Transcript H +1 - Milwaukie High

Acids/Bases
Notes One Unit Eleven
Pages 541-549
Properties of Acids
Properties of Bases
Structure of Bases
Neutralization Reactions
Lab Acid/Base Properties Next Class
Part I Effect of acids and base on indicators
• 1. Stand 3 clean, dry test tubes in the test tube rack. Add
about 2 mL of 6M HCl to one of the test tubes, 2 mL of 6M
HC2H3O2 to a second tube, and 2 mL of 0.5 M NaOH to the
third tube. CAUTION: Handle these chemicals with care. Test
each solution first with red litmus paper and then with blue
litmus paper.
• 2. Test each solution with pH paper.
• 3. Add 1 drop of phenolphthalein to each solution. Discard
the solutions as instructed. Clean and rinse the test tubes.
Part IV Neutralization
• 1. Using a clean dropper pipette, add 10 drops of
1.0M HCl and 1 drop of phenolphthalein to a test
tube. Test with pH paper.
• 2. Using a second dropper pipette, add 0.5M NaOH
drop-by-drop. After the addition of each drop, swirl the
test tube gently mix the contents. Count the total
number of drops of NaOH needed to cause a color
change. Once a color change is observed, test the
mixture with pH paper. Number of drops of 0.5M
NaOH to neutralize 10 drops of 1.0M HCl _________.
pH of neutral solution:_________
Lime Water Test For CO2
Zinc with hydrochloric acid
Metals With Acid
Questions on Lab
• 5. What type of reaction occurs between a metal and an acid? Write a
general equation for this type of reaction.
• Single
• M+ A------->H2+ salt
• 6. Explain the difference in reaction rates of a given metal with two
different acids.
• HCl is a strong acid. It makes more H3O+1. More H3O+1 means
faster reaction.
• 7. Write a balanced equation for the reaction between CO2 gas and lime
water, Ca(OH)2. What is the name of the milky precipitate that forms?
• Ca(OH) 2 + CO2 ------>CaCO3+ H2O
• 8. Explain the difference in the volumes (number of drops) of HCl and
NaOH required to produce a neutral solution in Part 4 of this
experiment.
• 10 drops to 20 drops means the base is half the concentration.
Notes One Unit Eleven
Properties of Acids
Properties of Bases
Structure of Bases
Neutralization Reactions
Properties of Acids
Sour taste.
warhead
React with “active” metals produce H2.
Al, Zn, and Fe
React with carbonates, producing CO2
and H2O.
Marble, baking soda, chalk, limestone.
Change color of vegetable dyes.
cabbage juice / On the top
cabbage juice / baking soda (left)
cabbage juice / vinegar (right).
Acids
acid - proton donor (H+1)
strong vs. weak acids
strong - lots of (H+ )
weak - very little (H+ )
HCl is a strong acid
Muriatic acid is
hydrochloric acid.
Making Bleach.
Making PVC pipe.
Making Table Salt.
Human stomach acid.
Cleaning steel.
Neutralize bases in
chemical plants.
Chrome tanning leather.
HNO3 is a strong acid
Explosives
Fertilizers etc.
Nitrate salts
To make H2SO4
Etching copper, brass,
bronze
Dyes, perfumes
Purification of Ag, Au, Pt
Properties of Bases
Also known as alkali.( Li, Na, K, Rb, Cs, Fr)
Taste bitter.
caffeine
often poisonous.
Solutions feel slippery.
Change color of vegetable dyes.
Different color than acid.
Red litmus turns blue.
React with acids to form salt and
water(Neutralization).
Acid + base salt +water
Common Base Features
Most ionic bases contain OH-1
ions.
NaOH (drain cleaner)
Some contain CO32- ions.
CaCO3(in Tums)
NaHCO3 (baking soda)
Molecular bases contain
structures that react with H+.
Mostly amine groups(-NH2).
Neutralization Reactions
acid + base  salt + water
Double-displacement
Some make CO2 and H2O
Acid Reactions
Acid+carbonateSalt +Water +Carbon Dioxide
Acid+Base Salt +Water
Acid+Metal Salt +Hydrogen
Acid and Carbonate
Acid+ CarbonateSalt+ Water+Carbon Dioxide
2 HNO3+1 Na2CO3 2 NaNO3 +1 H2O + 1CO2
H+1 NO3-1 Na+1 CO3-2
Salt ? =Sodium Nitrate
(Na+1)_(
1 NO -1 )_
1
3
NaNO3
Acid and Base
Water
Acid+ Base 
Salt +
1 H2SO4+ 2 KOH  1 K2SO4 + 2 H2O
H+1 SO4-2
K+1
OH-1
Salt ? =Potassium Sulfate
( K+1 )_(
2 SO -2 )_
1
4
K2SO4
Acid and Metal
Salt + Hydrogen
Acid+ Metal 
1 H2SO4 + 1 Mg(s) 1MgSO4 + 1 H2(g)
H+1 SO4-2 Mg(s)
Salt ? = Magnesium Sulfate
2 SO -2 )_
2
( Mg+2)_(
4
MgSO4
Titration Example One
What is the molarity of a NaOH solution, if
12.01 mL is required to titrate 5.99 mL of 0.100
M HCl?
