acids and bases
DESCRIPTION
Acids and Bases. L. Scheffler Lincoln High School. Acids and Bases. The concepts acids and bases were loosely defined as substances that change some properties of water. One of the criteria that was often used was taste. Substances were classified salty-tasting sour-tasting - PowerPoint PPT PresentationTRANSCRIPT
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Acids and Bases The concepts acids and bases were loosely defined as
substances that change some properties of water. One of the criteria that was often used was taste. Substances were classified
salty-tasting sour-tasting sweet-tasting bitter-tasting
Sour-tasting substances would give rise to the word 'acid', which is derived from the Greek word oxein, which mutated into the Latin verb acere, which means 'to make sour'
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Acids• React with certain metals to produce
hydrogen gas.• React with carbonates and bicarbonates
to produce carbon • dioxide gas• Vinegar is a solution of acetic acid.
Citrus fruits contain citric acid.
• Have a bitter taste• Feel slippery. • Many soaps contain bases.
Bases
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Properties of Acids Produce H+ (as H3O+) ions in water (the hydronium ion
is a hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
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Properties of Bases Generally produce OHGenerally produce OH-- ions in water ions in water
Taste bitter, chalkyTaste bitter, chalky
Are electrolytesAre electrolytes
Feel soapy, slipperyFeel soapy, slippery
React with acids to form salts and waterReact with acids to form salts and water
pH greater than 7pH greater than 7
Turns red litmus paper to blueTurns red litmus paper to blue “ “BBasicasic BBluelue””
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Arrhenius DefinitionArrhenius
Acid - Substances in water that increase the concentration of hydrogen ions (H+).
Base - Substances in water that increase concentration of hydroxide ions (OH-).
Categorical definition – easy to sort substances into acids and bases
Problem – many bases do not actually contain hydroxides
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Bronsted-Lowry Definition
Acid - neutral molecule, anion, or cation that donates a proton.
Base - neutral molecule, anion, or cation that accepts a proton. HA + :B HB+ + :A-
Ex HCl + H2O H3O+ + Cl-
Acid Base Conj Acid Conj Base
Operational definition - The classification depends on how the substance behaves in a chemical reaction
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Conjugate Base - The species remaining after an acid has transferred its proton.
Conjugate Acid - The species produced after base has accepted a proton.
HA & :A- - conjugate acid/base pair
:A- - conjugate base of acid HA
:B & HB+ - conjugate acid/base pair
HB+ - conjugate acid of base :B
Conjugate Acid Base Pairs
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Note: Water can act as acid or base
Acid Base Conjugate Acid Conjugate Base
HCl + H2O H3O+ + Cl-
H2PO4- + H2O
H3O+ + HPO4
2-
NH4+ + H2O
H3O+ + NH3
Base Acid Conjugate Acid Conjugate Base :NH3 + H2O
NH4+ + OH-
PO43- + H2O
HPO42- + OH-
Examples of Bronsted-Lowry Acid Base Systems
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pH and acidity1. Acidity or Acid Strength depends on Hydronium Ion
Concentration [H3O+]
2. The pH system is a logarithmic representation of the Hydrogen Ion concentration (or OH-) as a means of avoiding using large numbers and powers.
pH = - log [H3O+] = log(1 / [H3O+])
pOH = - log [OH-] = log(1 / OH-])
3. In pure water [H3O+] = 1 x 10-7 mol / L (at 20oC)
pH = - log(1 x 10-7) = - (0 - 7) = 7
4. pH range of solutions: 0 - 14
pH < 7 (Acidic) [H3O+] > 1 x 10-7 m / L
pH > 7 (Basic) [H3O+] < 1 x 10-7 m / L
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Calculating the pHpH = - log [H3O+]
Example 1: If [H3O+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example 2: If [H3O+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
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pH and acidity
The pH values of several common substances are shown at the right.
Many common foods are weak acids
Some medicines and many household cleaners are bases.
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Neutralization An acid will neutralize a base,
giving a salt and water as products
Examples HCl + NaOH NaCl + H2O
H2SO4 + 2 NaOH Na2SO4 + 2 H2O
H3PO4 + 3 KOH K3PO4 + 3 H2O
2 HCl + Ca(OH) 2 CaCl2 + 2 H2O
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Neutralization Calculations If the concentration of acid or base is
expressed in Molarity or mol dm-3
then:--The volume in dm3 multiplied by the
concentration yields moles. -- If the volume is expressed in cm3
the same product yields millimoles
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Neutralization Problems If an acid and a base combine in a 1 to 1 ratio, then
the volume of the acid multiplied by the concentration of the acid is equal to the volume of the base multiplied by the concentration of the base
Vacid C acid = V base C base
If any three of the variables are known it is possible to determine the fourth
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Neutralization Problems Example 1: Hydrochloric acid reacts with potassium hydroxide according to the following reaction:
HCl + KOH KCl + H2O If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of
KOH solution, what is the concentration of the KOH solution?
