metal + acid displacement. activity series of metals
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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