Acids and Bases Chapter 15
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Transcript of Acids and Bases Chapter 15
Acids and Bases Chapter 15
Notes One Unit TwelveProperties of Acids
Properties of Bases
Structure of Bases
Neutralization Reactions
Properties of AcidsSour 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).
Acidsacid - proton donor (H+1)
strong vs. weak acids
strong - lots of (H+ )
weak - very little (H+ )
HCl is a strong acidMuriatic 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, perfumesPurification of Ag, Au, Pt
Properties of BasesAlso known as alkali.( Li, Na, K, Rb, Cs, Fr)
Taste bitter.caffeineoften 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+carbonate Salt +Water+Carbon DioxideAcid+Base Salt +WaterAcid+Metal Salt +Hydrogen
Acid Reactions
Acid and CarbonateAcid+
Na2CO3 CarbonateSalt+ Water+Carbon Dioxide
HNO3+ NaNO3 + H2O + CO22 1 2 1 1
H+1 NO3-1 Na+1 CO3
-2
NaNONaNO33
=Sodium Nitrate=Sodium Nitrate( )_( )_NaNa+1+1 NONO33
-1-11111
Salt ?Salt ?
Acid and BaseAcid+
KOH Base Salt + Water
H2SO4+ KK22SOSO44 + H2O 1 2 1 2
H+1 SO4-2 K+1 OH-1
KK22SOSO44
=Potassium Sulfate=Potassium Sulfate( )_( )_KK+1+1 SOSO44
-2-21122
Salt ?Salt ?
Acid and MetalAcid+
Mg(s) Metal Salt + Hydrogen
H2SO4 + MgMgSO4 + H2(g)1 1 1 1
H+1 SO4-2 Mg(s)
MgMgSO4
= Magnesium Sulfate= Magnesium Sulfate
( )_( )_MgMg+2+2 SO4-2
2222
Salt ?Salt ?
Titration Example One
NaOH(aq)+ HCl(aq) NaCl(aq) + H2O(l)
0.000599 m HCl x1 m NaOH
1 m HCl=
1 1 1 1
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.
2. Find moles used of known solution.
3. Calculate moles used of unknown (titrant).
4. Calculate the molar concentration of the titrant.
MxV=n0.00599L = 0.100M x 0.000599m HCl
n/V=M0.000599m NaOH ÷0.01201L= 0.0499M NaOH
0.000599m NaOH
Notes Two Unit Twelve -Text Pages 550-558
• Self Ionization of Water
• Brönsted-Lowrey Acid-Base Theory
• Arrhenius Theory
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 • Base - proton acceptor• Acid-Conjugate Base / Base-Conjugate Acid
• HCl(aq) + H2O(l) Cl−1(aq) + H3O+1(aq)
• NaF(aq) + H2O(l) HF(aq) + NaOH(aq)
• OH−1(aq) + H2O(l) H2O(l)+ OH−1(aq)
• NH3(aq) + H2O(l) NH4+1(aq) + OH−1(aq)
A CB
A
A
CB
CB
CB
A
CAB
B
B
B
CA
CA
CA
H+1
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 Twelve
• Titration
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
NaOH + HC2H4OHCO2 NaC2H4OHCO2 + H2O
0.0899m NaOHx1m HC2H4OHCO2
1 m NaOH=
1 1 1 1
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.
2. Find moles used of known solution.
3. Calculate moles used of unknown (titrant).
4. Calculate the Molar Concentration of theTitrant.
MxV=n1.00M x 0.0899L = 0.0899m NaOH
n/V=M0.0899m HC2H4OHCO2÷0.01201L =
M=0.0899M HC2H4OHCO2
0.0899m HC2H4OHCO2
M
Titration Example Three
Ba(OH)2(q) + HCl(aq BaCl2(aq)+ H2O(l)
0.00463 m HCl x1 m Ba(OH)2
2 m HCl=
1 2 1 2
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.
2. Find moles used of known solution.
3. Calculate moles used of unknown (titrant).
4. Calculate volume of titrant.
MxV=n0.1057M x 0.0438L= 0.00463m HCl
n/M=V0.00234m Ba(OH)2 ÷0.0622M = 0.0376M Ba(OH)2
0.00232 m Ba(OH)2
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 Twelve Text Pages 559-567
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.
pH Scale
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Acid BaseNeutral
1M HCl 1M NaOHAmmonia Cleaner
Blood
WaterMilk
Stomach Acid
Lemon Juice
Vinegar
pH +pOH =14 pH +pOH =14
pH of strong acid
•Find the pH of a 0.15 M Find the pH of a 0.15 M solution of Hydrochloric solution of Hydrochloric acidacid
•pH = - log 0.15pH = - log 0.15•pH = - (- 0.82)pH = - (- 0.82)•pH = 0.82pH = 0.82
pH of a strong Base
•What is the pH of the What is the pH of the 0.0010 M NaOH solution?0.0010 M NaOH solution?
•pOH = - log (0.0010)pOH = - log (0.0010)
•pOH = 3pOH = 3
•pH = 14 – 3 = 11pH = 14 – 3 = 11pH +pOH =14 pH +pOH =14
Ionization Constants for Ionization Constants for Acids/Bases 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.
HNO2 NO2-1 + H+1
[HNO2 ] [NO2-1] [H+1]
Bef
ΔAt
[HNO2][H+1][NO2
-1]Ka=
-x +x +x0.15-x x x
0.15 0 0
Ka= [0.15][x][x]
= 4.5 x 10-4
X=[0.0082M]
1) Balanced Equation
2) Mass Action Expression3) What do we know?
4) Calculate the [H+1] and pH.
Very small
= [H+1] pH=-log[ ]0.0082M
pH=-log[H+1]
= 2.09
4.5 x 10-4
1.5 x 10-1
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.
C6H5COOH C6H5COO-1 + H+1
[C6H5COOH] [C6H5COO-1] [H+1]
Bef
ΔAt
[C6H5COOH][H+1][C6H5COO-1]
Ka=
-x +x +x
0.10-x x x
0.10 0 0
Ka= [0.10][x] [x]
= 6.5 x 10-5
X=[0.0025M]
1) Balanced Equation
2) Mass Action Expression3) What do we know?
4) Calculate the [H+1] and pH.
Very small
= [H+1]pH=-log[ ]0.0025M = 2.60
6.5 x 10-5
1.0 x 10-1
End