Solution:Vacid Cacid = Vbase Cbase
(15.00 cm3 )(0.500 M) = (24.00 cm3 ) Cbase
Cbase = (15.00 cm3 )(0.500 M) (24.00 cm3 ) Cbase = 0.313 M
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Neutralization Problems
Whenever an acid and a base do not combine in a 1 to 1 ratio, a mole factor must be added to the neutralization equation
n Vacid C acid = V base C base
The mole factor (n) is the number of times the moles the acid side of the above equation must be multiplied so as to equal the base side. (or vice versa)
Example
H2SO4 + 2 NaOH Na2SO4 + 2 H2O
The mole factor is 2 and goes on the acid side of the equation. The number of moles of H2SO4 is one half that of NaOH. Therefore the moles of H2SO4 are multiplied by 2 to equal the moles of NaOH.
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Neutralization Problems Example 3: Phosphoric acid reacts with potassium hydroxide according to the following reaction:
H3PO4 + 3 KOH K3PO4 + 3 H2O If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of
H3PO4 solution, what is the concentration of the H3PO4 solution?
Solution:In this case the mole factor is 3 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore
3 Vacid Cacid = Vbase Cbase
3 (15.00 cm3 )(Cacid) = (30.00 cm3 ) (0.300 M) Cacid = (30.00 cm3 )(0.300 M) (3) (15.00 cm3 ) Cacid = 0.200 M
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Neutralization Problems Example 2: Sulfuric acid reacts with sodium hydroxide according to the following reaction:
H2SO4 + 2 NaOH Na2SO4 + 2 H2O If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of
NaOH solution, what is the concentration of the NaOH solution?
Solution:In this case the mole factor is 2 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore
2 Vacid Cacid = Vbase Cbase
2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase
Cbase = (2) (20.00 cm3 )(0.400 M) (32.00 cm3 )
Cbase = 0.500 M
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Neutralization Problems Example 4: Hydrochloric acid reacts with calcium hydroxide according to the following reaction:
2 HCl + Ca(OH)2 CaCl2 + 2 H2O If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of
Ca(OH)2 solution, what is the concentration of the Ca(OH)2 solution?
Solution:In this case the mole factor is 2 and it goes on the base side, since the mole ratio of acid to base is 2 to 1. Therefore
Vacid Cacid = 2 Vbase Cbase
(25.00 cm3) (0.400) = (2) (20.00 cm3) (Cbase)
Cbase = (25.00 cm3 ) (0.400 M) (2) (20.00 cm3 )
Cbase = 0.250 M
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Acid Base DissociationAcid-base reactions are equilibrium processes.
The relationship between the relative concentrations of the reactants and products is a constant for a given temperature. It is known as the Acid or Base Dissociation Constant.
The stronger the acid or base, the larger the value of the dissociation constant.
]B[:
][OH [HB] K O][H K
[HA]
][H ]A[: K O][H K
constant. is solutions dilute in O][H
][H OH
:Note
O][H ]B:[
]OH][HB[ K
O][H HA][
]OH][ A[: K
waterin base a For waterin acid an For
-
-
b2eq
-
a2eq
2
3
2-
-
eq2
3-
eq
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Acid Strength Strong Acid - Transfers all of its protons to water;
- Completely ionized; - Strong electrolyte; - The conjugate base is weaker and has a negligible tendency to be protonated.
Weak Acid - Transfers only a fraction of its protons to water;
- Partly ionized; - Weak electrolyte; - The conjugate base is stronger, readily accepting protons from water
As acid strength decreases, base strength increases. The stronger the acid, the weaker its conjugate base The weaker the acid, the stronger its conjugate base
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Base Strength Strong Base - all molecules accept a proton; - completely ionizes; - strong electrolyte; - conjugate acid is very weak, negligible tendency to donate protons.
Weak Base - fraction of molecules accept proton; - partly ionized; - weak electrolyte; - the conjugate acid is stronger. It more readily donates protons.
