Transcript weak acid – strong base - Clayton State University

PRINCIPLES OF CHEMISTRY II
CHEM 1212
CHAPTER 16
DR. AUGUSTINE OFORI AGYEMAN
Assistant professor of chemistry
Department of natural sciences
Clayton state university
CHAPTER 16
REACTIONS BETWEEN
ACIDS AND BASES
VOLUMETRIC ANALYSIS
- Analysis by volume
Titration
- Increments of a known reagent solution (the titrant) are added
to an unknown solution (the analyte) until the reaction is complete
Titrant
- Usually in the buret
Analyte
- Usually in an erlenmeyer flask
VOLUMETRIC ANALYSIS
Common Titrations
- Acid-Base
- Oxidation-Reduction
- Complex Formation (Complexometric)
- Precipitation Reactions
Methods of Determining Analyte Consumption
- Color change
- Absorbance of light
- Sudden change in voltage or current between a pair of electrodes
VOLUMETRIC ANALYSIS
Equivalence Point
- The quantity of titrant added is the exact amount necessary for
stoichiometric reaction with analyte
End Point
- The quantity of titrant actually measured in an experiment
- Ideal end point is the equivalence point
TITRATIONS OF STRONG ACID – STRONG BASE
- Write balanced chemical equation between titrant and analyte
- Use a pH meter to record the pH after each addition of titrant
Or
- Calculate composition and pH after each addition of titrant
- Construct a graph of pH versus titrant added
TITRATIONS OF STRONG ACID – STRONG BASE
- Consider titration of base with acid
H+ + OH- → H2O
K = 1/Kw = 1/10-14 = 1014
- Equilibrium constant = 1014
- Reaction goes to completion
TITRATIONS OF STRONG ACID – STRONG BASE
H+ + OH- → H2O
At equivalence point
moles of titrant = moles of analyte
(V titrant)(M titrant) = (V analyte)(M analyte)
TITRATIONS OF STRONG ACID – STRONG BASE
Consider 50.00 mL of 0.100 M NaOH with 0.100 M HCl
- Three regions of titration curve exists
- Before the equivalence point where pH is determined by
excess OH- in the solution
- At the equivalence point where pH is determined by
dissociation of water (H+ ≈ OH-)
- After the equivalence point where pH is determined by
excess H+ in the solution
TITRATIONS OF STRONG ACID – STRONG BASE
First calculate the volume of HCl needed to
reach the equivalence point
(V HCl)(0.100 M) = (50.00 mL)(0.100 M)
Volume HCl = 50.00 mL
The sRfc table may be used as described in the textbook
TITRATIONS OF STRONG ACID – STRONG BASE
Before the equivalence point
Initial amount of analyte (NaOH)
= 50.00 mL x 0.100 M = 5.00 mmol
After adding 1.00 mL of HCl
mmol H+ added = mmol OH- consumed
mmol H+ = (1.00 mL)(0.100 M) = 0.100 mmol
mmol OH- remaining = 5.00 – 0.100 = 4.90 mmol
TITRATIONS OF STRONG ACID – STRONG BASE
Before the equivalence point
Total volume = 50.00 mL + 1.00 mL = 51.00 mL
[OH-] = 4.90 mmol/51.00 mL = 0.0961 M
pOH = - log(0.0961) = 1.017
pH = 14.000 - 1.017 = 12.983
TITRATIONS OF STRONG ACID – STRONG BASE
- Repeat calculations for all volumes added
- Increments can be large initially but must be reduced just before
and just after the equivalence point (around 50.00 mL in this case)
- Sudden change in pH occurs near the equivalence point
- Greatest slope at the equivalence point
TITRATIONS OF STRONG ACID – STRONG BASE
At the equivalence point
- pH is determined by the dissociation of water
H2O ↔ H+ + OHKw = x2 = 1.0 x 10-14
x = 1.0 x 10-7
pH = 7.00 (at 25 oC)
TITRATIONS OF STRONG ACID – STRONG BASE
After the equivalence point
- Excess H+ is present
After adding 51.00 mL of HCl
Excess HCl present = 51.00 – 50.00 = 1.00 mL
Excess H+ = (1.00 mL)(0.100 M) = 0.100 mmol
Total volume of solution = 50.00 + 51.00 = 101.00 mL
[H+] = 0.100 mmol/101.00 mL = 9.90 x 10-4 M
pH = -log(9.90 x 10-4) = 3.004
pH
TITRATION CURVE
Equivalence point
(maximum slope or point of inflection)
7
50.00
Volume of HCl added (mL)
pH
TITRATION CURVE
7
Equivalence point
(maximum slope or point
of inflection)
50.00
Volume of NaOH added (mL)
STOICHIOMETRY AND TITRATION CURVE
Compare titration of HCl with NaOH and H2SO4 with NaOH
(same volume and same concentration of acid)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Net ionic equation in both cases:
H+(aq) + OH-(aq) → H2O(l)
or
H3O+(aq) + OH-(aq) → 2H2O(l)
1 mol HCl conctributes 1 mol H3O+
1 mol H2SO4 contributes 2 mol H3O+
pH
STOICHIOMETRY AND TITRATION CURVE
HCl
H2SO4
Volume of NaOH added (mL)
BUFFER SOLUTION
- A mixture of a conjugate acid-base pair
- Tends to resist changes in pH upon addition of an acid or a base
- The resistive action is the result of equilibrium between the
weak acid (HA) and its conjugate base (A-)
HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
- Commonly used in biological systems
- Enzyme-catalyzed reactions depend on pH
BUFFER SOLUTION
Examples
HC2H3O2/C2H3O2-
HF/FNH3/NH4+
H2CO3/HCO3-
BUFFER SOLUTION
- When an acid is added, the conjugate base converts the excess
H3O+ ion into its acid (conjugate base removes excess H3O+)
H3O+(aq) + A-(aq) → HA(aq) + H2O(l)
- When a base is added, the acid converts the excess OH- ion into
its conjugate base and water (acid removes excess OH- ion)
HA(aq) + OH-(aq) → A-(aq) + H2O(l)
- These reactions go to completion (large equilibrium constants)
BUFFER SOLUTION
- In actual fact, the pH changes but very slightly
- Large amounts of added H3O+ or OH- may overcome the
buffer action and change pH of solutions
- Buffers are most effective when the
ratio of acid to conjugate base is 1:1
- Buffers are less efficient in handling acids if the
acid is more than the conjugate base
- Buffers are less efficient in handling bases if the
acid is less than the conjugate base
HENDERSON-HASSELBALCH
EQUATION
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
[H 3O ][A  ]
Ka 
[HA]
-If [HA] = [A-], then Ka = [H3O+]
Taking the negative logarithm on both sides gives
- logKa = - log [H3O+]
pKa = pH
HENDERSON-HASSELBALCH
EQUATION
- If [HA] ≠ [A-]
 [A - ] 

