Concept of & rsquo; acid and base

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Definitions :

– Ionization or ion dissociation : is the appearance of ions during the dissolution of substances called electrolytes: among which are acids and bases.

Acid : (selon Brönsted ).

It is a chemical entity, molecule or ion capable of releasing an ion H+ :

AH + H2O ⇔ A + H30+

Base :

It is a chemical entity, molecule or ion capable of capturing an ion H+ :

B + H20 ⇔ BH+ + OH


The acid-base reaction is a proton transfer.

Note :

There are compounds capable of capturing more protons H+ and others can give more protons H+ => polyacides, polybases.

Example : H2SO4 → HSO4 + H+
HSO4 → SO42- + H+

– amphoteric or ampholytic compounds : these are the compounds according to the nature of the medium can emit or capture protons, So have a behavior or acid, is basic.

Example: AH + H20base ⇔ A + H30+.
B + H20acid ⇔ BH+ + OH.

– Consequences of Rrôüsted-Lowry theory

1- Acid-base conjugate pair :

At any acid associated therewith a base and vice versa. The acid and the base define a torque acid / base or acid basic torque defined by the balanced half-reaction :

⇔ acid base + H+

Example :

HCIO2 ‌+ H+– H CIO2‘ => torque acid / base : HCIO2 / CIO2
acid base

2- acid-base reactions :

An acid-base reaction is a proton transfer reaction between acid-base couple noted 1 and another couple noted 2

3- Strength of acids and bases :

An acid is even stronger than its tendency to give H+ is higher. Conversely, a base is even stronger than its tendency to capture H+ is high.

Example: AH → A + H+ (HCI → H+ + CI)
AH A ⇔ + H+ ( CH3COOH ⇔ CH3COO + H+).

There are acids and bases of ” force ” average whose membership in one or the other of the two groups ( strong acid or base, weak acid or base ) is difficult to determine. In all cases, the force can be appreciated by applying the law of Guldberg and Waage.

HE HAS + H2O ⇔ H30++ AB + H2O ⇔ BH++ OH

The water, in very large excess, a concentration which does not vary during the reaction.

[H2O] = constant

then introduced, the notion of the constant acidity and alkalinity Next :

More K is large, more acid ( or base ) disassociated. In general, the quantity used :

pK = – logK

Conclusion : more pK is great, more acid (or base) is low and vis versa

Note : Ka and Kb are constants depending on :
– temperature
– the nature of the solvent

* Relationship between Ka and Kb :

PH AQUEOUS

Definition : It is a dimensionless number that measures the acidity or alkalinity of a medium. It is defined as follows :

PH = – log [H30+]

or : [H30+] is the concentration of ions
H30 in mol / l of the solution.

Similarly, we define : pOH = – log [OH].

Knowing that [H3O+] [OH] = 1014 => – log [H3O+] – log [OH+] = – log 1014

pH + pOH = 14 —> relationship between pH and pOH.

Scale acidity : It is measured relative to pure water ( neutral medium Ke = 1014 ).
A neutral medium : [H3O+] = [OH] = C et Ke = [H30+] [OH] = C2 => C = √Ke = 10-7
by definition : pH = – log [H30+] = – log 10-7 = 7 => pH of pure water = 7.

  • If an acid is dissolved in water, H ion concentration3O+ increases => [H30+] > 10-7 => pH<7.
  • On the other hand, if dissolving a base in water, dissolution will follow an increase in concentration in ions OH, therefore a decrease in H ions3O+ => [H3O+] < 10-7 => pH> 7.

Résumé :
medium pH acid < 7
neutral medium of pH = 7
medium basic pH > 7

Note : the scale of acidity is used 0 at 14.

