acids and bases

Acids and bases. Buffers

Brønsted-Lowry theory of acids and bases

acid base

HCl + H2O H3O+ + Cl- NH3 + H2O NH4

+ + OH-

ammonia

H2SO4 + H2O H3O+ + HSO4

- CH3NH2 + H2O CH3NH3+ + OH-

aminomethane methylammonium ion

HSO4- + H2O H3O

+ + SO42-

CH3COOH + H2O CH3COO- + H3O+

ethanoic acid acetate ion

(acetic acid)

Definitions:

A base is a substance that can

accept a hydrogen ion (H+).

An acid is a substance that can donate a hydrogen ion (H+).

Examples:


When an acid gives up its H+, the resulting anion is a base. This base is called conjugate base.

Acid H+ + Conjugate base

HCl + H2O H3O+ + Cl-

conjugate base

CH3COOH + H2O H3O++ CH3COO-

acetic acid acetate ion; conjugate base

An acid and its conjugate base form conjugate pair.

HCl/Cl- ;

CH3COOH /CH3COO-

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A base reacts by accepting H+. The species produced by

this reaction is its conjugate acid.

Base + H+ conjugate acid

NH3(aq) + H2O NH4+ + OH-

base conjugate acid

A base and its conjugate acid form conjugate pair.

NH3/NH4+.

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Substances that can either lose or gain a hydrogen ion are amphoteric substances.

NH3(aq) + H2O NH4+(aq) + OH-(aq)

H+ donor conjugate base

HCl + H2O Cl- + H3O+

H+ acceptor conjugate acid

Water acts as an acid when it donates hydrogen ions to ammonia.

It acts as a base in its reaction with hydrogen chloride.

Water is an amphoteric substance.

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Strength of acids and bases.

strong acids and bases

weak acids and bases

Strong acids are acids that ionize completely in water – HNO3,

HCl, H2SO4, HClO4.

HCl H+ + Cl-

Acids that are partially ionized in water are weak acids.

HCN – hydrocyanate

H2CO3 – carbonic acid

H3PO4, HF, CH3COOH etc.

Most organic acids are weak acids. The only H atom that does ionize to

some degree is the one, bonded to the oxygen atom of the COOH group.

CH3COOH CH3COO- + H+


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HCN H+ + CN- 1 mol acid produces 1 mol H+, this acid is

monoprotic acid

H2CO3 H+ + HCO3

- bicarbonate ion

HCO3- H+ + CO3

2- carbonate ion

1 mol acid produces 2 mols H+, this acid is diprotic acid

Polyprotic acids ionize stepwise.

Problem:

Write the equation for the stepwise ionization of the diprotic acid

H2SO3 (sulphurous acid) .

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Answer:

H2SO3 H+ + HSO3

- hydrogen sulfite ion (bisulfite ion)

HSO3- H+ + SO3

2- sulfite ion

Strong bases produce OH- ions in high

concentration in solutions.

NaOH → Na+ + OH-

Weak bases have low concentration of OH- ions

in their solutions.

NH3 + H2O NH4+ + OH-

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stronger acid than H3O+ weaker base than H2O

conjugate pair

HCl + H2O Cl- + H3O+

conjugate pair

stronger base weaker acid

than Cl- than HCl


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weaker acid than H3O+ stronger base than H2O

conjugate pair

CH3COOH + H2O CH3COO- + H3O+

conjugate pair

weaker base stronger acid

than CH3COO- than CH3COOH

CH3COO – H …….. CH3COO-


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Conjugate Acid-Base Pairs


Acid Base

strongest HClO4 HClO4- weakest

H2SO4 HSO4-

HCl Cl-

HNO3 NO3-

H3O+ H2O

HSO4- SO4

2-

H3PO4 H2PO4-

HF F-

CH3COOH CH3COO-

H2CO3 HCO3-

H2PO4- HPO4

2-

NH4+ NH3

HCO3- CO3

2-

HPO42- PO4

3-

weakest H2O OH- strongest

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stronger acid than H2O weaker base than OH- conjugate pair

CH3COOH + OH- CH3COO- + H2O

conjugate pair

stronger base weaker acid

than CH3COO- than CH3COOH

weaker base than OH- stronger acid than H2O

conjugate pair

NH3 + H2O NH4+ + OH-

conjugate pair

weaker acid stronger base

than NH4+ than NH3


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The strengths of acids are measured by their tendencies to transfer protons to a base, usually water. For an acid with the general formula HA,

the equilibrium constant for ionization is obtained by equation:

HA + H2O H3O+ + A-

K = [H3O+] [A-]

[HA] [H2O]

Water concentration - [H2O], is so large compared to the concentrations of the ions formed in the equilibrium, that is why [H2O] is included in another equilibrium constant, the acid ionization constant, Ka.

Ka = K [H2O] = [H3O+] [A-]

[HA]

The larger the value of Ka, the stronger the acid.

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To compare bases we use their tendency to accept protons from water. For the base with the general formula B, the base ionization reaction and the related equilibrium constant expression are:

B + H2O BH+ + OH-

K = [BH+] [OH-]

[B] [H2O]

As in the case of the acid ionization constant the [H2O] is included in the base ionization constant, Kb.

