Hydrolysis of salts

How is a salt formed? When an acid and a base react, H+ of the acid combines with OH- of the base and they neutralise each other, while remaining anion of the acid and cation of the base combine to form the salt. So we can form different categories of salts on the basis of parent acid-base:

• Salt of strong acid and strong base
• Salt of strong acid and weak base
• Salt of weak acid and strong base
• Salt of weak acid and weak base

When we put a salt in water, it dissolves in water. Polar force of water weakens the electrostatic attraction which binds the anion and cation of the salt and separates them into ions. Water molecules hydrate these ions and keep them separated. Sometimes these ions react with water molecules, anions try to polarise water molecule and if they succeed to create enough polarization it may cause the breakage of O-H bond of water. This process is called the hydrolysis.

Hydrolysis of salt

In this post we will discuss different categories of salts and see what happens when they are dissolved in water.

Let’s discuss the first category: Salt of strong acid and strong base, like NaCl. NaCl is made by the reaction of HCl and NaOH. When it is dissoved in water it dissociates to form Na⁺ and Cl^-which then react with water molecule and reproduce their parent acid and base. Since both parent acid and base are equally strong, they neutralise each other and the pH of water remains unchanged. So it may seem that such salts don’t cause hydrolysis.

NaOH + HCl ⟶ NaCl + H2O
Na⁺_(aq) + H2O ⇌ NaOH + H⁺
Cl^-_(aq)  + H⁺ ⇌ HCl

Salt of strong acid and weak base: NH4Cl is made from acid HCl and base NH₄OH. In water NH₄Cl dissociates completely in NH4⁺ and Cl^- ions. NH₄⁺ ions successfully polarize water molecule and cause hydrolysis.

NH₄OH + HCl ⟶ NH₄Cl + H₂O
NH₄⁺_(aq)  + H₂O ⟶ NH₄OH + H⁺

Ammonium ion (NH₄⁺) reacts with water and form ammonium hydroxide (NH₄OH) and H⁺. NH₄OH is a weak base which dissociates a little. Thus OH^- ions of water are consumed by NH₄⁺ and H⁺ reacts with Cl^-ions.

Cl^- _(aq) + H⁺ ⇌ HCl

HCl is a strong acid and it dissociates completely. That means H⁺ions produced by hydrolysis of NH₄Cl salt remain in solution while OH^- ions get trapped by NH₄⁺. This higher concentration of H⁺ ions makes the solution acidic.

Salt of weak acid and strong base: CH₃COONa is a salt of weak acid CH₃COOH and strong base NaOH. When it is dissolved in water it dissociates completely.  

CH₃COONa + H₂O ⟶ CH₃COO^-+ Na⁺
Na⁺ + H₂O ⇌ NaOH + H⁺
CH₃COO^- + H⁺ ⟶ CH₃COOH

Na⁺ ions cause hydrolysis and produce NaOH and H⁺ions. NaOH is a strong base and it dissociates completely thus OH^-ions of water remain in the solution. And H⁺ ions react with acetate ion (CH3COO^-) and form acetic acid, which is a weak acid. It dissociates partially so H⁺ ions of water get trapped and the pH of solution increases.

Salt of weak acid and weak base: CH₃COONH₄is a salt of weak acid CH₃COOH and weak base NH₄OH. When it is dissolved in water, it dissociates completely.  

CH₃COOH + NH₄OH ⟶ CH₃COONH₄+ H₂O
CH₃COONH₄ + H₂O ⟶ CH₃COO^-+ NH₄⁺

NH₄⁺ ions cause hydrolysis and produce weak base NH₄OH and H⁺ ions. And when H⁺ ions react with CH₃COO^- ions, they produce weak acid CH₃COOH. The resulting acid and base both are weak and dissociate partially. It means, the concentration of both H⁺ and OH^- gets affected and to calculate the pH of such solution we have to know the values of pK_aand pK_b.

pH = 7+ ½ (pK_a - pK_b)

And by using the above equation we can calculate the pH of such solutions. We have learnt how salts affect the pH of water. But not everything is completely soluble in water, there are a number of substances which are slightly soluble or insoluble in water. So what are the factors which decide the solubility of any substance? In the next post we will learn about solubility and see if it is possible to increase or decrease the solubility of any substance. 