pH  pK a  log 
 [HA] 
or
 [B] 
pH  pK a  log 


 [BH ] 
- pH changes by 1 if the ratio changes by a factor of 10
HENDERSON-HASSELBALCH
EQUATION
In general
 Cb 
pH  pK a  log  
 Ca 
or
 nb 
pH  pK a  log  
 na 
- Ca and Cb are the analytical concentrations of the acid
and the conjugate base, respectively
- na and nb are the number of moles of the acid
and the conjugate base, respectively
Ca = na/V and Cb = nb/V (concentration ratio equals mole ratio)
HENDERSON-HASSELBALCH
EQUATION
Addition of a Strong Acid
- Some buffer base is converted to the conjugate acid
Addition of a Strong Base
- Some buffer acid is converted to the conjugate base
PREPARING BUFFERS
- Measure the amount of weak acid (HA) to be used
[or weak base (B)]
- Calculate the amount of strong base (OH-) to be added
[or strong acid (H+)]
- This makes a mixture of HA and A- which is a buffer
[or B and BH+]
Or
- Add the correct proportions of the acid and its conjugate base,
then check the pH
PREPARING BUFFERS
Prepare 1.00 L 0f 0.100 M tris buffer solution at pH 8.40
- Convert mole to gram tris hydrochloride (MM = 157.60 g/mol)
- Weigh out sample and dissolve in a beaker with ~ 800 mL H2O
- Add NaOH solution until pH is exactly 8.40 (continuous stirring)
- Quantitatively transfer solution to a 1000 mL volumetric flask
- Dilute to the mark and mix
BUFFER CAPACITY
- A measure of how well a solution resists changes in pH
- Increases with increasing concentration of buffer
- Maximum when pH = pKa
- The greater the buffer capacity the less the pH changes
upon addition of H+ or OH-
- Choose a buffer whose pKa is closest to the desired pH
- pH should be within pKa ± 1
WEAK ACID – STRONG BASE
Consider 50.00 mL 0f 0.0100 M acetic acid with 0.100 M NaOH
pKa of acetic acid = 4.76
HC2H3O2 + OH- → C2H3O2- + H2O
For the reverse reaction
Kb = Kw/Ka = 5.8 x 10-10
Equilibrium constant = 1/Kb = 1.7 x 109
So large that we can assume the reaction goes to completion
WEAK ACID – STRONG BASE
Determine volume of base at equivalence point
mmol HC2H3O2 ≈ mmol OH-
(V NaOH)(0.100 M) = (50.00 mL)(0.0100 M)
Volume NaOH = 5.00 mL
WEAK ACID – STRONG BASE
Four types of calculations to be considered
Before OH- is added
- pH is determined by equilibrium of weak acid
HA
↔
H+ + A-
[x 2 ]
[x 2 ]
Ka 

