1. Determine what reaction if any occurs. If a reaction ...kalmerscience.weebly.com/uploads/1/3/4/3/13437780/4.7.pdf · Stoichiometry of Precipitation Reactions ... into a solution

Section 4.7

Stoichiometry of Precipitation Reactions

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1. Determine what reaction if any occurs. If a reaction

occurs write a balanced molecular equation.

2. Write the balanced net ionic equation for the

reaction.

3. Calculate the moles of reactants.

4. Determine which reactant is limiting. (if given 2 reactants)

5. Convert to moles of product & then desired unit

Solving Stoichiometry Problems for Reactions in Solution

Section 4.7


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Stoichiometry of Precipitation Reactions:

Example 1: Calculate the mass of barium sulfate

formed when 250.0 mL of 0.0500 M BaCl2 and 150.0 mL

of 0.100 M Na2SO4 are mixed.

Step 1: The Molecular equation:

BaCl2 + Na2SO4 BaSO4 + 2NaCl

Step 2: The net Ionic equation

Ba2+ + SO42- BaSO4


Section 4.7


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Step 3: Calculate the moles of reactants

Step 4: Determine the limiting reagent

Step 5: Determine the moles of product and then

convert to other units as needed


2 21 0.0500 250.0 0.0125

1000

L molmol Ba mL mol Ba

mL L

2 2

4 4

1 0.100 150.0 0.0150

1000

L molmol SO mL mol SO

mL L

2

40.0150 0.0150

1

mol SO

20.0125

0.01251

mol Ba

2 4 442

4

1 233.4 0.0125 2.92

1 1

mol BaSO g BaSOmol Ba g BaSO

mol Ba mol BaSO

Ba2+ is limiting

Section 4.7


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• Where are we going?

Find the mass of solid Pb3(PO4)2 formed.

• How do we get there?

Write the balanced molecular & net ionic equation

What are the moles of reactants present in the solution?

Which reactant is limiting?

Convert to moles of product & then to the desired unit

What mass of Pb3(PO4)2 will be formed?

Let’s Think About It

Example 2: 10.0 mL of a 0.30 M sodium phosphate

solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate

solution (assume no volume change) What mass of

precipitate will form?

Section 4.7


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solution (assume no volume change) What mass of

precipitate will form?

Step 1: The Molecular equation: (what ppt forms)

2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s)

Step 2: The net Ionic equation

3Pb2+ + 2PO43- Pb3(PO4)2

Section 4.7


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Step 3: Calculate the moles of reactants

Step 4: Determine the limiting reagent

Step 5: Determine moles of product & convert to the

desired unit


3 41 0.30

10.0 .00301000 1

L molmLX x molNa PO

mL L

3 21 0.20

20.0 .0040 (N )1000 1

L molmLX x molPb O

mL L

3 4.00300.0015

2

molNa PO 3 2.0040 (N )

.00133

molPb O

3 4 2 3 4 23 4 2

3 2 3 4 2

1 ( ) 811.5 ( ).0040 1.1g ( )

3 (N ) 1 ( )

molPb PO gPb POx x Pb PO

molPb O molPb PO

Pb2+ is limiting.

Section 4.7


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To find the concentration of nitrate ions left in

solution after the reaction is complete.


What are the moles of nitrate ions present in the

combined solution?

What is the total volume of the combined solution?




solution (assume no volume change). What is the

concentration of nitrate ions left in solution after the

reaction is complete?

Section 4.7


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Example 3:

• Nitrate ions are spectator ions & don’t participate in

the reaction.

• We know from the earlier problem that Pb+2 was

limiting therefore: .0040M


3 3.0040N 2 .0080 NO x mol O

33

.0080 N.27 N

.0300

mol OM O

L

• The total volume of the solution is 30.0mL

Section 4.7


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Find the concentration of phosphate ions left after the reaction

is complete.


What are the moles of phosphate ions present in the solution at

the start of the reaction?

How many moles of phosphate ions were used up in the

reaction to make the solid Pb3(PO4)2?

How many moles of phosphate ions are left over after the

reaction is complete?

What is the total volume of the combined solution?


Example 4: 10.0 mL of a 0.30 M sodium phosphate solution

reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume

no volume change). What is the concentration of phosphate ions

left in solution after the reaction is complete?

Section 4.7


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Example 4:

• Take moles of limiting and turn to excess; this

determines the amount of phosphate used

during the reaction.


4 34

2.0040 +2 .0027 3

3 2

molPOPb x molPO

molPb

0.0030 mol– 0.0027 mol= 0.00033 mol of PO4-3 left

• There were 0.0030 mol of Na3PO4 at the start,

which means there are 0.0030 mol PO4-3

Section 4.7


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Example 5: Calculate the mass of Ag2S produced

when 125ml of 0.200MAgNO3 is added to an excess

of Na2S solution.


