Home ADVANCED LEVEL TOPIC 4: ACIDS, BASES AND SALTS ~ CHEMISTRY FORM 6

TOPIC 4: ACIDS, BASES AND SALTS ~ CHEMISTRY FORM 6

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ACIDS, BASES AND SALTS

TOPIC 4: ACIDS, BASES AND SALTS ~ CHEMISTRY FORM 6

Arrhenius concept of acids and bases .

What is an acid? (according to Arrhenius concept of acids and bases)

ACIDS, BASES AND SALTS:– Arrhenius considered that an acid is a substance which when dissolved in water dissociate to produce Hions as the only positively charged ions i.e.

e.g
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– He considered a base to be a substance which produce hydroxyl ions when dissolved in water as the only negatively charge ions i.e

e.g.
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The neutralization of acid with a base yields to a salt and water.

e.g.
+  +

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Ionic equation

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According to Arrhenius, neutralization reaction is all about formation of water.

Weakness of Arrhenius equation

i) This concept is limited to water. It refers to Hand OH ions derived from water. A true general concept of acid and base should be appropriate to other solvent like liquid NH3 and alcohols.

ii) The concept does not provide the room for acids and bases which do not contain Hions and HO  ions.

Bronsted – Lowry concept of acids and bases


Bronsted and Lowry proposed a theory of acids and bases applicable to all solvents.

–    They proposed that, an acid is any substance that can donate a proton to any other substance.

e.g
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–    A base is a substance that can accept a proton from any other  substance.

E.g.

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             Base         Acid

– They are called a monoprotic acid an acid which donates only one proton.

e.g. HNO3, HCl

– Diprotic acid can donate two protons.

E.g. H2SO4

– A polyprotic acid is an acid that can donate more than one proton.

E.g. H2SO4, H3PO3, H2C2O4

–    A polyprotic base is one which can accept more than one proton.

E.g.

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– A monoprotic base is one which can accept only one proton.

E.g.,

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NOTE: HCl and Cl– are acid-base conjugate pair. Another example is HNO3 and

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Amphoteric (amphiprotic) acids and bases.

These behave as bronsted – Lowry acids or bases

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Conjugate acid base pair

-For every acid, there is a corresponding (conjugate) base to accept a proton

E.g.

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Acid                         base

(Proton donor)   (Proton acceptor)

HA and A are conjugate pair i.e.

HA is a conjugate acid of A and A is a conjugate base of HA.

®             In a solution, there must be a base to accept a proton

E.g.

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Acid        Base       Acid        Base

E.g. 

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A1, B1, and A2, B2, are acid base conjugate.

NOTE:

From Bronsted–Lowry concept of acid and bases, the stronger the acid, the weaker it’s conjugate base and the stronger the base, the weaker its conjugate acid.

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CH3COOH is a weak acid, but its conjugate base i.e.  is a strong base.

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–                         H2O is a weak base, but  is strong acid.

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Advantage of Bronsted–Lowry concept over Arrhenius

®                 It can apply to any solvent not necessarily. Here the definition of bases is much wider.

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Weakness:

Since the concept is based on proton transfer, it does not consider other compounds which do not contain hydrogen i.e AlCl3, BF3, SO3.

In contrast to Arrhenius theory, acid and bases are no longer related to salts (by neutralization).

Question 1:

A.    Define

i)        Conjugate acid-bases pair.

ii)      Conjugate base.

B.     For the following pairs, write down the equation to show the conjugate acid-bases pair.

i)        /

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ii)

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iii)

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iv)    /

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Question 2:

a.       Write the formula and give the name of the conjugate base for each of the following acids.

i)

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ii)

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b.      Write the name and formula of the conjugate acids for each of the following bases.

i)        NH3

ii)      Br 

iii)    HS

Question3:
In each of the following acidbase reaction. Identify the acid and the base on the left and their conjugate partners on the right.

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a)

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b)

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ANSWERS:

a)

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 Acid         Base     Acid      Base

HCN, CN is a conjugate pair and NH3, NH4+   is a conjugate pair

b)

HSO4, SO4-2, is a conjugate pair

c)

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Acid                            Base                        Base                            Acid

[Al(H2O] 3+, [Al(H2O) 5 OH] 2+ is a conjugate pair.

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OH– and O is a conjugate pair.