1. Balance the equation.
1NaOH(aq)+ 1 HCl(aq) 1 NaCl(aq) + 1 H2O(l)
2. Find moles used of known solution.
MxV=n
0.100M x 0.00599L = 0.000599m HCl
3. Calculate moles used of unknown (titrant).
1 m NaOH
0.000599 m HCl x
= 0.000599m NaOH
1 m HCl
4. Calculate the molar concentration of the titrant.
n/V=M
0.000599m NaOH ÷0.01201L = 0.0499M NaOH
Notes Two Unit Eleven -Text Pages 550-558
Self Ionization of Water
Brönsted-Lowrey Acid-Base Theory
Arrhenius Theory
Pages 550-558
Self Ionization of Water
• Water is and acid and a base at the same
time…amphoteric
• H2O(l) + H2O(l)  H3O+ + OH• Mass Action Expression
• Kw = [H3O+][OH-]
• Or Kw = [H+][OH-]
• Kw = 1.0 x 10-14
Brönsted-Lowrey Acid-Base Theory
•
•
•
•
Acid - proton donor
H+1
Base - proton acceptor
Acid-Conjugate Base / Base-Conjugate Acid
HCl(aq) + H2O(l)  Cl−1(aq) + H3O+1(aq)
A
B
CB
CA
• NaF(aq) + H2O(l)  HF(aq) + NaOH(aq)
B
A
CA
CB
• OH−1(aq) + H2O(l)  H2O(l)+ OH−1(aq)
B
A
B
A
CA
CB
• NH3(aq) + H2O(l)  NH4+1(aq) + OH−1(aq)
CA
CB
Arrhenius Theory
• Bases form OH- ions in water.
• Acids form H+ ions in water.
Arrhenius theory
•
•
•
•
HCl(aq)  H+1(aq) + Cl−1(aq)
HF(aq)  H+1(aq) + F−1(aq)
NaOH(aq)  Na+1(aq) + OH−1(aq)
NH3(aq) +H2O(aq) NH4+1(aq) + OH−1(aq)
Polyprotic Acids
• More than one proton to donate.
• H2CO3(aq)  H+1(aq) + HCO3-1(aq)
• HCO3-1(aq)  H+1(aq) + CO3-2(aq)
• H2SO4  H+1(aq) + HSO4−1(aq)
• HSO4−1  H+1(aq) + SO4−2(aq)
Notes Three Unit Eleven
Titration
Pages 574-578
Titration
• Titration is a technique to determine
the concentration of an unknown
solution.
• Titrant (unknown solution)
• Phenolphthalein identifies the
Endpoint.
Titration Endpoint
• Add 10mL of HCl and three drops phenolphthalein to the
flask.
• Add about 8mL base, swirl and add the rest of the base
using increasingly faster spins of the valve.(?????)
Titration-Acid Volume
•
•
•
•
•
•
•
Acid Burette
Initial Reading?
1.98mL
Final Reading?
7.97mL
Volume Used?
5.99mL
Titration-Volume of Base Used.
•
•
•
•
•
•
•
Base Burette
Initial Reading?
0.00mL
Final Reading?
12.01mL
Volume Used?
12.01mL
Titration Example Two
• Lactic acid Concentration of Sauerkraut
• For Joe’s final in chemistry, he was asked to
find the concentration of lactic acid in
homemade sauerkraut.
• He did not have any home made sauerkraut.
• Therefore, Joe was left to make the home made
sauerkraut.
• He looked on line and found the following
recipe.
Clean and Quarter 35lb of Fresh Cabbage
Shred the Cabbage into a Crock
Add 3 Tbsp salt per 5 pounds of cabbage.
Mix salt and cabbage.
Pack the
cabbage into a
crock and
weight it down.
It should be
fermented in
one month.
Titration Example Two
What Is the molarity of a sauerkraut juice if
10.0 mL is titrated using 89.9 mL of 1.00 M
NaOH?