[F - x]
[0.0100 - x]
x = 4.1 x 10-4
pH = 3.39
WEAK ACID – STRONG BASE
Before equivalence point
By adding OH- a buffer solution of HA and A- is formed
After adding 0.100 mL OHHA + OH- → A- + H2O
Initial mmol
Final mmol
0.500
0.490
0.0100
0
0
0.0100
WEAK ACID – STRONG BASE
Before equivalence point
 [A - ] 

pH  pK a  log 
 [HA] 
 [0.0100] 
  3.07
pH  4.76  log 
 [0.490] 
WEAK ACID – STRONG BASE
Before equivalence point
If volume of OH- added is half the volume at equivalence point
HA = A- = 0.250 mmol
 [0.250] 

pH  4.76  log 
 [0.250] 
pH = pKa = 4.76
WEAK ACID – STRONG BASE
At equivalence point
Volume of OH- = 5.00 mL
mmol OH- = (5.00 mL)(0.100 M) = 0.500 mmol
HA is used up and [HA] = 0
mmol A- = 0.500 mmol
WEAK ACID – STRONG BASE
At equivalence point
Only A- is present in solution
A- + H2O ↔ HA + OH[A-] = (0.500 mmol)/(50.00 mL + 5.00 mL) = 0.00909 M
WEAK ACID – STRONG BASE
At equivalence point
[x 2 ]
Kb 
[F - x]
Kb = Kw/Ka = 5.8 x 10-10
x = [OH-] = 2.3 x 10-6
pOH = 5.64
pH = 14.00 – 5.64 = 8.36
pH is not 7.00 but greater than 7.00
(pH at equivalence point increases with decreasing strength of acid)
WEAK ACID – STRONG BASE
After equivalence point
- Strong base (OH-) being added to weak base (A-)
- pH is determined by the excess [OH-] (approximation)
- After adding 5.10 mL OH[OH-] = (0.10 mL)(0.100 M)/(50.00 mL + 5.10 mL)
= 1.81 x 10-4
pH = 14.00 – pOH = 14.00 – 3.74 = 10.26
WEAK ACID – STRONG BASE
pH
Minimum slope
8.36
pH = pKa
Equivalence point
(maximum slope or point of inflection)
pKa
2.50
5.00
Volume of NaOH added (mL)
STRONG ACID – WEAK BASE
- The reverse of weak acid and strong base
B + H+ → BH+
- Similarly assume reaction goes to completion
STRONG ACID – WEAK BASE
Consider 50.00 mL of 0.0100 M pyridine with 0.100 M HCl
Kb of pyridine = 1.6 x 10-9
Determine volume of acid at equivalence point
mmol pyridine ≈ mmol H+
(V HCl)(0.100 M) = (50.00 mL)(0.010 M)
Volume HCl = 5.00 mL
STRONG ACID – WEAK BASE
Four types of calculations to be considered
Before H+ is added
- pH is determined by equilibrium of weak base
(determined using Kb)
B + H2O ↔ BH+ + OH-
STRONG ACID – WEAK BASE
Before equivalence point
- By adding H+ a buffer solution of B and BH+ is formed
 [B] 
pH  pK a  log 