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Q: Conjugate acid-base pair is a pair which shows that for every proton lost by acid, there is a corresponding base to accept it.

a)                            Conjugate base.

b)

i)

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ii)

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iii)

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LEWIS CONCEPT OF ACIDS AND BASES.

·
– Lewis proposed an even broader concept of acids and bases focusing on electron transfer rather than total transfer.

– According to Lewis an acid is a substance that can accept a pair of electrons. Therefore, an acid is an electron pair acceptor.

– A base is a substance that can donate a pair of electrons i.e base is an electrons pair donor.

NOTE

Bronsted–Lowry acids e.g. HCl, H2SO4, HNO3, are not Lewis acid.
Thus, an acid-base reaction can occur when a base provide a pair of electrons to share with an acid resulting into coordinate compound or complex.

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Therefore, ammonia chloride ions (AlCl3) are Lewis base, while H+, BF3 are Lewis acids.

NOTE:

A Bronsted–Lowry base (like NH3) reacts by donating electron pair to a proton. Therefore Bronsted–Lowry bases are also Lewis bases

Reason:

This is because upon donating a pair of electron, it would have accepted a proton. Therefore, Bronsted – Lowry bases are also Lewis bases.

i . F fluoride ionSO2-

ii.  Sulphate ion

iii. NH+4  ammonium ion

iv.  HBr  Hydrogen bromid

v.   H2S Hydrogen sulphide.

Reason:

These cannot accept a lone pair of electrons hence the Lewis concept of acid is usually used in special cases.

IONIC EQUILIBRIUM  OF ACIDS AND BASES.

Most acids and bases are weak i.e does not ionize fully when dissolve in water. Thus a part from water equilibrium, they also establish equilibrium.

e.g.

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Ammonia which is a typical weak base ionizes as follows:

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But the ionization of weak acid/ bases generally occurs to a greater extent than that of water.

STRENGTH OF WEAK ACIDS AND BASES.

The position of equilibrium of a reaction between the acid and water varies from one weak acid to another. The further to the left it lies, the weaker the acid is   

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The equilibrium constant is written as;

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But H20 is constant at constant temperature.

Putting the constant on the same side

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Where Ka = dissociation/ ionization constant of an acid.

for
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Similarly for weak base, the position of equilibrium varies from base to base. The further to the left it lies, the weaker the base is

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Where Kb = dissociation/ ionization constant of a base.

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The Kand Kb values are used to determine the strength of acids and bases i.e  Ka and Kvalues are quite small for a very weak acid/ base reflecting very title ionization of these acids/ base in solution.

Example:

The Kb value for C6H5NH2  is 4.17 x 10-10   ,NH3 is 1.78 x 10-5   Indicate which base is stronger than the other.

NH3 is stronger than C6H5NH2

The Kb value for C6H5NH2  is smaller than C6H5NH2NH3

The strength of weak acids and bases can also be determined from its degree of dissociation (Ostwald’s dilution law)

Since Ka and Kb values are inconvenient to handle usually pKa and pKb are used.

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For example;

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The lower the value for the stronger the acid base respectively and vice versa

THE RELATIONSHIP BETWEEN pka AND pkθ FOR A CONJUGATE ACID–BASE PAIR.

Consider the equilibrium.
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The product of Ka and Kgives.
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Example
Formic acid (HCOOH)  has a Kof 1.78 x 10-14  moles. Calculate the [H3O+]   and the pH of 0.1M. Solution of HCOOH.

  Solution

Since Ka value is small, the expression

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NOTE:
The approximation is done when [HA]O is greater than 100 Ka

But if initial concentration [HA]O is less than 100 Ka, then the exact expression formed must be solved.
Calculate [H3O+] and pH in which has Ka value of moles/ dm3.

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Solution:

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Question set 1:
Formic acid (HCOOH)  has Ka of (HCOOH) 1.8 x 10-14   moles dm3. If you have0.001 m a solution of the acids. What is the pH of this solution, what is the concentration of HCOOH at equilibrium?

 

2. If the acid HA is 2% ionized in solution of concentration 0.01m, calculate

a)      Ka                                                                                                         

b)      pKa

3.Calculate the degree of ionization   is 9.37 in a 0.1M aqueous solution

Word Image 1735 Word Image 17364.   Calculate pH of a 0.052m acetic acid solution if Ka is 

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5. For a 0.1M solution of benzoic acid, calculate.

i)    Concentration of ions and molecules in solution

ii)    The degree of ionization of the acid.

iii)   pH of the solution

6). A hypothetical weak base (MOH) has Kb of  for the reaction.