1. Balance the equation.
1NaOH + 1 HC2H4OHCO2  1 NaC2H4OHCO2 + 1 H2O
2. Find moles used of known solution.
MxV=n
1.00M x 0.0899L = 0.0899m NaOH
3. Calculate moles used of unknown (titrant).
1m HC2H4OHCO2
0.0899m NaOHx
= 0.0899m HC2H4OHCO2
1 m NaOH
4. Calculate the Molar Concentration of theTitrant.
n/V=M
0.0899m HC2H4OHCO2 ÷0.01201L = M
M=0.0899M HC2H4OHCO2
Titration Example Three
What is the volume of a 0.0622M Ba(OH)2
solution, if it is titrated using 43.8 mL of
0.1057 M HCl?
1. Balance the equation.
1Ba(OH)2(q) + 2 HCl(aq 1 BaCl2(aq)+ 2H2O(l)
2. Find moles used of known solution.
MxV=n
0.1057M x0.0438L= 0.00463m HCl
3. Calculate moles used of unknown (titrant).
1 m Ba(OH)2
0.00463 m HCl x
= 0.00232 m Ba(OH)2
2 m HCl
4. Calculate volume of titrant.
n/M=V
0.00234m Ba(OH)2 ÷0.0622M = 0.0376M Ba(OH)2
Lab B Titration Results
Trial #1
HCl
NaOH
Trial #2
HCl
NaOH
Trial #3
Trial #4
HCl
NaOH
HCl
NaOH
Intial reading
0.01
0.21
11.06
21.10
21.21
31.35
31.34
42.00
Final reading
11.05
11.27
21.00 31.54
31.33
42.07
38.95
49.99
Volume used
11.04
11.60
9.94
10.21
10.72
7.61
7.99
10.44
Table a
Calculations
• For each trial, calculate the molarity of the NaOH
solution using the relationship
• Conclusion and Question
• 1. How reproducible were the results of your four
trials?
Conclusions continued
• 2. Define these terms: standard solution; titration;
endpoint.
• Standard Solution: When the concentration of a
solution is known to a high degree of accuracy and
precision.
• titration: When the concentration of an acid or base is
determined by neutralizing it.
• endpoint: The point where you actually stop a titration,
usually because an indicator has changed color. This is
different than the "equivalence point" because the
indicator might not change colors at the exact instant
that the solution is neutral.
Notes Three Unit Eleven
Weak Acid/Base
pH Scale
Calculating pH
Pages 559-567
Which acid is stronger?
6M HCl
zinc
magnesium
iron
copper
6M HC2H3O2
Weak Acids and Bases
•
•
•
•
•
A weak acid
little H+1
A weak base
little OH-1
[H+] or [OH-] from a Keq.
pH = -log[H+1]
pH is a measure of the amount of hydrogen in
a solution. It is based on the water.
Stomach Vinegar
pH Scale
Acid
Blood
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Neutral
Acid
Base
1M HCl
Ammonia 1M NaOH
Lemon
Juice
Milk Water
pH +pOH =14
Cleaner
pH of strong acid
• Find the pH of a 0.15 M solution of
Hydrochloric acid
• pH = - log 0.15
• pH = - (- 0.82)
• pH = 0.82
pH of a strong Base
• What is the pH of the 0.0010 M
NaOH solution?
• pOH = - log (0.0010)
• pOH = 3
pH +pOH =14
• pH = 14 – 3 = 11
Ionization Constants for
Acids/Bases
Calculating pH (pH=-log[H+1])
The Ka for nitrous acid is 4.5 x 10-4. Calculate the
[H+1] and pH in 0.15 M nitrous acid solution.
1) Balanced Equation
HNO2  NO2-1 + H+1
2) Mass Action Expression
3) What do we know?
[HNO2 ]
[NO2-1]
Bef
0.15
0
Very small
Δ
+x
-x
At
0.15-x
x
[NO2-1] [H+1]
Ka= [HNO ]
2
[H+1]
0
4.5 x 10-4
+x
-1
1.5
x
10
x
4) Calculate the [H+1] and pH.
[x][x]
Ka= [0.15] = 4.5 x 10-4 pH=-log[H+1]
pH=-log[0.0082M ]= 2.09
X=[0.0082M] = [H+1]
Calculating Concentrations Using Ka
The Ka for benzoic acid is 6.5 x 10-5. (a) Calculate the
concentrations of C6H5COO-1 and H+ in a 0.10 M
benzoic acid solution. (b) Calculate pH.
1) Balanced Equation
C6H5COOH  C6H5COO-1 + H+1 [C H COO-1][H+1]
6 5
Ka=
[C6H5COOH]
2) Mass Action Expression
3) What do we know?
[C6H5COOH] [C6H5COO-1] [H+1]
Bef
0.10
0
small
Δ Very -x
+x
At
0.10-x
x
4) Calculate the [H+1] and pH.
[x] [x]
Ka=
[0.10]
0
+x
x
6.5 x 10-5
1.0 x 10-1
X=[0.0025M] = [H+1]
= 6.5 x 10-5 pH=-log[0.0025M]= 2.60
End