As base strength decreases, acid strength increases. The stronger the base, the weaker its conjugate acid. The weaker the base the stronger its conjugate acid.
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Common Strong Acids/Bases
Strong BasesStrong BasesSodium Hydroxide
Potassium Hydroxide
*Barium Hydroxide
*Calcium Hydroxide
*While strong bases they are not very soluble
Strong AcidsStrong AcidsHydrochloric Acid
Nitric Acid
Sulfuric Acid
Perchloric Acid
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Water has the ability to act as either a Bronsted- Lowry acid or base.
Autoionization – spontaneous formation of low concentrations of [H+] and OH-] ions by proton transfer from one molecule to another.
Equilibrium Constant for Water
7--
o14--w
o14--3w
-3
22c
22
-3
c
10 x 0.1 ][OH ][H
: WaterPure In
C)25(at 10 x 0.1 ][OH ][H K
C)25(at 10 x 0.1 ][OH ]O[H K
][OH ]O[H O]H[K
O]H[
][OH ]O[H K
Water as an Equilibrium System
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Weak Acid EquilibriaA weak acid is only partially ionized.Both the ion form and the unionized form exist at equilibrium HA + H2O H3O+ + A-
The acid equilibrium constant is
Ka = [H3O+ ] [A-] [HA]
Ka values are relatively small for most weak acids. The greatest part of the weak acid is in the unionized form
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Weak Acid Equilibrium Constants
Sample problem . A certain weak acid dissociates in water as follows: HA + H2O H3O+ + A-
If the initial concentration of HA is 1.5 M and the equilibrium concentration of H3O+ is 0.0014 M. Calculate Ka for this acidSolutionKa = [H3O+ ] [A-]
[HA] I C E Substituting
[HA] 1.5 -x 1.5-x Ka = (0.0014)2 = 1.31 x 10-6 [A-] 0 +x x 1.4986[H3O+ ] 0 +x x
x = 0.0014 1.5-x = 1.4986
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Weak Base EquilibriaWeak bases, like weak acids, are partially ionized. The degree to which ionization occurs depends on the value of the base dissociation constantGeneral form: B + H2O BH+ + OH-
Kb = [BH+][OH-] [B]
ExampleNH3 + H2O NH4
+ + OH-
Kb = [NH4+][ OH-]
[NH3]
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Weak Base Equilibrium Constants
Sample problem . A certain weak base dissociates in water as follows: B + H2O BH+ + OH-
If the initial concentration of B is 1.2 M and the equilibrium concentration of OH- is 0.0011 M. Calculate Kb for this baseSolutionKb = [BH+ ] [OH-]
[B]
I C E Substituting[B] 1.2 -x 1.2-x Kb = (0.0011)2 = 1.01 x 10-6 [OH-] 0 +x x 1.1989[BH+ ] 0 +x x
x = 0.0011 1.2-x = 1.1989
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Weak Acid Equilibria Concentration Problems
Problem 1. A certain weak acid dissociates in water as follows: HA + H2O H3O+ + A-
The Ka for this acid is 2.0 x 10-6. Calculate the [HA] [A-], [H3O+ ] and pH of a 2.0 M solutionSolutionKa = [H3O+ ] [A-] = 2.0 x 10-6
[HA] I C E Substituting
[HA] 2.0 -x 2.0-x Ka = x2 = 2.0 x 10-6
[A-] 0 +x x 2.0-x[H3O+ ] 0 +x x If x <<< 2.0 it can be dropped
from the denominator
The x2 = (2.0 x10-6)(2.0) = 4.0 x10-6 x = 2.0 x 10-3
[A-] = [H3O+ ] = 2.0 x10-3 [HA] = 2.0 - 0.002= 1.998
pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7
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Weak Acid Equilibria Concentration Problems
Problem 2. Acetic acid is a weak acid that dissociates in water as follows: CH3COOH + H2O H3O+ + CH3COO-
The Ka for this acid is 1.8 x 10-5. Calculate the [CH3COOH],[CH3COO-] [H3O+ ] and pH of a 0.100 M solutionSolution
Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5
[CH3COOH] I C E Substituting
[CH3COOH] 0.100 -x 0.100-x Ka = x2 = 1.8 x 10-5
[CH3COO- ] 0 +x x 0.100-x[H3O+] 0 +x x If x <<< 0.100 it can be dropped
from the denominatorThe x2 = (1.8 x10-5)(0.100) = 1.8 x10-6 x = 1.3 x 10-3
[CH3COO--] = [H3O+ ] = 1.3 x10-3 [CH3COOH ] = 0.100 - 0.0013 = 0.0987
pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88
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Weak Base EquilibriaExample1. Ammonia dissociates in water according to the following equilibrium
NH3 + H2O NH4+ + OH-
Kb = [NH4+][ OH-] = 1.8 x 10-5
[NH3]Calculate the concentration of [NH4
+][ OH-] [NH3 ]and the pH of a 2.0M solution.