Kb = K [H2O] = [BH+] [OH-]

[B]

The larger the value of Kb, the stronger the base.

Ka and Kb are related in this way:

Ka . Kb = Kw Kw = 1.10 -14

Therefore, if you know one of them, you can calculate the other.

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Problem

Lactic acid is a monoprotic organic acid produced in metabolic reactions. Its Ka is 1.10-4. Calculate the Kb of its conjugate base.

Solution:

Lactic acid – HL,

than formation of its conjugate base can be represent with the

equation:

HL + H2O L- + H3O+

acid conjugate base

Ka.Kb = Kw

1.10-4.Kb = 1.10-14

Kb =1.10 -14 = 1.10 -10

1.10-4


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Values of Ka and Kb are very small that is way it is convenient

to define pKa and pKb.

pKa = -logKa

pKb = -logKb

pKa + pKb = pKw = 14

pKa and pKb are used to determine if the acid or base is

strong or weak.

The larger the value of pKa (pKb) , the weaker the acid

(base).

Problem

For CH3COOH/CH3COO─ pKa + pKb = 14. Calculate the

pKb(CH3COO¯ ) if pKa(CH3COOH) = 4,76. Is the acetate ion

(CH3COO¯ ) a strong base?


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Ionic product of water

Water is amphoteric and can act as an acid and a base with itself.

Water self ionizes at 250C to produce hydronium ions and hydroxide ions.

H2O + H2O H3O+ + OH-

acid base conjugate conjugate

acid base

One water molecule behaves as an acid, and another behaves as a base in the reaction.

The equilibrium constant expression is

K = [H3O+].[OH-]

[H2O]2

The concentration of water is so large that it remains virtually constant. It is included in an alternate equilibrium constant, ion product constant of water, Kw.

[H2O] = constant Kw = K.[H2O]2 = [H3O+].[OH-]

Kw = [H3O+].[OH-]

Kw – ion product constant of water

Kw = 1.10-14 at 25°C, it depends on temperature.

In all water solutions Kw = 1.10-14

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In pure water [H+] = [OH-] = 1.10-7M.

Depending on [H+] solutions are acidic, neutral and basic.

Concentration of H+ pH

Acidic solution [H+] > 1.10-7M < 7

Neutral solution [H+] = 1.10-7M = 7

Basic solution [H+] < 1.10-7M >7

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pH value

• pH is defined as the negative logarithm of the molarity of the hydrogen ions (hydronium ions).

pH = log [H+]

• for [H+] = 1.10-7M

pH = -log(1.10-7) = - (-7) = 7

This is the pH of neutral solution,

• pH of acidic solution is <7

and pH of basic solution is >7

• pOH = -log[OH-] pKw = - logKw

• as you know Kw = [H+].[OH-] = 1.10 -14 then

pKw = pH + pOH = 14

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Problem

Blood sample contains 4,5 . 10-8 M hydronium ions. What is

the pH?

Solution

[H3O+] = 4,5.10-8 M

pH = log [H3O+]

pH = log (4,5.10-8) = -log 4,5 + (-log 10-8)= -0,65 + 8 = 7,35

Problem

A 0,0050 M morphine solution has [OH-] = 8,8.10-5 M. What is

the pH of the solution?

Solution

[OH-] = 8,8.10-4 M

pOH = log [OH-]

pOH = log (8,8.10-4) = -log 8,8 + (-log 10-4)= -0,94 + 4 = 3,06

pH + pOH = 14

pH = 14 – pOH = 14 – 3,06 = 10,94

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pH

2 3 4 5 6 7 8 9 10 11 12

neutral @ 25oC

(H+) = (OH-)

distilled water

acidic

(H+) > (OH-)

basic or alkaline

(H+) < (OH-)

natural waters

pH = 6.5 - 8.5

normal rain (CO2)

pH = 5.3 – 5.7

acid rain (NOx, SOx)

pH of 4.2 - 4.4

0-14 scale for the chemists

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pH

1 2 3 4 5 6 8 9 10 11

The biological view in the human body

acidic basic/alkaline

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Methods to measure pH

Using indicators

Hind H+ + Ind ¯

colour A colour B

Paper tests like litmus paper and pH paper

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Using pH meter

• Tests the voltage of the electrolyte

• Converts the voltage to pH

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Buffer – a solution that prevents a drastic pH change when

either H+ or OH- is added to it.

Each buffer consists of a weak acid and its conjugate base

(salt of that acid)

or weak base and its conjugate acid (salt of that base)

dissolved in water.

Examples:

acetic acid and sodium acetate dissolved in water

CH3COOH / CH3COONa;

ammonia and ammonium chloride dissolved in water

NH3 / NH4Cl

Buffers

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Acetate buffer - CH3COOH / CH3COONa

determines pH of the buffer solution

CH3COOH CH3COO- + H+

CH3COONa CH3COO- + Na+

CH3COOH / CH3COO-

The buffer reacts to neutralize any strong acid or base that is

added to the solution.

could alter pH

HCl H+ + Cl-

H+ + Cl- + CH3COO- + Na+ CH3COOH + Na+ + Cl-

For short: H+ + CH3COO- CH3COOH

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NaOH Na+ + OH- could change pH

Na+ + OH- + CH3COOH CH3COO- + Na+ + H2O

Short equation: OH- + CH3COOH CH3COO- + H2O

! In general:

Added acid reacts with the base that present into a solution.

