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HOMEWORK• Do in this order • 51,52,53,55,61,65,67,23,25,27,29,31,33, 37a&e, 39,41,43a,45,47, and 49a&b
Acid-Base Equilibria
REMEMBERWhen working problems in this chapter, it will
help if you always……..
Strong Acid/Strong Base Titration
Excess Base
Excess Acid
moles of acid =moles of base
WHAT IS THE DIFFERENCE BETWEEN END POINT AND
EQUIVALENCE POINT?
Strong Acid/Strong Base Titration•Determine the pH if 50 ml of 0.25 M HCl is mixedwith 30 ml of 0.25 M NaOH
Strong Acid/Strong Base TitrationDetermine the pH when 30.0 ml of 0.500 M HClis added to 60 ml of 0.250 M NaOH.
Strong Acid/Strong Base TitrationDetermine the pH when 25.0 ml of 0.400 M HClis added to 85.0 ml of 0.300 M NaOH.
Strong Acid/Strong Base Titrationa. Determine the pH if enough 0.700 M NaOH is
added to 30.00 ml of .500 M HCl to reach the equivalence point.
b. Determine the volume of NaOH required to reach the equivalence point.
Weak Acid/Strong Base Titration
All strong base is converted to CB
Use Kb = X2
moles base – x total volume
Weak Acid/Strong Base Titration
Buffer RegionUse [H+
] = Ka [ moles acid ] [ moles base]You are in in the buffer region when the moles of acid are greater than base.
Excess base regionMoles of base is greater than acid
Determine the moles ofstrong base left over . Divide by total volume gives [OH-
] concentration.
All strong base is converted to CBat the equivalence point
Use Kb = X2
moles base – x total volume
HENDERSON-HASSELBALCH EQUATION (Buffers)
pH pKa + log[Conj. base]
[Acid]
Derivative is
H+ = Ka [HB] [B- ]
If you have a buffer , use the aboveequation. Makes the math easier.
HENDERSON-HASSELBALCH EQUATION (Buffers)
pH pKa + log[Conj. base]
[Acid]Derivative ispKa = -log10Ka
pH = pka when moles of acid = moles of base
H+ = Ka [HB] [B- ]
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH.1. Before the addition of any base (initial pH).2 After the addition of 8.00 mL of 0.500 M NaOH (buffer region).3. After the addition of 10.00 mL of 0.500 M NaOH
(half-neutralization point).4. After the addition of 20.00 mL of 0.500 M NaOH
(equivalence point).5. After the addition of 21.00 mL of 0.500 M NaOH
(beyond the equivalence point).
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH.
1. Before the addition of any base .
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH 2. after the addition of 8.00 mL of 0.500 M NaOH
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH 3. after the addition of 10.00 mL of 0.500 M NaOH
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH 4. after the addition of 20.00 mL of 0.500 M NaOH
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH 5. After the addition of 21.00 mL of 0.500 M NaOH
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Determine the pH of a solution when 25.0 ml of 2.00 M HC2H3O2 is titrated halfway to the equivalence point with NaOH
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Determine the pH of a solution when 25.0 ml of 2.00 M HC2H3O2 is titrated with 1.50 M NaOH to the equivalence point,
Neutralization ReactionWeak Acid - Strong Base
5 Positions on the Titration Curve
Buffer Region
Excess AcidRegion
Use [H+ ] = Ka [ moles acid ]
[ moles base]
You are in in the buffer region when the moles of weak is greater than strong
All strong acid is converted to CA ate the equivalence point Use
Ka = X2
moles base – x total volume
Moles of strong are greater than weak Determine the moles of acid left over then divide by total volume. Gives H+ concentration
WEAK BASE STRONG ACID TITRATION
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia with 0.500 M HCl1. Before the addition of any acid (initial pH).2 After the addition of 8.00 mL of 0.500 M HCl (buffer region).3. After the addition of 10.00 mL of 0.500 M HCl
(half-neutralization point).4. After the addition of 20.00 mL of 0.500 M HCl
(equivalence point).5. After the addition of 21.00 mL of 0.500 M HCl
(beyond the equivalence point).