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Calculate the equilibrium Concentration of MOH, M+ and OH– in a  solution MOH.

The weak base methylamine has Kb of 5 x 10– 4 .It reacts with water according to the equation.

Calculate the equilibrium concentration of OH in a solution of base. What are the of the solution?

Hydroxyl amine has a Kb of What are the  of the base ?

A 0.1m  solution of chloroacetic acid (ClCH2COOH) has a of 1.95. Calculate Ka for the acid.

IONIC PRODUCT OF WATER AND PH.

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–   Water auto–ionizes i.e transfers a proton from one water molecule to another producing H3O and OH

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Base          Acid

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But the concentration of H2O is much larger than the two ions and is constant at constant temperature. Since the concentration of is constant it is made part of the constant.

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Where = ionic product of water constant.

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In pure water, Kw = 1 x 10-14 at 25°c. Since every one H3O+ ion formed also one OH ion is formed, thus the concentrations are equal.

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Since the concentrations are equal, this implies that pure water is neutral.

Scale

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–   Is a scale which shows degree of acidity or alkalinity of a solution:

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=

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= 7

I.e pH of 7 is neutral point.

    Is the negative logarithm of base 10 hydrogen ion.

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    Is the negative logarithm of base 10 of hydroxyl ion concentration.

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For pure water

        = 7   pOH= 7
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NOTE:

In acidic solution, the concentration of   H3O+ is greater than [0H –].

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i.e

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In basic solution, the concentration of

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i.e [

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From

, Introducing negative log on both sides

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Variation of of pure water with temperature.

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The formation of H3O+ and OH ions from water is an endothermic process i.e the forward reaction absorbs heat.

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-According to Le Chatelier`s principle, when you increase the temp, the forward reaction is favoured, thus concentration of  H3Oand OH– ions will increase but in equal amounts.

Thus, the pH will drop but the water will not be acidic the pH scale will also change. It won`t remain as 1 to 14 and the neutral point will also shift. The direct effect on increasing temperature is to increase Kw.

The table below shows the effect of temperature on Kw and each value of Kw a new pH must be calculated.

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Word Image 5034 Word Image 5035                       6.1449

From the table, pH of water fall as temperature increases. This does not mean that water becomes more acidic at higher temperature. This solution is only acidic, If the concentration of is great than

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Example:

1.      The value of Kw at physiological temperature of a body is . What is the pH at neutral point of water at this temperature.

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Solution:

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2.      At , is 15. Find at O°c, if the [H+] is

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Solution:

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STRONG ACIDS AND BASES.

When an acid is added to water as an aqueous solution of HCl in addition to self ionization of water, the acid also ionizes.

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Due to common ion effect of hydroxonium ion as HCl is fully ionized, it suppresses the ionization of water hence and [O ion from water will be less than

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It is generally acceptable to consider the ionization of HCl to be the sole source of hydroxonium ions. This is also applicable in strong bases for OH ions.

Example1:

If 0.001m of NaOH is added to enough amount of 1L of water, what is the concentration of OH – and H3O+ ions?

Solution:

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  Example2:

Calculate the concentration of, and in a 100 cm3 sample of 0.015 m

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NOTE:

HCl is the main source of H3O+

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NOTE: Molar concentrations are independent of solution volume i.e [H3O+] in 0.015M HCl is the same whether we are describing 1L, 10L or 100cm3 thus the volume of acid is not involved in this calculation.

e.g. Calculate the [], and in 50 cm3 of

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0.01M          0.01M     2 X 0.01=0.02M

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From:

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Word Image 5095NEUTRALIZATION REACTION.

This is the reaction between H3Ofrom an acid and OH – ions from a base to form water. Therefore when solutions of acids and bases are mixed together, the chemical reaction must occur in which and combine to form water.

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This reaction occur in order to maintain the required value of equilibrium constant Kw. The final solution can be acidic basic or neutral depending on the and after neutralization reaction.

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Example:

What is the [H3O+] obtained by mixing 100cm3 of 0.015m HCl and 50cm3 of 0.01m Ba(OH)2 solution, Is the final solution acidic or basic?