Added base reacts with the acid that present into a solution.

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Problem

How would the ammonium buffer (NH4OH / NH4Cl) neutralize

the effect of the addition of NaOH and HCl so that the pH value

remains the constant?

Solution:

NH4OH is the base

NH4+ from NH4Cl is its conjugate acid

So, added NaOH must react with the acid

NaOH + NH4Cl NH4OH + NaCl

For short: OH- + NH4+ NH4OH

HCl added reacts with the base

HCl + NH4OH NH4Cl + H2O

H+ + OH- H-OH

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Phosphate buffer:

H2PO4- / HPO4

2-

How this buffer reacts to neutralize addition of

HCl?

HPO42- + HCl H2PO4

- + Cl-

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pH of the buffer

What does the pH of buffer depend on? How to calculate

pH of buffer?

First we have to know [H+] in buffer solution. If the buffer

consists of weak acid HA and its salt MA obviously H+ are

produced from HA.

HA H+ + A-

Ka = [H+][A-] [H+] = Ka [HA]

[HA] [A-]

Thus pH=-log[H+] pH=-logKa + log [A-]

[HA]

pH = pKa + log [A-] (1)

[HA]

HA - acid

A- - conjugate base

pH depends on the strength of acids and the ratio [A-]

[HA]

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In the case of buffer consisting of a base and its salt the

equation used for pH calculating is:

pH = 14 – pKb + log [B]___ (2)

[HB+]

pH depends on strength of acids or bases

and on the ratio [A-]

[HA]

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The equations ( 1 ) and ( 2 )

are called Henderson – Hasselbalch’s equations.

pH = pKa + log [A-] (1)

[HA]

pH = 14 – pKb + log [B]___ (2)

[HB+]

They are used to calculate pH of buffers.

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Buffer capacity is defined as a moles of strong acid or

strong base which must be added into 1 L of buffer solution

in order to change its initial pH with 1 pH unit.

= number of moles of acid (base)

pH

Buffer capacity is higher when solutions are concentrated.

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Problem

Express the Henderson-Hasselbalch’s equation of

carbonate buffer (H2СO3/HСO3-). Calculate the pH value of

carbonate buffer with [H2СO3] = 0,02M and [HСO3-] = 0,02M,

pKa = 6,36.

Solution

pH = pKa + log [A-] pKa = 6,36

[HA] H2СO3 is the acid

HСO3- - is its conjugate base, so

pH = pKa + log [HCO3-]

[H2CO3]

pH = 6,36 + log 0,02 = 6,36

0,02

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H2CO3 / HCO3- - bicarbonate buffer; in the blood stream

H2PO4- / HPO4

2- - phosphate buffer; involved in kidney

functions

protein buffer: plasma protein and hemoglobin

How do these buffers work?

Bicarbonate buffer

1. HCO3- + H3O

+ H2CO3 + H2O

(aq) (aq) (aq) (l )

H2CO3 H2O + CO2

(aq) (aq) (g)

2. H2CO3 + OH- HCO3- + H2O

(aq) (aq) (aq) (l )

[H2CO3 ] / [HCO3- ] = 1:20

Buffers in the Body

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Phosphate buffer – H2PO4- / HPO4

2-

H2PO4- + OH- HPO4

2- + H2O

HPO42- + H3O

+ H2PO4- + H2O

[ H2PO4- ]/ [HPO4

2- ]=1:4

Hemoglobin buffer: HHb / Hb-O2-

HHb + O2 ↔ HbO2- + H+

oxyhemoglobin

HCO3- + H+ → H2CO3

H2CO3 → CO2 + H2O

Net oxygenation reaction is

HHb + O2 + HCO3- + H+ + H2CO3 ↔ HbO2

- +H+ + H2CO3 +CO2 + H2O

HHb + O2 + HCO3- ↔ HbO2

- + CO2 + H2O

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HHb-NH2 + CO2 → HHb-NH-COO- + H+

carbaminohemoglobin

H2O + CO2 ↔ H2CO3

H2CO3 → HCO3- + H+

Hb-O2- + H+ → HHb + O2

The net equation for removal of carbon dioxide is obtained by

summing the three equations.

Hb-O2- + CO2 + H2O ↔ HHb + O2 + HCO3

-

deoxygenation reaction

This reaction illustrates the effect of hemoglobin buffer.

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Acidosis – pH of blood lower than 7.28 for a period

of time

Alkalosis - pH of blood higher than 7.40 for a period

of time

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Terms

Acid

Base

Amphoteric

Conjugate pair (asid-base pair)

Acid ionization constant Ka

Base ionization constant Kb

Ion product constant of water Kw

pKa, pKb, pKw

pH, pOH, pH of biological fluids

Indicators

Buffers

How to calculate pH of the buffers

Buffer capacity

Buffers in the human body


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acids and bases

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