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 1. Before the addition of any acid.
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 2. After the addition of 8.00 mL of 0.500 M HCl
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 3. After the addition of 10.00 mL of 0.500 M HCl
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration Curve
4. After the addition of 20.00 mL of 0.500 M HCl
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration Curve
Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia
5. After the addition of 21.00 mL of 0.500 M HCl
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration CurveCalculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia
Determine the pH of a solution when 25.0 ml of 2.00 M NH3 is titrated halfway to the equivalence point with HCl
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration Curve
Determine the pH of a solution when 25.0 ml of 2.00 M NH3 is titrated to the equivalence point with 0.125 M HCl.
Neutralization ReactionWeak Base - Strong Acid
5 Positions on the Titration Curve
Titration of a Polyprotic Weak Acid
Titration of a Polyprotic Strong Acid
2nd equivalence point
1st equivalence point
37
pH Review
• pH = - log [H+]• H+ is really a proton• Range is from 0 - 14• If [H+] is high, the solution is acidic; pH < 7• If [H+] is low, the solution is basic or alkaline ;
pH > 7
38
The Body and pH
• Homeostasis of pH is tightly controlled• Extracellular fluid = 7.4• Blood = 7.35 – 7.45• < 6.8 or > 8.0 death occurs• Acidosis (acidemia) below 7.35• Alkalosis (alkalemia) above 7.45
39
40
Small changes in pH can produce major disturbances
• Most enzymes function only with narrow pH ranges
• Acid-base balance can also affect electrolytes (Na+, K+, Cl-)
• Can also affect hormones
41
Acidosis• Principal effect of acidosis is depression of the CNS
through ↓ in synaptic transmission.• Generalized weakness• Deranged CNS function the greatest threat• Severe acidosis causes
– Disorientation– coma – death
42
Alkalosis• Alkalosis causes over excitability of the central and
peripheral nervous systems.• Numbness• Lightheadedness• It can cause :
– Nervousness– muscle spasms or tetany – Convulsions – Loss of consciousness– Death
43
The function of a buffer is to resist changes in the pH of a solution.
Buffers are just a special case of the common ion effect. Buffer CompositionWeak Acid + Conj. BaseHOAc + OAc-
H2PO4- + HPO4
2-
Weak Base + Conj. AcidNH3 + NH4
+
Buffer Solutions
ACID USES UP ADDED OH-
CONJ. BASE USES UP ADDED H+
OAc- + H2O HOAc + OH-
LeChatelier's Principle
NOTE: THE CONJUGATE ACID/BASE PAIRS ARE ON OPPOSITE SIDES OF THE EQUATION.
Buffer Solutions
Preparing a BufferPreparing a BufferA buffer can be prepared by 1. Using a weak acid and ½ the moles of a strong
base or less2. Using a weak base and ½ the moles of a strong
acid or less3. A weak acid and the salt of a weak base4. A weak base and the salt of a weak acid
Preparing a BufferPreparing a Buffer Using a weak acid and ½ the moles of a strong
base or less
Why does this work? 30.0 ml of 0.200 M acetic acid and 15.0 ml of 0.200 M sodium hydroxide C2H3O2
- + OH- H C2H3O2 + H2O
I C E
Preparing a BufferPreparing a Buffer Using a weak base and ½ the moles of a strong
base or less
Why does this work? 30.0 ml of 0.200 M ammonia and 15.0 ml
of 0.200 M HCl NH3 + H2O NH4
+ + H2O
I C E
Derivative isH+ = Ka [HB] [B- ] If you have a buffer , use the aboveequation. Makes the math easier.