2AgNO3+ Na2S Ag2S +2NaNO3

3 2 23 2

3 3 2

1 .200 1 247.93.10

1000 2 1125

1

L molAgNO molAg S gAg SAgNO x x x x gAg S

mL LAgNO molAgNO molL

Ag Sm

Section 4.7


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Done for Today….


Section 4.8

Acid–Base Reactions

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• Acid—proton donor

• Base—proton acceptor

• For a strong acid and base reaction:

H+(aq) + OH–(aq) H2O(l)

Acid–Base Reactions (Brønsted–Lowry)

Section 4.8


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Neutralization of a Strong Acid by a Strong Base

Section 4.8


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1. List the species present in the combined

solution before any reaction occurs, and decide

what reaction will occur.

2. Write the balanced net ionic equation for this

reaction.

3. Calculate moles of reactants.

4. Determine the limiting reactant, where

appropriate.

5. Calculate the moles of the required reactant or

product.

6. Convert to grams or volume (of solution), as

required.

Performing Calculations for Acid–Base Reactions

Section 4.8


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• Titration – delivery of a measured volume of

a solution of known concentration (the titrant)

into a solution containing the substance

being analyzed (the analyte).

• Equivalence point – enough titrant added to

react exactly with the analyte.

• Endpoint – the indicator changes color so

you can tell the equivalence point has been

reached.

Acid–Base Titrations

Section 4.8


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Concept Check

For the titration of sulfuric acid (H2SO4) with

sodium hydroxide (NaOH), how many moles

of sodium hydroxide would be required to

react with 1.00 L of 0.500 M sulfuric acid to

reach the endpoint?

1.00 mol NaOH

Section 4.8


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To find the moles of NaOH required for the

reaction.


What are the ions present in the combined

solution? What is the reaction?

What is the balanced net ionic equation for the

reaction?

What are the moles of H+ present in the solution?

How much OH– is required to react with all of the

H+ present?


Section 4.9

Oxidation–Reduction Reactions

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• Reactions in which one or more electrons

are transferred.

Redox Reactions

Section 4.9


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Reaction of Sodium and Chlorine

Section 4.9


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1. Oxidation state of an atom in an element = 0

2. Oxidation state of monatomic ion = charge of

the ion

3. Oxygen = 2 in covalent compounds (except in

peroxides where it = 1)

4. Hydrogen = +1 in covalent compounds

5. Fluorine = 1 in compounds

6. Sum of oxidation states = 0 in compounds

7. Sum of oxidation states = charge of the ion in

ions

Rules for Assigning Oxidation States

Section 4.9


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Exercise

Find the oxidation states for each of the

elements in each of the following

compounds:

• K2Cr2O7

• CO32-

• MnO2

• PCl5

• SF4

K = +1; Cr = +6; O = –2

C = +4; O = –2

Mn = +4; O = –2

P = +5; Cl = –1

S = +4; F = –1

Section 4.9


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• Transfer of electrons

• Transfer may occur to form ions

• Oxidation – increase in oxidation state

(loss of electrons); reducing agent

• Reduction – decrease in oxidation state

(gain of electrons); oxidizing agent

Redox Characteristics

Section 4.9


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Concept Check

Which of the following are oxidation-reduction

reactions? Identify the oxidizing agent and the

reducing agent.

a)Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

b)Cr2O72-(aq) + 2OH-(aq) 2CrO4

2-(aq) + H2O(l)

c)2CuCl(aq) CuCl2(aq) + Cu(s)

Section 4.10

Balancing Oxidation–Reduction Equations

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1. Write the unbalanced equation.

2. Determine the oxidation states of all atoms in

the reactants and products.

3. Show electrons gained and lost using “tie

lines.”

4. Use coefficients to equalize the electrons

gained and lost.

5. Balance the rest of the equation by

inspection.

6. Add appropriate states.

Balancing Oxidation–Reduction Reactions by Oxidation States

Section 4.10


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• Balance the reaction between solid zinc

and aqueous hydrochloric acid to

produce aqueous zinc(II) chloride and

hydrogen gas.

Section 4.10


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• Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g)

1. What is the unbalanced equation?

Section 4.10


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0 +1 –1 +2 –1 0

2. What are the oxidation states for each atom?

Section 4.10


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1 e– gained (each atom)


0 +1 –1 +2 –1 0

2 e– lost

• The oxidation state of chlorine remains unchanged.

3. How are electrons gained and lost?

Section 4.10


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1 e– gained (each atom) × 2


0 +1 –1 +2 –1 0

2 e– lost

• Zn(s) + 2HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g)

4. What coefficients are needed to equalize the

electrons gained and lost?

Section 4.10


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• Zn(s) + 2HCl(aq) Zn2+(aq) + 2Cl–(aq) + H2(g)

5. What coefficients are needed to balance the

remaining elements?

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1. Determine what reaction if any occurs. If a reaction ...kalmerscience.weebly.com/uploads/1/3/4/3/13437780/4.7.pdf · Stoichiometry of Precipitation Reactions ... into a solution