I C E Substituting[NH3] 2.0 -x 2.0-x Kb = x2 = 1.8x 10-5
[OH-] 0 +x x 2.0-x[NH4
+] 0 +x x If x <<< 2.0 it can be dropped from the denominator
The x2 = (1.8 x10-5)(2.0) = 3.6 x10-5 x = 6.0 x 10-3 [OH-] = [NH4
+] = 6.0 x10-3 [NH3] = 2.0- 0.006= 1.994pOH = - log [OH-] =-log (6.0 x10-3) = 2.22pH = 14-pOH = 14-2.22 = 11.78
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Amphoteric Solutions A chemical compound able to react with both an
acid or a base is amphoteric. Water is amphoteric. The two acid-base couples
of water are H3O+/H2O and H2O/OH-
It behaves sometimes like an acid, for example
And sometimes like a base :
Hydrogen carbonate ion HCO3- is also amphoteric,
it belongs to the two acid-base couples H2CO3/HCO3
- and HCO3-/CO3
2-
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Common Ion EffectThe common ion effect is a consequence of Le Chatelier’s Principle When the salt with the anion (i.e. the conjugate base) of a weak acid is added to that acid,
It reverses the dissociation of the acid.Lowers the percent dissociation of the
acid.A similar process happens when the salt with the cation (i.e, conjugate acid) is added to a weak base.These solutions are known as Buffer Solutions.
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Buffer Solutions - Characteristics A solution that resists a change in pH.
It is pH stable. A weak acid and its conjugate base
form an acid buffer. A weak base and its conjugate acid
form a base buffer. We can make a buffer of any pH by
varying the concentrations of the acid/base and its conjugate.
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Buffer Solution CalculationsCalculate the pH of a solution that is 0.50 M CH3COOH and 0.25 M NaCH3COO.
CH3COOH + H2O H3O+ + CH3COO- (Ka = 1.8 x 10-5)
Solution
Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5
[CH3COOH] I C E . Substituting
[CH3COOH] 0.50 -x 0.50-x Ka = x (0.25+x) = 1.8 x 10-5
[CH3COO-] 0.25 +x 0.25+x (0.50-x)[H3O+] 0 +x x If x <<< 0.25 it can be dropped from both
expressions in ( ) since adding or subtracting a small amount will not significantly change the value of the ratio
Then the expression becomes x(0.25)/(0.50) = 1.8 x 10-5
x = 3.6 x 10-5 = [H3O+]
pH = - log [H3O+] =-log(3.6 x 10-5 ) = 4.44
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Buffer Solutions - Equations
1. Ka = [H3O+] [A-] [HA]
2. [H3O+] = Ka [HA] [A-]
The [H3O+] depends on the ratio [HA]/[A-] Taking the negative log of both sides of equation
2 above
pH = -log(Ka [HA]/[A-])
pH = -log(Ka) - log([HA]/[A-])
pH = pKa + log([A-]/[HA])
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Henderson Hasselbach Equation
pH = pKa + log([A-]/[HA]) pH = pKa + log(Base/Acid) This expression is known as the Henderson-
Hasselbach equation. It provides a shortcut from using the I.C.E. model for buffer solutions where the concentration of both [A-] and [HA] are significantly greater than zero.
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Using the Henderson -Hasselbach Equation
pH = pKa + log([A-]/[HA]) Example
Calculate the pH of the following of a mixture that contains 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
HC3H5O3 + H2O H3O+ + C3H5O3-
Solution
Using the Henderson-Hasselbach equation
pH = - log (1.4 x 10-4) + log ( 0.25/0.75 )
= 3.85 + (-0.477) = 3.37
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Henderson-Hasselbach Equation and Base BuffersFor a base a similar expression can be written
pOH = pKb + log ([BH+] / [B])
pOH = pKb + log ([Acid] / [Base])
Example: Calculate the pH of a solution that contains 0.25 M NH3 and 0.40 M NH4Cl
(Kb = 1.8 x 10-5)
Solution
pOH = - log(1.8 x 10-5) + log (0.40/0.25)
= 4.74 + 0.204 = 4.94
pH = 14 - pOH = 14 - 4.94 = 9.06
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Henderson-Hasselbach Equation & Base Buffers Methyl amine is a weak base with a Kb or 4.38 x 10-4
CH3NH2 + H2O CH3NH3+ + OH-
Calculate the pH of a solution that is 0.10 M in methyl amine and 0.20 M in methylamine hydrochloride.
