation

Step 6 Check that the sum of the individual
atoms the charges is the same on each side of
the equation
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Write a net ionic equation for the following
exchange reaction
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Write a net ionic equation for the following
exchange reaction
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Write a net ionic equation for the following
exchange reaction
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3. Formation of a Molecular Compound

• An exchange reaction will also occur if two of
  the ions react to form a molecular compound that
  does not precipitate. Water is a common example.
• Hydrogen ion (H) hydroxide ion (OH) combine
  to form water (H2O) a molecular compound that
  does not readily dissociate back into ions.
• This is also called an acid-base reaction.

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a) Acids Bases

• Acids Bases are two important classes of
  molecules that form water when they react with
  one another in an exchange reaction.
• Acids bases can be defined in a number of ways
  - we will start with a very general definition.

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a) Acids Bases

• An acid is any substance that increases the
  concentration of hydrogen ions (H - also called
  protons) in solution when dissolved in pure
  water.
• Strong acids dissociate completely into their
  ions when dissolved in water.
• eg. HCl is a strong acid.

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a) Acids Bases

• When dissolved in water, H ions combine with
  water molecules to form the more stable hydronium
  ion (H3O).
• H (aq) actually represents H3O (aq).

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a) Acids Bases

• A base is any substance that increases the
  concentration of hydroxide ions (OH-) when
  dissolved in pure water.
• Strong bases dissociate completely into their
  ions when dissolved in water.
• eg. Sodium hydroxide is a strong base

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b) Neutralization Reactions

• When an aqueous solution containing a strong acid
  is mixed with an aqueous solution containing a
  strong base, an acid-base exchange reaction takes
  place.
• This reaction neutralizes the acid the base.
• In other words, H (the acid) OH (the base)
  react to form H2O (neutral).

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b) Neutralization Reactions

• Predict the products of the following acid-base
  reaction.

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b) Neutralization Reactions

• Predict the products of the following acid-base
  reaction.

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b) Neutralization Reactions

• Predict the products of the following acid-base
  reaction.

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b) Neutralization Reactions

• Net ionic equations can be written for acid-base
  exchange reactions.
• For now, we will focus only on neutralization
  reactions between strong acids bases - dont
  worry about weak acids bases.
• Write a net ionic equation for the following
  reaction

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b) Neutralization Reactions

• Net ionic equations can be written for acid-base
  exchange reactions.
• For now, we will focus only on neutralization
  reactions between strong acids bases - dont
  worry about weak acids bases.
• Write a net ionic equation for the following
  reaction

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5. Formation of a Gas

• Finally, an exchange reaction will occur if it
  results in the formation of a gaseous compound
  that can escape from the solution.
• Some gases that are commonly formed in exchange
  reactions are H2, H2S, CO2 SO2
• CO2 is formed by the decomposition of carbonic
  acid - H2CO3
• SO2 is formed by the decomposition of sulfurous
  acid - H2SO3

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5. Formation of a Gas

• Metal carbonates react with acids to form
• carbonic acid (H2CO3).

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5. Formation of a Gas

• Metal carbonates react with acids to form
• carbonic acid (H2CO3).
• Carbonic acid rapidly decomposes to form water
  and carbon dioxide - a gas.

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5. Formation of a Gas

• Metal carbonates react with acids to form
• carbonic acid (H2CO3).
• Carbonic acid rapidly decomposes to form water
  and carbon dioxide - a gas.

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5. Formation of an Gas

• Metal hydrogen carbonates (bicarbonates) also
  react with acids to form carbon dioxide gas.
• Write a balanced equation a net ionic equation
  for the reaction of sodium bicarbonate (baking
  soda - NaHCO3) with HCl.

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B. Oxidation-Reduction Reactions

• When tin (IV) oxide is heated with carbon, an
  oxidation-reduction (redox) reaction takes place.
• An oxidation-reduction reaction involves the
  transfer of electrons from one reactant to
  another.
• When one reactant is oxidized, the other must
  simultaneously be reduced.

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1. Reduction

• When a substance gains electrons, it is said to
  be reduced.
• In a reduction, there is a decrease (reduction)
  in the real or apparent electric charge on the
  atom.
• In the following reaction, tin is being reduced.
• A reactant that is reduced in a redox reaction is
  called an oxidizing agent because it is
  simultaneously oxidizing the other reactant.

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2. Oxidation

• When a substance loses electrons, it is said to
  be oxidized.
• In an oxidation, there is an increase in the real
  or apparent electric charge on the atom.
• In the following reaction carbon is being
  oxidized.
• A reactant that is oxidized in a redox reaction
  is called a reducing agent because it is
  simultaneously reducing the other reactant.