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How to Prepare Buffer Solutions?

In the last post we have learnt that a weak acid and its conjugate base are the main constituents of a buffer solution . There are three ways to prepare a buffer solution.

1.      Weak acid and its salt
2.      Weak acid and strong base
3.      Salt of weak base and strong acid

Each of these ways will give us a pair of weak acid and its conjugate base. Let us see how.

Method I: Prepare a Buffer solution by weak acid and salt: for example we will take acetic acid (pK_{a ­­­}= 4.7) and salt sodium acetate.

CH₃COOH  ⟶  H⁺ + CH₃COO^-

Acetic acid is a weak acid so it dissociates less. It makes the acidic part of buffer.

CH₃COONa  ⟶  Na⁺ + CH₃COO^-

Sodium acetate is a salt and it completely ionizes in water and makes the basic part of buffer which is conjugate base of acetic acid.

Method II: Prepare a Buffer solution by Weak acid and strong base: for example we will take acetic acid (pK_{a ­­­}= 4.7) and NaOH.

As you have seen in the previous example, acetic acid makes the acidic part of buffer. NaOH is a strong base and it dissociates completely.

NaOH  ⟶  Na⁺ + OH^-

When it reacts with acetic acid it neutralised it and produces water and the salt sodium acetate.

CH₃COOH + Na⁺ + OH^-  ⟶  CH₃COONa + H₂O

Salt sodium acetate ionises completely in water and makes the basic part of buffer which is conjugate base of acetic acid.

Method III: Prepare a Buffer solution by Salt of weak base and strong acid: if we take sodium acetate salt and HCl let’s see what happens.

CH₃COONa  ⟶  Na⁺ + CH₃COO^-

Sodium acetate is a salt and it ionises completely in water and makes conjugate base of acetic acid.

CH₃COO^- + HCl  ⟶  CH₃COOH + Cl^-

And when HCl reacts with the conjugate base, the base accepts proton given by HCl and produces acetic acid. This makes the acidic part of buffer and so on.

You have seen that how every way reaches to the one common point. Let’s try to prepare 1M buffer of acetic acid/ sodium acetate with pH 4, pK_a of acetic acid is 4.76.

To prepare this buffer we have to find out the desired concentration of acidic form - acetic acid, and basic form - acetate ions.

By using Henderson-Hasselbalch equation:

pK_a  = pH + log [acidic form]/ [Basic form]

4.76 = 4 + log [CH₃COOH]/ [CH₃COO^-]

0.76 = log [CH₃COOH]/ [CH₃COO^-]

10^0.76 = [CH₃COOH]/ [CH₃COO^-]

5.75 = [CH₃COOH]/ [CH₃COO^-]

5.75[CH₃COO^-] = [CH₃COOH]

Concentration of acetic acid/sodium acetate buffer solution is 1M, that means:

[CH₃COOH] + [CH₃COO^-] = 1M

5.75[CH₃COO^-] + [CH₃COO^-] = 1M

6.75[CH₃COO^-] =1M

[CH₃COO^-] = 0.15M = 15mmol

So,

[CH₃COOH] = 1-0.15 = 0.85M = 85mmol

Now we have the desired concentrations of both constituents. Lets see how to prepare the buffer by each of the methods:

Methods to Prepare Buffer Solutions

I. Weak acid and its salt: For this method, we will use 85mmol of acetic acid and 15 mmol of sodium acetate to prepare buffer with pH 4.

II. Weak acid and strong base: For this method, we will need acetic acid and NaOH. When acetic acid reacts with NaOH, it gets neutralized by it and produces conjugate base

CH₃COOH + NaOH  ⟶  CH₃COONa + H₂O

CH₃COONa  ⟶  CH₃COO^- + Na⁺

That means here NaOH acts as a source of conjugate base so, the concentration of NaOH will be equal to the desired concentration of conjugate base (acetate ion). We will need 15mmol NaOH and 85mmol acetic acid to prepare this buffer.

