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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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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 H2 from acids top to bottom
Potassium + Water
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)
Metal + Metal Salt 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)
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
Cola DrinksSolvent
• water
Solutes
• carbon dioxide (gas)
• sweetener (solid)
• phosphoric acid (liquid)
• caramel color (solid)
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
Preparation of 1.00 L of 0.0100 M
KMnO4 solution from solid
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
Solution Preparation by Dilution
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
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
Stoichiometric Relationships
Titrationsat equivalence mol H+ = mol OH-
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)
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-
• 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
• 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
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