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B. Oxidation-Reduction Reactions

• In every redox reaction, one reactant is reduced
  one reactant is oxidized.
• In the reaction above, the charges on the atoms
  indicate which reactant is being reduced and
  which is being oxidized.

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B. Oxidation-Reduction Reactions

• LEO the lion goes GER.
• Loses Electrons - Oxidized
• Gains Electrons - Reduced

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B. Oxidation-Reduction Reactions

• LEO the lion goes GER.
• Loses Electrons - Oxidized
• Gains Electrons - Reduced

Without any charges, how can you tell which
reactant is being reduced which is being
oxidized?
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a) Common Oxidizing Reducing Agents

• One way to identify the oxidizing reducing
  agents in a redox reaction is to recognize common
  oxidizing reducing agents.
• Reducing Agents Lose Electrons
• Hydrogen (H2) hydrogen-containing compounds are
  reducing agents
• Carbon (C) is a reducing agent commonly used to
  reduce metal oxides to neutral metals
• Certain metals (Na, K, Fe, Al) are good at giving
  up electrons - strong reducing agents

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a) Common Oxidizing Reducing Agents

• Oxidizing Agents Gain Electrons
• Oxygen (O2) hydrogen peroxide (H2O2) are always
  oxidizing agents
• The halogens (F2, Cl2, Br2, I2) are almost always
  oxidizing agents
• Nitric acid (HNO3), dichromate ion (Cr2O72)
  permanganate ion (MnO4) are strong oxidizing
  agents

Magnesium metal is oxidized in the presence of
oxygen.
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a) Common Oxidizing Reducing Agents
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a) Common Oxidizing Reducing Agents

• There are a few different ways to think about
  oxidation and reduction.

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3. Oxidation Numbers

• Oxidation numbers allow you to keep track of the
  electron transfers in a redox reaction.
• The oxidation number of an element compares the
  charge of a neutral uncombined atom with its
  actual charge or its relative charge in a
  compound.
• Oxidation numbers always change in a redox
  reaction.

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3. Oxidation Numbers

• Rule 1 The oxidation number of an atom of a pure
  element is zero.
• eg. Neutral atoms (Na, Cu, Al, etc.) have an
  equal number of protons electrons - no net
  charge
• Eg. In molecules containing only one type of atom
  (O2, H2, Cl2, etc.), bonding electrons are shared
  equally - no net charge

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3. Oxidation Numbers

• Rule 2 The oxidation number of a monatomic ion
  equals its charge.
• eg. Cations have fewer electrons than protons -
  positive net charge (Na 1, Ca2 2)
• eg. Anions have more electrons than protons -
  negative net charge (O2 2, Cl 1)
• How do you calculate oxidation numbers of
  elements in compounds?

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3. Oxidation Numbers

• Rule 3 Some elements have the same oxidation
  number in almost all compounds - can be used as
  reference points.
• Hydrogen 1 unless attached to metal (eg. NaH) -
  then -1
• Halogens -1 except in interhalogen compounds
  (eg. BrCl) then 0 F is always -1
• Oxygen -2 except in peroxides (eg. H2O2) then
  -1
• Group 6A (O, S, Se, Te) -2 in most compounds

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3. Oxidation Numbers

• Rule 4 The sum of the oxidation umbers in a
  neutral compound is zero. The sum of the
  oxidation numbers in a polyatomic ion equals the
  overall charge on the ion.
• Lets practice assigning oxidation numbers to
  some elements

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Assign an oxidation number to each atom in the
following species.
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0
0
2
-2
1
1
-3
-2
6
-1
4
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For the following reaction, assign oxidation
numbers to each element label the oxidizing
agent the reducing agent.
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For the following reaction, assign oxidation
numbers to each element label the oxidizing
agent the reducing agent.
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C. Solution Concentrations

• As we have seen in this chapter, many chemical
  reactions occur in solution.
• The reactants (solutes) are dissolved in a liquid
  (solvent).
• When the solvent is water, it is an aqueous
  solution.
• To know the quantity of solute in a given volume
  of liquid, you must know the concentration of the
  solution.

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1. Molarity

• The most commonly used unit of concentration is
  molarity (M).
• Molarity moles of solute
• liters of solution
• Note that the denominator specifies liters of
  solution, not liters of solvent.
• The solution includes both the solvent the
  solute, once it is dissolved.