III. Salt of weak base and strong acid: For this method, we will need sodium acetate and HCl. When they reacts:

CH₃COONa  ⟶  CH₃COO^- + Na⁺

CH₃COO^- + HCl  ⟶  CH₃COOH + Cl^-

HCl gives proton to acetate ion and produces acetic acid. Thus HCl acts as the source of acidic part of the buffer. That means we will need HCl equal to the desired concentration of acetic acid. To prepare this buffer we will need 85mmol HCl and 15mmol sodium acetate.

Now you have learnt how to prepare buffer solution by different methods. Let’s solve some problems to build better understanding.

How would you make 100 ml buffer solution with a pH 4 that is 0.3M in acetic acid and 0.2M sodium acetate using a 1M acetic acid solution and 2M CH₃COONa solution.

First we have to calculate how many moles are present in 100ml buffer solution, when concentration of acetic acid is 0.3M

M = mmol/ ml

0.3M = mmol/ 100ml

= 30mmol

Now we know that we need 30 mmol of acetic acid to prepare desired buffer. Now calculate the amount of 1M acetic acid solution to get 30mmol.

M = mmol/ ml

1M = 30mmol/ ml

= 30ml

Similarly we have to do calculations for sodium acetate. Find out the number of mmol present in 100ml buffer, when its concentration is 0.2M

M = mmol/ ml

0.2M = mmol/ 100ml

= 20mmol

Now calculate the amount of 2M sodium acetate solution to get 20mmol.

M = mmol/ ml

2M = 20mmol/ ml

= 10ml

That means to prepare desired buffer we need to mix 30ml acetic acid solution and 10 ml sodium acetate solution and make it up to 100 ml by mixing remaining 60ml water.

I hope these posts have helped you to understand the mystery of buffers but if you have any doubts, please feel free to leave a comment. 

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Buffer solutions

In previous posts you have learnt that dissociation constant (K_a and K_b) tells us the strength of acid or base while the strength of solution is determined by pH. Do you know that milk, water, blood, fruit juices all have a particular pH. Can you guess what happens if it gets disturbed? It will spoil the nature and taste. And it can be fatal if pH change occurs to blood. Blood transports oxygen to all the cells of the body and its normal pH is 7.35 to 7.45. If its pH decreases or increases even for few seconds, it will cause death of that person. But blood has a system to deal with any pH change and this system maintains the pH of blood. This system is called Buffer.

A buffer solution maintains the pH of a solution even if you add H⁺ or OH^- ions to it. In this post we will see how it works? 

Buffer solution is made up of weak acid and its conjugate base. If we add OH^- to the solution, weak acid gives proton to neutralize them and if we add H⁺ to the solution, conjugate base accept those protons to nullify the excess H⁺concentration. Thus, the addition of small amount of H⁺ or OH^- ions doesn’t affect the pH of the solution. Mostly buffer solutions can maintain the pH by ±1. Let’s take an example of Brönsted acid-base couple to understand the working of a buffer solution:

HA ⇌ H⁺ + A^-

Let’s write the equation for acid dissociation constant:

K_a = [H⁺] [A^-] / [HA]

Taking log of both side of equation,

Log K_a = log [H⁺] + log [A^-] / [HA]

On multiplying by -1,

-log K_a = -log [H⁺] - log [A^-] / [HA]

Or

-log K_a = -log [H⁺] + log [HA]/ [A^-]

pK_a  = pH + log [HA]/ [A^-]

here [HA] represents the acid concentration and [A^-] represents the concentration of basic form. This is the Henderson-Hasselbalch equation. It can be used to determine the pH of buffer solution. It also tells us whether a compound will exist in its acidic form or basic form at a particular pH. 

Lets take the example of CH₃COOH, its pK_a is 4.7. Let’s find out in which form it will exist at pH 5.

CH₃COOH ⇌ H⁺ + CH₃COO^-

pK_a  = pH + log [acidic form]/ [Basic form]

4.7 = 5 + log [CH₃COOH]/ [CH₃COO^-]

-0.3 = log [CH₃COOH]/ [CH₃COO^-]

10^-0.3 = [CH₃COOH]/ [CH₃COO^-]

0.5 = [CH₃COOH]/ [CH₃COO^-]

It means that the basic form of acetic acid [CH₃COO^-] will be present in excess at pH 5 or you can say that at pH 5 acetic acid exists in its basic form. There is another way to get your answer fast:

pK_a  = pH + log [acidic form]/ [Basic form]

pK_a  - pH = log [acidic form]/ [Basic form]

pK_a  - pH = log [acidic form] - log [Basic form]

On comparing both side,

pK_a ≈ log [acidic form]

And

pH ≈ log [Basic form]

So; if (pH > pK_a) that means basic form > acidic form

If (pH < pK_a) that means basic form < acidic form.

