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Metal + Acid Displacement

Metal + Acid Displacement

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Metal + Acid Displacement. Activity Series of Metals. Activity Series of Metals. metals higher in series react with compounds of those below metals become less reactive to water top to bottom metals become less able to displace H 2 from acids top to bottom. Potassium + Water. - PowerPoint PPT Presentation

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Page 1: Metal + Acid Displacement

Metal + Acid Displacement

Page 2: Metal + Acid Displacement

Activity Series of Metals

Page 3: Metal + Acid Displacement

Activity Series of Metals

• metals higher in series react with compounds of those below

• metals become less reactive to water top to bottom

• metals become less able to displace H2 from acids top to bottom

Page 4: Metal + Acid Displacement

Potassium + Water

Page 5: Metal + Acid Displacement

Activity Series of Metals

Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g)

Zn(s) + 2HBr(aq) ZnBr2(aq) + H2(g)

Page 6: Metal + Acid Displacement

Metal + Metal Salt Displacement

Page 7: Metal + Acid Displacement

Which of the following reactions does NOT happen?

100

0

1300% 0% 0%0%0%

1. Cu(s)+H2SO4(aq)CuSO4(aq)+H2(g)

2. 2HNO3(aq)+2K(s)2KNO3(aq)+H2(g)

3. FeCl2(aq)+Zn(s)ZnCl2(s)+Fe(s)

4. Ca(s)+2H2O(l)Ca(OH)2(aq)+H2(g)

5. Cu(s)+2AgNO3(aq)2Ag(s)+Cu(NO3)2(aq)

Page 8: Metal + Acid Displacement

Solution

A homogeneous mixture of two or more substances comprising the solvent which is the majority of the mixture and one or more solutes which are the smaller fraction.

Concentration – how much solute is present in a given amount of solution

Page 9: Metal + Acid Displacement

Cola DrinksSolvent

• water

Solutes

• carbon dioxide (gas)

• sweetener (solid)

• phosphoric acid (liquid)

• caramel color (solid)

Page 10: Metal + Acid Displacement

Molarity – a measure of concentration

The number of moles of solute per liter of solution.

molarity M

moles of soluteM =

liters of solution

units molar = mol/L = M

Page 11: Metal + Acid Displacement

Preparation of 1.00 L of 0.0100 M

KMnO4 solution from solid

Page 12: Metal + Acid Displacement

Which value do you NOT need to determine the molarity of a solution?

100

0

130

Mas

s of s

olute

Mola

r mas

s of s

olute

Volum

e of s

olve

nt a

dded

Total v

olum

e of s

olutio

n

0% 0%0%0%

1. Mass of solute

2. Molar mass of solute

3. Volume of solvent added

4. Total volume of solution

Page 13: Metal + Acid Displacement

Solution Preparation by Dilution

Page 14: Metal + Acid Displacement

Dilution

• Molarity (mol/L) × Volume (L) = Moles

• If we take a sample (say 25.0 mL) from a solution (say 0.372 M) and add extra water (say to a total volume of 500. mL) the moles of solute are unchanged

• Thus M1V1 = moles = M2V2

• 0.372 M × 25.0 mL = M2 × 500. mL

• M2 = 0.0186 M

Page 15: Metal + Acid Displacement

How many L of conc HNO3 (16.0 M) are needed to prepare 0.500 L of 0.250 M nitric acid?

100

0

130

32.0

L

16.0

L

0.50

0 L

0.25

0 L

0.00

781

L

0% 0% 0%0%0%

1. 32.0 L2. 16.0 L3. 0.500 L4. 0.250 L5. 0.00781 L

Page 16: Metal + Acid Displacement

Stoichiometric Relationships

Page 17: Metal + Acid Displacement

Titrationsat equivalence mol H+ = mol OH-

Page 18: Metal + Acid Displacement

EXAMPLE: A sample of lye, sodium hydroxide, is neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?

2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O

(25.0 mL NaOH) #mL H2SO4 =

(0.400 mol NaOH)

(1 L NaOH)

(1 L)

(1000 mL)

(1 mol H2SO4)

(2 mol NaOH)

(1 L H2SO4)

(0.200 mol H2SO4)

= 25.0 mL H2SO4

(1000 mL)

(1 L)

Page 19: Metal + Acid Displacement

Ion Concentrations

• 0.100 M NaCl

• NaCl(aq) Na+(aq) + Cl-

(aq)

• Is 0.100 M in both Na+ and Cl-

• 0.100 M Al2(SO4)3

• Al2(SO4)3(aq) 2 Al3+(aq) + 3 SO4

2-(aq)

• Is 0.200 M in Al3+ and 0.300 M in SO42-

Page 20: Metal + Acid Displacement

• Consider the NaOH + H2SO4 reaction. What are the final concentrations of Na+ and SO4

2-?

• Stoichiometric reaction, can use either reactant to determine moles of product

• 25.0 mL NaOH × 0.400 mol NaOH × 1 L 1 L 1000 mL0.0100 mol NaOH × 1 mol Na2SO4 = 0.0050 mol Na2SO4

2 mol NaOH

Final Ion Concentrations

Page 21: Metal + Acid Displacement

• Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 0.0500 L

• [Na2SO4] = 0.00500 mol = 0.100 M 0.0500 L

• [Na+] = 2 × 0.100 M = 0.200 M

• [SO42-] = 0.100 M

Page 22: Metal + Acid Displacement

Which solution has the highest concentration of SO4

2-?

100

0

130

0.20

M C

uSO4

0.15

M N

a2SO

4

0.07

0 M

Fe2

(SO4)

3

0.10

M C

e(SO4)

2

0% 0%0%0%

1. 0.20 M CuSO4

2. 0.15 M Na2SO4

3. 0.070 M Fe2(SO4)3

4. 0.10 M Ce(SO4)2