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How many moles of NaOH are there in 275 mL of a
3.00 M solution?
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How many moles of NaOH are there in 275 mL of a
3.00 M solution?

• 275 mL 0.275 L
• 0.275 L x 3.00 mole 0.825 mol NaOH
• 1 L

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If 5.28 g of NaCl is placed in a 500-mL
volumetric flask and water is added until the
solution is exactly 500.0 mL, what is the
molarity of the resulting solution?
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If 5.28 g of NaCl is placed in a 500-mL
volumetric flask and water is added until the
solution is exactly 500.0 mL, what is the
molarity of the resulting solution?

• First, convert mass of NaCl into moles
• 5.28 g NaCl x 1 mol NaCl 9.034x10-2 mol
• 58.443 g
• Now, calculate molarity
• 9.034x10-2 mol 0.181 M
• 0.5000 L

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What is the concentration of Na ions in a 2.0 M
solution of Na2SO4?
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What is the concentration of Na ions in a 2.0 M
solution of Na2SO4?

• In this case, you have to look at the formula of
  the solute to answer the question
• 2.0 mol Na2SO4 x 2 mol Na 4.0
  M 1 L
• 1 mol Na2SO4

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a) Preparing Solutions of Known Molarity

• Solutions of known concentration can be made by
  dissolving a known mass of a solid in a liquid.
• Remember, the final volume includes the solvent
  and the dissolved solute.

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a) Preparing Solutions of Known Molarity

• When preparing a solution from a pure solute, you
  must use moles molar mass to calculate the
  correct mass of solute to use.
• eg. How would you prepare 100.0 mL of 0.0500 M
  Na2CO3 from solid Na2CO3?

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a) Preparing Solutions of Known Molarity

• When preparing a solution from a pure solute, you
  must use moles molar mass to calculate the
  correct mass of solute to use.
• eg. How would you prepare 100.0 mL of 0.0500 M
  Na2CO3 from solid Na2CO3?
• (0.0500 M)(0.1000 L) 0.00500 mols
• (0.00500 mol)(105.989 g/mol) 0.530 g
• Dissolve 0.530 g of Na2CO3 in water and add
  enough water to bring the total volume to 100.0
  mL.

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a) Preparing Solutions of Known Molarity

• Solutions of known concentration can also be made
  by the dilution of more concentrated solutions.
• In this case, it helps to remember that the
  number of moles in the original sample does not
  change when it is diluted. The number of moles
  of solute in the solution remains constant
  throughout the dilution operation.
• Moles Molarity x Volume, so MV is a constant
  throughout the dilution.

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a) Preparing Solutions of Known Molarity

• When youre doing a dilution
• M1V1 M2V2
• eg. (0.100 M)(0.1000 L) (0.0100 M)(1.000 L)
• 0.0100 mol 0.0100 mol

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• 100.0 mL of a 6.0 M HCl solution is diluted with
  water to yield a 500.0-mL solution. What is the
  molarity of the new solution?

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• 100.0 mL of a 6.0 M HCl solution is diluted with
  water to yield a 500.0-mL solution. What is the
  molarity of the new solution?
• (0.1000 L)(6.0 mols/L) (0.5000 L)M2
• M2 1.2 M

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How would you prepare 400.0 mL of 2.00 M HNO3
from concentrated nitric acid (16.0 M)?
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How would you prepare 400.0 mL of 2.00 M HNO3
from concentrated nitric acid (16.0 M)?

• M1V1 M2V2
• (0.4000 L)(2.00 M) (16.0 M)V2
• V2 0.0500 L
• Dilute 50.0 mL of concentrated nitric acid to a
  final volume of 400.0 mL.

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2. Molarity Aqueous Reactions

• Molarity is another tool for analyzing reaction
  stoichiometry (similar to molar mass).

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What is the maximum mass (in g) of AgCl that can
be precipitated by mixing 50.0 mL of 0.025 M
AgNO3 solution with 75.0 mL of 0.015 M NaCl
solution?
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What is the maximum mass (in g) of AgCl that can
be precipitated by mixing 50.0 mL of 0.025 M
AgNO3 solution with 75.0 mL of 0.015 M NaCl
solution?
AgNO3 (0.0500 L)(0.025 mol AgNO3 /L)(1 mol
AgCl/1 mol AgNO3) 0.00125 mol AgCl NaCl
(0.0750 L)(0.015 mol NaCl/L)(1 mol AgCl/1 mol
NaCl) 0.001125 mol AgCl (0.001125 mol
AgCl)(143.32 g/mol) 0.161 g AgCl