Now you can calculate pH of a buffer solution by using Henderson-Hasselbalch equation. Buffer solution is made up of weak acid and its conjugate base. 

Let’s calculate the pH of buffer solution prepared by mixing 20ml of 0.1 M formic acid and 15ml of 0.5 M sodium formate, pK_a of formic acid is 3.75.

HCOOH ⥃ H⁺ + HCOO^-

Formic acid is a weak acid so it dissociates less. It makes the acidic part of buffer.

HCOONa ⟶ Na⁺ + HCOO^-

Sodium formate is a salt and it ionises completely in water and makes the basic part of buffer.

First we have to calculate the concentration of formic acid and sodium formate in the buffer.

The relation between M, mmol and ml is as follows and we can calculate the number of mmoles or mls by simply putting the values of all the other variables in it and solving the equation.

M = Mole/ Lit = mmol/ml

So how many moles are present in 20ml of 0.1 M formic acid?

0.1M = mmol/ 20ml = 0.1 × 20

=2mmol of formic acid

And how many moles are present in 15ml of 0.5 M sodium formate?

M = mmol/ml

0.5 = mmol/15ml = 0.5 × 15

= 7.5mmol of sodium formate

Now place the values in Henderson-Hasselbalch equation to calculate the pH of buffer solution:

pK_a  = pH + log [acidic form]/ [Basic form]

3.75 = pH + log [2]/[7.5]

3.75+0.57 = pH

pH =4.32

as pH is greater than pK_a , basic form will exist in majority.

There are three ways to prepare a buffer solution.

1.      Weak acid and its salt.
2.      Weak acid and strong base

3.      Salt of weak base and strong acid

In the next post we will discuss all these ways to prepare buffer solutions.

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Inductive effect

In the previous post I gave you some compounds and asked you to arrange them in the order of their acidity. They were CH₃COOH, Br CH₂COOH, ICH₂COOH, ClCH₂COOH and FCH₂COOH, let’s examine them.

To compare their acidity, we have to look inside their structure. Let’s draw the Kekulé structure of these compounds. Now you must be able to find out the acidic H in these molecules. When you compare all these molecules you will find that each has one different electronegative element. This electronegative element is connected to acidic H through a series of sigma bonds. As you know that the electron density of a sigma bond is maximum, along its axis, so it will be easier for electronegative element to pull electron density from O-H bond. This withdrawal of electron density is called the inductive effect. This makes the O-H bond electron deficit and so molecule easily gives off its proton (H⁺).

Inductive Effect and Acidity of molecule

 The same inductive effect shares the burden of extra negative charge of O when conjugate base is which increases the stability of the base. Stable bases are weak bases which rarely accept proton, so its conjugate acid is stronger. Or as we say; the stable structures are formed faster so the acid gives off proton faster because it will produce a more stable conjugate base.

Inductive effect

Strength of Inductive effect depends on the power of the element to pull electron density. It means more electronegative element creates stronger inductive effect which increases the acidity of molecule.

I hope you have understood the Inductive effect and how it affects the acidity of the molecule. Let’s compare the acidity of CH₃CH₂CHClCOOH and CH₃CHClCH₂COOH.

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The effect of structure on pKa

We have learnt that the strength of an acid is determined by its readiness to gives off proton (H⁺). The secret of strength lies in the structure of the acid; a structure which encourages it to give off proton readily makes the acid stronger. Everything in nature prefers to go in the direction of stability; similarly an acid gives off proton to make a more stable base. Therefore, the stability of its conjugate base decides the strength of a particular acid. Let’s try to understand it:

Let’s compare the acidic strength of CH₃OH and NH₃. First, draw the Kekulé structures of these compounds:

Acidic Hydrogen

In CH₃OH, there are 4 H atoms but only one can be given by the molecule. H atom directly attached to the electronegative atom is the acidic H, which can be given by the molecule. If CH₃OH gives off a proton its conjugate base will be CH₃O^- and similarly NH₂^- will be the conjugate base of NH₃. Now we will draw the structure of these conjugate bases and try to compare their stability.