PH calculation acidic solutions

1- Case of strong acids : Dissolving a strong acid is complete

AH + H2O —► H3O+ + A

The concentration of ions [H3O+] is therefore equal to the concentration of the acid :

[H3O+] = C => pH = – log C

2- For weak acids : dissociation is balanced

AH + H2O ⇔ H3O++ A+

Initially, starting from a concentration C AH. At equilibrium it will form A’ so that :

[AH] + [A] = C => [AH] = C – [A] ——-(1)

Ka = [A] [H3O+] / [AH] ——–(2)

According to the reaction equation : [H3O+] = [A] and seen the equation (1); equation (2) will become:

Ka = [H3O+]² / [AH] = [H3O+]² / C – [A] = [H3O+]² / C – [H3O+]

[H3O+]² + Ka [H3O+] – KaC = 0

The positive root gives [H3O+]. It is simple to calculate the pH, if the solution is very low ; that is to say [A] <<< [AH] therefore :

[AH] + [A] = C => [AH] = C et Ka = [H3O+]² / C

=> KaC = [H3O+]2 => [ H3O+] = ( KaC )1/2
=> – log [H3O+] = 1/2 (pKa – logC)

The pH is equal to : pH = 1/2 ( pKa – logC )

PH calculating basic solutions

1- If strong bases : Dissolving a strong base is total

B + H2O -> BH+ + OH

The concentration of ions [OH] is equal to the concentration of the base:

and as Ke = [OH][H3O+] = 10-14

so : [H3O+] = Ke / [OH] = 10-14/[OH] = 10-14/C

=> pH = -log [H3O+] = -log(10-14/C ) => pH = 14 + log C

2- If weak bases :

B + H2O ⇔ BH++ OH

Initially, starting from a concentration C B. A balance will be formed BH+ so that :

[B] + [BH+] = C => [B] = C- [BH+] ——-(1)

Kb = [BH+][OH] / [B] ———(2)

According to the reaction equation : [OH] = [BH+] and seen the equation (1); equation (2) will become :

Kb = [OH]² / [B] = [OH]² / C – [BH+] = [OH]² / C – [OH]² => [OH]² + Kb [OH] – KbC = 0

The positive root gives [OH]. It is simple to calculate the pH, if the solution is very low i.e. [BH+] <<< [B] therefore :

[B] + [BH+] = C=>[B]≅C and Kb= [OH]² / C
=> KbC = [OH]2 => [OH] = (KbC)1/2 => – log [OH] = 1/2 ( pKb – logC )
=> pOH = 1/2 ( pKb – logC )

as : pH + pOH = 14 a pKa + pKb, = 14.

For the pH of the solution, it is sufficient to replace : pOH by 14 – pH et pKb through 14- pKa and finally will :

pH = 7 + 1/2 ( pKa + logC )

PH of buffer solution

1- Definition : a buffer solution is a mixture of weak acid and its conjugate base.

2- pH calculation : Consider a mixture of a weak acid HA concentration CA and conjugate base of concnetration CB

HE HAS + H2O ⇔ A + H3O+ (ionization of the acid)

A + H2O ⇔ OH + HE HAS+ (hydrolysis of the conjugate base)

so Ka = [A][H3O+] / [AH] => [H3O+] = Ka [HE HAS] / [A]

=> PH = pKa + log [A] = pKa + log [base conj] / [acid]

PH = pKa + log [base conj] / [acid]

PH acid-base solutions :

1- Mixing strong acid + strong base :

if a = concentration of the acid after mixing
and b = concentration of the base after mixing

How to calculate t- brought to pH ?

2- Mixing strong acid + low base :

Let a = concentration of the acid after mixing
and b = concentration of the base after mixing

H3O+ + B ⇔ BH+ + H2O.

2.1 – If > b : H3O+ + B ⇔ BH+ + H2O excess strong acid

pH = – log [H3O+]= – log [a-b].

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2.2 – a b > a : H3O+ + B ⇔ BH+ + H2O

2.3 – A and b = : H30++ B ^ Btf + H20

3- weak acid – strong base :

Let a = concentration of the acid after mixing
and b = concentration of the base after mixing

AH + OH’ A’ + H2O

a- If > b : excess weak acid.

AH + OH ⇔ A + H2O+


b- a b > a : excess strong base.

AH + OH ⇔ A + H2O+


c- A and b = :

AH + OH ⇔ A + H2O+


Note : Can be experimentally draw the pH curve = f (b); said neutralization curve (in fact, this curve is obtained by a gradual addition in cm3 to a solution of, b cm3 d & rsquo; a basic solution and vice versa).

Course of Dr Tayeb Benmachiche Akila – Faculty of Constantine