Acids and Bases презентация

Vocabulary
Acidity, alkalinity, aqueous
Donor, acceptor
Dissociation
Indicator

Learning Objectives

Acids and Bases
Arrhenius definition: Classified in terms of formula and behaviour in water Acid:
Base:
Contains H in formula and produces H+ (or H3O+) in water. e.g., HCl, H2SO4, HCN, HNO3
Contains OH in formula and produces OH- in water.
e.g., NaOH, Mg(OH)2, KOH

Acids and Bases
Brønsted-Lowry definition: An acid-base reaction is a proton transfer process
Acid:
Base:
Proton donor during acid-base reaction.
Must contain H in formula.
e.g., NH4+, HCl, H2SO4, HCN, HNO3, HAc (CH3COOH)
Proton acceptor during acid-base reaction.
Must contain a lone pair of electrons in formula (to make a bond with H+).
e.g., NH3, Cl-, HSO4-, CN-, NO3-, Ac- (CH3COO-)

Brønsted-Lowry Acid-Bases
Conjugate acid-base pairs:
Every acid-base reaction has two conjugate acid-base pairs.
NH4+ is the conjugate acid of NH3:
Is formed when NH3 acts as a base and accepts a proton
HCl/Cl-
H2O/OH-
H3O+/H2O
NH3 is the conjugate base of NH4+:
Is formed when NH4+ acts as an acid and donates a proton
NH4+/NH3
Can donate or accept protons.
Can act as either acid or base.
e.g., H2O, HSO4-
Amphiprotic (amphoteric):

Proton-Transfer using a Weak Base:

Proton-Transfer using a Weak Acid:

pH
pH is defined as the negative base-10 logarithm of the hydronium ion concentration.
pH = −log [H3O+]

Autoionization of Water
As we have seen, water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
This is referred to as autoionization.

Ion-Product Constant
The equilibrium expression for this process is
Kc = [H3O+] [OH−]
This special equilibrium constant is referred to as the ion-product constant for water, Kw.
At 25 °C, Kw = 1.0 × 10−14

pH
In pure water,
Kw = [H3O+] [OH−] = 1.0 × 10−14
Because in pure water [H3O+] = [OH−],
[H3O+] = (1.0 × 10−14)1/2 = 1.0 × 10−7

pH
Therefore, in pure water,
pH = −log (1.0 × 10−7) = 7.00
An acid has a higher [H3O+] than pure water, so its pH is <7
A base has a lower [H3O+] than pure water, so its pH is >7.

Some common
pH values:

Other “p” Scales
The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).
Some similar examples are
pOH −log [OH−]
pKw −log Kw

Because
[H3O+] [OH−] = Kw = 1.0 × 10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00

How Do We Measure pH?
For less accurate measurements, one can use
Litmus paper
“Red” paper turns blue above ~pH = 8
“Blue” paper turns red below ~pH = 5
An indicator

How Do We Measure pH?
For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

How much is the equilibrium displaced towards the
formation of the products (ionization)
What is the difference between a strong and a weak acid?
Strong and Weak Acids
Strong acid
Strong electrolyte
Close to 100% products
(dissociation)
Weak acid
Weak electrolyte
Very little products
(not much dissociation)
Similar for strong/weak bases

Strong Acids
You will recall that the six strong acids are HCl, HBr, HI, HNO3, H2SO4 and HClO4.
These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.
For the monoprotic strong acids,
[H3O+] = [acid].

Strong Bases
Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).
Again, these substances dissociate completely in aqueous solution.

Example problem 1:
What is the pH of a 7.52 x 10-4 M CsOH solution?
Is the solution neutral, acidic or basic?
b) What is the pOH of a 1.59 x 10-3 M HClO4 solution?
Is the solution neutral, acidic or basic?
Strong electrolyte, dissociates completely: [OH-] = [CsOH] = 7.52 x 10-4 M
pOH = - log [OH-] = 3.124
pH = 14 – pOH = 10.877
Basic solution
Strong electrolyte, dissociates completely: [H+] = [HClO4] = 1.59 x 10-3 M
pH = - log [H+] = 2.799
pOH = 14 – pH = 11.201
Acidic solution

Example problem 2:
What is the [H3O+] and [OH-] of a solution with a pH of 2.56?
b) What is the [H3O+] and [OH-] of a solution with a pOH of 3.78?
[H3O+] = 10pH = 10(-2.56) = 2.8 x 10-3
[OH-] = 1.00 x 10-14 / 2.75 x 10-3 = 3.6 x 10-12
[OH-] = 10pOH = 10(-3.78) = 1.7 x 10-4
[H3O+] = 1.00 x 10-14 / 1.66 x 10-4 = 6.0 x 10-11

Dissociation Constants
For a generalized acid dissociation,
the equilibrium expression would be
This equilibrium constant is called the acid dissociation constant, Ka.