pOH = pKb + log ([BH+] / [B])
Solution pOH = -log (4.38 x 10-4) + log (0.20 / 0.10)
= 3.36 + 0.30 = 3.66
pH = 14- 3.66 = 10.34
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Additional Buffer ProblemsHow many grams of sodium formate, NaCHOO, would have to
be dissolved in 1.0 dm3 of 0.12 M formic acid, CHOOH, to
make the solution a buffer of pH 3.80? Ka= 1.78 x 10-4
pH = pKa + Log ([A-]/[HA])
Solution 3.80 = -log (1.78 x 10-4) + Log [A-] - Log [0.12]
3.80 = 3.75 + Log [A-] - (-0.92)
Log [A-] = 3.80 - 3.75 - 0.92 = - 0.87
[A-] = 10-0.87 = 0.135 mol dm-3
The molar mass of NaCHOO = 23+12+1+2(16) = 58.0 gmol-1
So (0.135 mol dm-3)(58.0 gmol-1 ) = 7.8 grams per dm-3
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Relationship of Ka, Kb & Kw HA weak acid. Its acid ionization is
A- is the conjugate base Its base ionization is
Multiplying Ka and Kb and canceling like terms
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Titration Curves A graph showing pH vs volume of acid
or base added The pH shows a sudden change near
the equivalence point The Equivalence point is the point at
which the moles of OH- are equal to the moles of H3O+
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Strong acid-strong base Titration Curve
At equivalence point, Veq:
Moles of H3O+ = Moles of OH-
There is a sharp rise in the pH as one approaches the equivalence point
With a strong acid and a strong base, the equivalence point is at pH =7
Neither the conjugate base or conjugate acid is strong enough to affect the pH
15_327
01.0
Vol NaOH added (mL)
50.0
7.0
13.0
pH
100.0
Equivalencepoint
pH
cm3 base added
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Weak acid-strong base Titration Curve
The increase in pH is more gradual as one approaches the equivalence point
With a weak acid and a strong base, the equivalence point is higher than pH = 7
The strength of the conjugate base of the weak acid is strong enough to affect the pH of the equivalence point
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Buffered Weak Acid-Strong Base Titration Curve
The initial pH is higher than the unbuffered acid
As with a weak acid and a strong base, the equivalence point for a buffered weak acid is higher than pH =7
The conjugate base is strong enough to affect the pH
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Polyprotic Weak Acids Polyprotic acids have more than one hydrogen
that can be neutralized Phosphoric Acid has three hydrogen ions.
H3PO4 + H2O H3O+ + H2PO4-
H2PO4- +H2O H3O+ + HPO4
2-
HPO42- +H2O H3O+ + PO4
3-
At given pH only one acid form and one conjugate base predominate
pH 0-4.7: H3PO4 and H2PO4-
pH 4.7-9.7: H2PO4- and HPO4
2-
pH 9.7-14: HPO42- and PO4
3-
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Polyprotic Weak Acid-Strong Base Titration Curve
Phosphoric Acid has three hydrogen ions.
There are three equivalence points
H3P04 + H2O H3O+ + H2PO4-
H2PO4- +H2O H3O+ + HPO4
2-
HPO42- +H2O H3O+ + PO4
3-
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Salts A salt is the neutralization product of an acid and
a base. The anion comes from the acid and the cation
from the base. Examples
HCl + NaOH NaCl + H2O.
H2SO4 + 2 KOH K2SO4 + H2O.
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Salts If a salt is the result of a
-- Strong acid and a strong base, the pH is near neutral.
HCl + NaOH NaCl + H2O.
-- Weak acid and a strong base, the pH will be greater than 7.
CH3COOH + NaOH NaCH3COO + H2O
-- Strong acid and a weak base, the pH will be lower than 7.
NH4OH + HCl NH4Cl +H2O
-- Weak acid and a weak base, the pH depends on whether the acid or the base is stronger.