You have a buffer if have a conjugate acid base pair or the moles of weak (acid/base)is greater than the strong (acid/base)
Buffer SolutionspH pKa + log
[Conj. base][Acid]
Buffer SolutionsWhich of the following mixtures would result in a buffered solution when 1.0 L of eachof the two solutions are mixed?
a. 0.2 M HNO3 and 0.4 M NaNO3 b. 0.2 M HNO3 and 0.4 M HFc. 0.2 M HNO3 and 0.4 M NaFd. 0.2 M HNO3 and 0.4 M NaOH
H+ = Ka [HB] [B- ]
Buffer SolutionsHow does adding water to a buffer system affect the pH?
H+ = Ka [HB] [B- ]
Buffer SolutionsHow does adding water to a buffer system affect the pH?
CONCENTRATION of the acid and conjugate base are not important.It is the RATIO OF THE NUMBER OF MOLES of each. The volumes cancel. You can ignore the volume and use
H+ = Ka HB moles B- moles
Buffer SolutionsProblem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? Ka = 1.8 x 10-5
Adding an Acid to a BufferProblem: What is the pH when 1.00 mL of 1.00 M HCl
is added to 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M
Adding an Base to a Buffer What is the pH when 1.00 mL of 1.00 M NaOH is
added to 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M .
Buffer SolutionsHow many moles of HCl must be added to 1.0 L of 1.0 M sodium acetate to producea solution buffered ata. pH= pKab. pH =4.20
Buffer SolutionsConsider a solution that contains both ammonia and ammonium chloride. Calculate the ratio of [NH3 ]/[NH4 Cl ] at a pH of 5.23.
Preparing a BufferPreparing a BufferYou want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH pKa).
—then you get the exact [H3O+] by adjusting the ratio of acid to conjugate base.
Preparing a Buffer SolutionBuffer prepared fromHCO3
- weak acid
CO32- conjugate base
HCO3- + H2O H3O+ + CO3
2-
Preparing a BufferYou want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M
POSSIBLE ACIDS Ka
HSO4- / SO4
2- 1.2 x 10-2
HOAc / OAc- 1.8 x 10-5
HCN / CN- 4.0 x 10-10
Best choice is acetic acid / acetate.
Remember pKa = -log10 Ka
pH = pKa when moles acid = moles of base
You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M
Preparing a Buffer
You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M
[H3O] 5.0 x 10-5 = [HOAc]
[OAc- ] (1.8 x 10-5 )
Preparing a Buffer
You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M
Solve for [HOAc]/[OAc-] ratio = 2.78/ 1Therefore, if you use 0.100 mol of NaOAc and 0.278
mol of HOAc, you will have pH = 4.30.
[H3O] 5.0 x 10-5 = [HOAc]
[OAc- ] (1.8 x 10-5 )
Preparing a Buffer
Buffer capacity
The buffer capacity is the measure of the
amount of acid or base that the solution can
absorb without significant change in pH
Depend on how many moles of the conjugate
acid-base pair are present
- for solutions having the same concentration,
the greater the volume greater buffer
capacity
Buffer capacity
To increase the buffering capacity of a buffer
system, increase the moles of acid and moles of base
• CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each.
Buffer Capacity And Buffer Range
• There is a limit to the capacity of a buffer solution to neutralize added acid or base, and this limit is reached before all of one of the buffer components has been consumed.
• In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize.
• As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal (pH=pKa).
Acid-Base Indicators
• An acid-base indicator is a weak acid having one color and the conjugate base of the acid having a different color. One of the “colors” may be colorless.
HIn + H2O º H3O+ + In-
color 1 color 2• Acid-base indicators are often used for
applications in which a precise pH reading isn’t necessary.
• A common indicator used in introductory chemistry laboratories is litmus.
17-3 Acid-Base Indicators
• Color of some substances depends on the pH.
HIn + H2O In- + H3O+
In the acid form the color appears to be the acid color.In the base form the color appears to be the base color.Intermediate color is seen in between these two states.The complete color change occurs over about 2 pH units.
Remember….LeChatelier’s Principle
Slide 69 of 45
Indicator Colors and Ranges
Slide 27 of 45 General Chemistry: Chapter 17 Prentice-Hall © 2007
Titration Curve ForStrong Acid - Strong Base
Acid-Base Titrations
Indicators should change color close to the equivalence point.