Solution:

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+2

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1       1

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=

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=[][]

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Question:

1)      What is the pH of a solution obtained by dissolving 312 cm3 of HCl, measured at 30°c at 340mmHg in 3.25lL of water?

Solution:

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From

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[] =1.7626

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2)      Calculate the pH of neutralization point when 40cm3 of 0.1m NaOH is mixed with 60cm3 of 0.1m HCl.

Solution:

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1

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+

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3)      Calculate the pH of the solution obtained when.

a) 1cm3 of 0.1m NaOH is added to 100cm3 of 0.001m HCl

b) 1cm3 of 0.1m NaOH is added to 100cm3 of 0.1m HCl

Solution:

a)

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X

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b)

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Again

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=

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=

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BUFFER SOLUTIONS:

A buffer solution is a solution which maintains its pH when small amount of an acid or alkali is added to it.
OR

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Is the one that resist a change in pH when small amount of acid or alkali is added to the solution.

A buffer solution usually consists of a weak acid and one of its salt or a weak base and one of its salts.

Types of buffer solutions.

i)  Acidic buffer solution.

This is the buffer solution which keeps the pH below 7.

They are formed by mixing a weak acid and its salt (of a strong base)

e.g.
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How does the buffer system work?

Consider

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O

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( )

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Since the salts is strong, it dissociates completely into increases the concentration of shifting the equilibrium to the left hand side suppressing the dissociation of acetic  acid due to common ion effect. Hence is equal to the salt concentration and due to the common ion effect becomes equal to the initial concentration of the acid.

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Therefore the solution will contain these important species.

i)A lot of unionized acid

ii)  A lot of acetate ions from

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iii)  Enough to make the solution acidic.

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When little Hare added, the following reaction occurs.

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®          The acetate ions concentration from the salt are large enough to consume the added hydrogen ions therefore there will be no accumulation of H+ in the solution.

®           If OH are added the following reaction occur ;

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This decrease in the solution, shifting the equilibrium to the right hand side to replace used to neutralize added. Therefore no       accumulation of in the solution.

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NOTE:

Addition of to acidic buffer increases the acid concentration, but decreases the salt concentration by the same amount of added.

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Addition of to acidic buffer decreases the acid concentration but increases the i.e salt by the same amount of added.

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Where

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According to Henderson Hesselbach

Example:

1.      buffer solution containing 1M of acid has a of 4.742.

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a)      Determine the salt concentration in buffer given

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b)      To 1 dm3 of a buffer, is added. Calculate the of the resulting buffer solution.

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c)      Calculate the when are added to 1dm-3 of the buffer.

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d)     Calculate the of the solution when are added to 1dmof water sample.

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e)      Calculate the of the solution when are added to 1dmof water sample.

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ANSWER:

1.      Solution

a)      From Henderson Hesselbach equation.

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]

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b)

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From

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c)

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From

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d)

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From

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e)

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SALTS

Salts

SALT HYDROLYSIS 

A salt is a compound which contain metallic or radicle or positive radicle rather than hydrogen (H+) and acidic or negative or anion radicle rather than hydroxyl ion (OH‾)

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CLASSIFICATION OF SALTS

Salts are categorized into four major classes.

These are:

i. Normal salt (strong salts).

ii. Salts with strong cation and weak anion.

iii. Salts with weak cation and strong anion.

iv. Salts with weak cation and weak Anion.

i) STRONG SALTS

These are salts with strong cation and strong anions.

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  ii).SALTS WITH STRONG CATION AN WEAK ANIONS.

These are also known as basic salts mostly are organic salts or salts of carboxylic acid

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    iii) SALTS WITH WEAK CATION AND STRONG ANION

These salts are termed as acidic salts

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iv.) SALT WITH WEAK CATION AND WEAK ANION

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Salt hydrolysis is the reaction between salt and water to give acid and base it is a reversible reaction of neutralization reaction.

In salt hydrolysis only weak ions react with water to give the respective products.

HYDROLYSIS OF CLASSES OF SALTS

1.     Hydrolysis of normal salts

The salts having strong cation and strong anion do not undergo Salt hydrolysis process, rather than ionizing in solution to give free ions.

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2.     Hydrolysis of basic salts

The salt having strong cation and weak anion undergo the type of hydrolysis termed as Anionic salt hydrolysis

Definition:

Anionic salt hydrolysis is the reaction between water and salt with strong cation and weak anion where by the weak anion react with water to give acid and base.