 [BH ] 
- When volume of H+ added = half the volume at equivalent point
pH = pKa (for BH+)
STRONG ACID – WEAK BASE
At equivalence point
- B has been converted into BH+
- B is used up and [B] = 0
pH is calculated by considering BH+
BH+ ↔ B + H+
pH is not 7.00 but less than 7.00
STRONG ACID – WEAK BASE
After equivalence point
- Strong acid (H+) is being added to weak acid (BH+)
pH is determined by the excess [H+] (approximation)
STRONG ACID – WEAK BASE
Minimum slope
pKa
pH
pH = pKa
Equivalence point
(maximum slope or point of inflection)
2.50
5.00
Volume of HCl added (mL)
ESTIMATING THE pH OF MIXTURES
Chemical System
pH Range
pH Estimate
Strong Acid
0–2
1
Weak Acid
2–5
3
Buffer
3–6
4
Neutral
7
7
Weak base
8 – 11
10
Strong Base
12 – 14
13
END POINT
Use of Indicators
- Indicators are acids or bases so a few drops of
dilute solutions are used to minimize indicator errors
- Acidic color if pH ≤ pKHIn - 1
- Basic color if pH ≥ pKHIn + 1
- A mixture of both colors if pKHIn - 1 ≤ pH ≤ pKHIn + 1
- Use an indicator whose transition range overlaps the steepest
part of the titration curve
END POINT
Use of pH Electrodes
- The end point is where the slope of the curve is greatest
- The end point is the volume at which the first
derivative of a titration curve is maximum
- The end point is the volume at which the second
derivative of a titration curve is zero
INDICATORS
- Acid-base indicators are highly colored weak acids or bases
- The various protonated forms have different colors
- Have very low concentrations in order not to
interfere with analytes
INDICATORS
HIn + H2O ↔ H3O+ + In- Indicators exist predominantly in the protonated
form (HIn) in acidic solutions
- Indicators exist predominantly in the deprotonated
form (In-) in basic solutions
- The result is color change
 [In - ] 

pH  pK HIn  log 
 [HIn] 
POLYPROTIC ACIDS
- Have more than one acidic proton
- Produce more than one proton per molecule when ionized
Examples
Phosphoric acid (H3PO4)
Carbonic acid (H2CO3)
Sulfuric acid (H2SO4)
Amino acids
POLYPROTIC ACIDS
Amino Acids
- Building blocks of proteins
- Have acidic carboxylic acid group and basic amino group
- The acidic proton resides on the N of the amino group
- Have positive site (amino group) and negative site (acid group)
- Called zwitterion
- Both groups are protonated at low pH and depotonated at high pH
DIPROTIC SYSTEMS
- Contain two acidic protons
H2A ↔ HA- + H+
HA- ↔ A2- + H+
(Ka1)
(Ka2)
- Acid dissociation constants: Ka1 > Ka2
A2- + H2O ↔ HA- + OHHA- + H2O ↔ H2A + OH-
(Kb1)
(Kb2)
- Base association constants: Kb1 > Kb2
DIPROTIC SYSTEMS
H2A ↔ HA- + H+ (Ka1)
+
HA- + H2O ↔ H2A + OH- (Kb2)
=
H2O ↔ H+ + OHKa1 x Kb2 = Kw
Ka2 x Kb1 = Kw
DIPROTIC SYSTEMS
If Ka1 >> Ka2
- A solution of a diprotic acid behaves like a solution of a
monoprotic acid with Ka = Ka1
If Kb1 >> Kb2
- The fully basic form of a diprotic acid can be
considered as monobasic with Kb = Kb1
DIPROTIC SYSTEMS
Amphoteric Substances
- Substances that can react as either an acid or a base
Amphiprotic Substances
- Substances that can either donate or accept a proton
DIPROTIC SYSTEMS
The Intermediate Form
- Is both an acid and a base
- Can donate or accept a proton
- Called amphiprotic
pH 
1
pK a1  pK a2 
2
TITRATION CURVES
pH
Diprotic acids (two equivalence points)
H2A/HA-
HA-/A2-
Excess OH-
pKa2
pKa1
volume of OH- added
TRIPROTIC SYSTEMS
Ka1 x Kb3 = Kw
Ka2 x Kb2 = Kw
Ka3 x Kb1 = Kw
First Intermediate (H2A-)
Second Intermediate (HA2-)
1
pH  pK a1  pK a2 
2
1
pH  pK a2  pK a3 
2
FACTORS AFFECTING SOLUBILITY
- Solubility is influenced by the acid-base properties of the
constituent cations and anions
- Solubility is influenced by pH if ions react with H+ or OH-
Simultaneous Equilibria
Solubility equilibrium and acid-base reaction
occur at the same time
FACTORS AFFECTING SOLUBILITY
Salts of Anions of Weak Acids
- More soluble than predicted by Ksp calculations
if the anion reacts with H+
- Anion produced from dissociation is consumed by
H3O+ causing additional salt to dissociate
(Le Chatelier’s principle)
MXn(s) ↔ Mn+(aq) + nX-(aq)
X-(aq) + H3O+(aq) ↔ HX(aq) + H2O(aq)
where M is a metal
FACTORS AFFECTING SOLUBILITY
Salts of Transition-Metal Cations
- Transition-metal ions react with substances that donate electrons
- Species donating electrons is known as a ligands
(NH3, H2O, OH-)
- These species are Lewis bases
- Free metal ion is consumed by complex formation
causing additional salt to dissociate