Dissociation Constants
For a generalized base dissociation,
the equilibrium expression would be
This equilibrium constant is called the base dissociation constant, Kb.

For the example equation:
CH3COOH ⇌ CH3COO- + H+
CH3COOH is the acid and CH3COO- is its conjugate base.
Combining Ka and Kb gives:
= [H3O+] [OH−] = Kw
=> Kw = Ka ×Kb
=> pKw = pKa + pKb

Dissociation Constants
The greater the value of Ka, the stronger the acid.

Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.
We know that

Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.
To calculate Ka, we need the equilibrium concentrations of all three components.
We can find [H3O+], which is the same as [HCOO−], from the pH.

Calculating Ka from the pH
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
4.2 × 10−3 mol/dm3 = [H3O+] = [HCOO−]
[HCOOH] = 0.10 − 4.2 × 10−3
= 0.0958 = 0.10 mol/dm3

Calculating Ka from pH
= 1.8 × 10−4

Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C.
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Ka for acetic acid at 25°C is 1.8 × 10−5.

Calculating pH from Ka
The equilibrium constant expression is

Calculating pH from Ka
Now,x = [H3O+] = [C2H3O2−]
(1.8 × 10−5) (0.30) = x2
5.4 × 10−6 = x2
2.3 × 10−3 = x

Calculating pH from Ka
pH = −log [H3O+]
pH = −log (2.3 × 10−3)
pH = 2.64

Titration
A known concentration of base (or acid) is slowly added to a solution of acid (or base).

Titration
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

acid
alkali
end-point
METHYL ORANGE

acid
alkali
end-point
acid
alkali
PHENOLPHTHALEIN

Acid-Base Indicators
indicate the equivalence point of a titration.
are weak organic acids for which weak acid and conjugate base are different colors.
the color of the solution depends on the ratio of the In- to the HIn forms:
HIn(aq) + H2O(l) H3O+(aq) + In- (aq)
the indicators change color in specific pH ranges close to the pKa value of the indicator, pKin.

Acid-Base Indicators
The sharp change in color of the indicator signals the end point of the titration.
For the end point of an indicator to be useful,
i.e. indicate the equivalence point accurately:
it must occur at a volume of titrant very close to that for the equivalence point of the titration
the color change of the indicator must be dramatic enough to be detected.

Indicator Colors and Ranges

Titration of a Strong Base with a Strong Acid
The pH at the equivalence point in these titrations is ~ 7.
Bromothymol Blue can be used as its color change from blue to yellow is in the pH=7 range.
Also, phenolphthalein can be used because of the high slope between pH 3.5 to 10.5

Strong acid – Strong base
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Strong acid – Strong base
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Strong acid – Strong base
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Titration of a Weak Base with a Strong Acid
The pH at the equivalence point in these titrations is < 7.
Methyl red is the indicator of choice.

Strong acid – Weak base
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Strong acid – Weak base
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Strong acid – Weak base
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Titration of a Weak Acid with a Strong Base
The pH at the equivalence point in these titrations is ~9.
Phenolphthalein would be the indicator of choice.

Weak acid – Strong base
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Weak acid – Strong base
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Weak acid – Strong base
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Weak acid – Weak base
pH at equivalence
depends on relative strength of acid and base
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Weak acid – Weak base
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Weak acid – Weak base
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SUMMARY
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SUMMARY
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SUMMARY
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Buffers:
Solutions of a weak conjugate acid-base pair.
They are particularly resistant to pH changes, even when strong acid or base is added.

Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

Buffers
If acid is added, the F− reacts to form HF and water.

Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

Buffer Calculations
Rearranging slightly, this becomes
Taking the negative log of both side, we get

Buffer Calculations
So
Rearranging, this becomes
This is the Henderson–Hasselbalch equation.

Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is
1.4 × 10−4.

Henderson–Hasselbalch Equation
pH = 3.85 + (−0.08)
pH = 3.77

Buffer Uses
Electroplating
Manufacture of Dyes
Calibrating pH meters
Buffering blood using combinations of:
HCO3- ; hemoglobin ; H2PO4- ; HPO42-

When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.

Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Calculating pH Changes in Buffers
A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.
Before the reaction, since
n (CH3COOH) = n (CH3COO−)
pH = pKa = −log (1.8 × 10−5) = 4.74

Calculating pH Changes in Buffers
The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:
CH3COOH(aq) + OH−(aq) ⎯⎯→ CH3COO−(aq) + H2O(l)