Example

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Initially the salt will ionize

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Anion will react with H2O

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Example:

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Ionization

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Anion will react with H2O

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iii) Hydrolysis of salts with weak cation and strong anion. This process is also known as cationic salt hydrolysis
Definition:

Cationic salt hydrolysis is the reaction between water and salt with weak cation and strong anion in which weak cation react with water to give acid and base.

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iv) Salt with weak cation and weak anion (weak salt )

This salt undergo both cationic and anionic salt hydrolysis because both weak ions will react with water to give acid and base.

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Hydrolysis constant for anionic salts hydrolysis

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From hydrolysis equation Kh can be obtained

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HYDROLYSIS CONSTANT (Kh)

Kh is the ratio of  product of molar concentration of the products to that concentration reactant raised to their power which is equal to the balancing number in a hydrolysis equation.

During hydrolysis the weak acid formed and water molecules will also dissociate.

The weak acid dissociation

where
ka = Dissociation constant of acid (CH3COOH)
kw = Dissociation constant of water
kh = Hydrolysis constant.

Word Image 293Hydrolysis constant for cationic salt hydrolysis

Consider the hydrolysis of NH4Br
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During hydrolysis the weak base is formed and water molecules will dissociate.

Weak base dissociation

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where
kb = Dissociation constant of Base

Example

With examples in each briefly write short note on the following

i)  Cationic salt hydrolysis

It is the reaction between water and salt with strong anion and weak cation in which the weak cation react with water to give acid and base.

Example

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ii)    Anionic salt hydrolysis

It is the reaction between water and salt with strong cation and weak anion in which the weak anion react with water to give acid and base.

Example

The above chemical reaction is Anionic salt hydrolysis
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(iii) Hydrolysis constant

Hydrolysis constant is the ratio of the products molar concentration to the reactant concentration raised to their powers which is equal to the balancing number in the hydrolysis equation.

Normally it is denoted by kh

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For anionic salt hydrolysis.


(iv) Acidic salt:

Acidic salt is the type of salt which contain a strong anion and weak cation.

Example:

(NH4)2SO4 , NH4Br and NH4I.


(V) Basic salt

Basic salt is the type of salt which contain a strong cation and a weak anion.

Example:

CH3COOK, Na2C2O4 and CH3CH2COONa.


(vi) Salt hydrolysis :

Salt hydrolysis is the reaction between salt and water to produce base and acid.

Under this process only weak radicals/ions are associated in the reaction.

Example

C2O42- + H2O …….. H2C2O4 + OH

NH4+ + H2O …………………. NH4OH + H+

Example

1. Briefly differentiate cationic salt hydrolysis and anionic salt hydrolysis.

Solution

Cationic salt hydrolysis is a reaction between water and salt with weak cation and strong anion where by the weak cation react with water to produce acid and base while

anionic salt hydrolysis is a reaction between water and salt with weak anion and strong cation in which the weak anion react with water to produce acid and base.

1. With an example of salt and type of hydrolysis derive the relationship between ka, kw and kh.

Solution

Consider the hydrolysis of Sodium Oxalate.

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During hydrolysis acid formed together with water molecules, ionize in a solution.

Acid ionization.

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Example

During Hydrolysis the base and water molecule also ionize.

Base ionization

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Water ionization

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Example 4:

Calculate the Hydrolysis constant for the hydrolysis of 0.05 M NH4NO3.if its Kb is 6.67 x10-6

Solution

Word Image 304 Word Image 305Word Image 306DEGREE OF HYDROLYSIS

Definition:

Is the fraction or percentage of the salts that has reacted with water to form acid and base

The degree of Hydrolysis is denoted by h.

Consider the anionic salt hydrolysis below:

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Hydrolysis Reaction:

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But

(1 – h) 1

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But c = 1/v.

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But

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Then

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This is for anionic salt hydrolysis

Where
h is the hydrolysis constant

Kw is the water dissociation constant

Ka is the acid dissociation constant

THE HYDROLYSIS OF SALT WITH WEAK CATION AND ANION

Consider the salt of weak cation and weak anion as AB. During hydrolysis, the salt will ionize.

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During the hydrolysis process there is formation of

(i) Weak acid

(ii) Weak base

Ionization of weak acid

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Ionization of weak base

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Ionization of water

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For salt that undergo both cationic and anionic salt hydrolysis (weak salt)

Example

Consider the hydrolysis of ammonium acetate, then derive the relationship between Kh,Kw,Ka and Kb for such hydrolysis.

Solution

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During the hydrolysis, weak acid and weak base is formed ionization of weak acid

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PH EQUATIONS FOR SALT HYDROLYSIS

a) FOR ANIONIC HYDROLYSIS

Consider the hydrolysis of CH3COONa

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Salts Salts

But, from water dissociation

Salts Salts

note
pH = -LOG[H+]

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This is the equation for anionic salt hydrolysis

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b) FOR CATIONIC HYDROLYSIS

Salts Salts Salts Salts

This is for the salt undergoing cationic salt hydrolysis

OR

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Example

a)     What do you understand by

I.         Salt hydrolysis

Is the reaction between water and saltl to produce acid and base.

II.       Cationic hydrolysis

Is the reaction between water and salt with strong anion and weak cation at which the weak cation react with water to give acid and base.

III.       Anionic hydrolysis

Is the reaction between water and salt with strong cation and weak anion in which the weak anion react with water to produce acid and base.

IV.       Conjugate pair

Is the pair of conjugate acid or conjugate base or acid respectively.

b. For cationic hydrolysis show that

Salts

Solution. Consider the hydrolysis of NH4CL

NH4H20 NHOH + H+

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1 – h h h

At equilibrium (1 – h)c hc hc

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But

Assume that 1 – h 1 (h is very small)

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calculate the conc. of H₃O⁺ present in 0.0001M HCOOK ka. HCOOK = 2.4 X 10-4

Kw = 10-14

Solution

HCOOK HCOO⁻ + K⁺

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Salts

Example 2:

a.     Define

pH is the negative logarithm of hydrogen molar concentration present in the solution.

Acidic salt is the type of salt that consist of strong anion and weak cation.

Basic salt is the type of salt that consist of strong cation and weak anion.

b.     Normal salt do not undergo anionic or cationic salt hydrolysis. Briefly account for this statement.

Soln.

Normal salt do not undergo neither anionic nor cationic salt hydrolysis because they have strong acidic and basic radicle and the hydrolysis take place to the weak radicles only. Hence they just ionize in the water only.

c.      Calculate the pH of the solution with 0.0152M. CH3COOK if KaCH3COOH =2.18 X 10-8 at a particular temperature.

Soln.

From pH equation.

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Alternatively

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d.     Calculate the pH of the solution with 0.001M of NH4NO3 if KNH4OH is 1.8 x 10-4.

Soln.

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Example 3:

Calculate the hydrolysis constant and the pH of 0.625M solution of CH3COONa. Kaof CH3C-OOH= 1.754 X 10-5 Kw =1 x 10-14.

Soln.

Salts

Salts

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E-xample 4.

a.     -What is the pof 0.2M solution of NaCN? Ka for HCN =4 X 10 -10 ionic product of -water =10-14.

Soln.-

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b.     -Calculate the pH of 0.2M NH4Cl. Kb = 1.8×10-5.

S-oln.

Salts Salts

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Also

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Amount dissociated = 0.0248

% hydrolysed = 0.0248 × 100%

= 2.48%

The percentage hydrolysis is 2.48%

 

B. calculate the hydrolysis constant and degree of hydrolysis of NH4Cl in 0.001m solution kb=1.8 × 10-5

Solution.

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The hydrolysis constant is 5.55 × 10-10

Salts

The hydrolysis constant is 7.45 × 10-4

C. calculate the degree of hydrolysis of ammonium acetate if the dissociation constant of ammonium hydroxide is 1.8 × 10-5 that of acetic acid is 1.8 × 10-4 and the ionic product of H2ois 1 × 10-14 (0.55 × 10-2 ).

Solution

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Data

Ka = 1.8 × 10-5

Kn = 1.8 × 10-5

Kw = 1 × 10-14

From hydrolysis equation

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Example 6.

What is hydrolysis constant of salt?


Answer

Is the ratio of products concentration to reactant concentration in moles/dm3 raised to their powers which is equal to the balancing number in a hydrolysis equation

Why aqueous solution of sodium carbonate is alkaline derive an expression for the hydrolysis constant and pH of this solution.

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