You can see that the negative charge is on the electronegative elements of both the conjugate bases. You know that negative charge is best carried by an electronegative element and so the efficiency of electronegative atoms to carry this negative charge decides the stability of conjugate bases. Therefore, we have to compare electronegativities of these elements and since Oxygen is more electronegative than Nitrogen, the conjugate base CH₃O^- will be more stable than NH₂^-.

“Stable structure is formed faster”, that’s why CH₃OH gives off proton more readily than NH₃ and is stronger acid than NH₃.

Now we can summarize our findings as:

More electronegative element stabilizes the conjugate base and results in higher acidity of the compound.

To compare the acidity of different acids, follow these steps:

1. Find the Hydrogen directly attached to the electronegative element
2. Compare the electronegativities of electronegative elements
3. Stability of conjugate base is the same as the order of electronegativity. A base with higher electronegative element is more stable.
4. More stable conjugate base will give the stronger acid. Because, Stable bases are less likely to accept proton that means these are weaker bases and we know that weak conjugate base has corresponding strong acid.

Let’s compare acidity of CH₄, NH₃, HF and H₂O.

Electronegativity order will be: C < N < O < F

Stability of conjugate base will be: CH₃^- < NH₂^- < OH^- < F^-

Therefore, the order of acidity will be: CH₄ < NH₃ < H₂O < HF

We have learnt to compare the strength of acids by comparing their electronegativity. Let’s try to arrange the following in the order of their acidity of HF, HCl, HBr and HI.

If you compare their electronegativity you will find:

I < Br < Cl < F

But when you compare their charge carrier ability, you have to consider their size too. Because when there is a question to balance the extra charge, size matters. The bigger the element, the better it is to accommodate the charge. So this time the order of stability of conjugate base will be:

F^- < Cl^- < Br ^- < I^-

Hence the order of acidity will be:

HF < HCl < HBr < HI

In today's post, you have understood how electronegativity and size of the element contribute in the stability of conjugate base and affect the acidity of the molecule. Now try to compare the acidity of CH₃COOH, BrCH₂COOH, ICH₂COOH, ClCH₂COOH and FCH₂COOH. I will give you the answer in the next post.

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Relation between Ka and Kb

In the last post we have seen that dissociation constant of acid K_a is directly proportional to the H⁺ concentration and  dissociation constant of base K_b is directly proportional to the OH^-concentration. As we know that H⁺  and OH^- are related to each other too, so there must be some relation between K_a and K_b.

To find out their relation we have to study a reaction in which we can get both K_a and K_b, so that we can compare them with each other. And such a reaction can be provided by Brönsted acid base pair, so let’s take an example of Brönsted acid base pair:

NH_3(aq) + H₂O_(l) ⇌ NH₄⁺_(aq)  + OH^- _(aq)

In this reaction NH₃ acts as base and it’s conjugate acid is NH₄⁺. If we consider forward reaction, we can get the equation for K_b:

NH_3(aq) + H₂O_(l) ⟶ NH₄⁺_(aq)  + OH^- _(aq)        ---------------(1)

K_b = [NH₄⁺] [OH^-]/ [NH₃]

If we write a reaction for dissociation of acid NH₄⁺, we can get the following equation for K_a:

NH₄⁺_(aq)  + H₂O_(l) ⟶ H₃O⁺_(aq) + NH_3(aq)       ---------------(2)

K_a = [H₃O⁺] [NH₃]/ [NH₄⁺]

Relation between pKa and pKb

If we add equation 1 and 2, we will get a new equation:

2H₂O_(l) ⟶ H₃O⁺_(aq) + OH^- _(aq)

This is the dissociation reaction of water we have studied before and we know that:

K_w = [H₃O⁺][OH^-]

Now you can see that if we multiply K_a and K_b we will get K_w

K_a × K_b = {[H₃O⁺] [NH₃]/ [NH₄⁺]}{[NH₄⁺] [OH^-]/ [NH₃]}

K_a × K_b = [H₃O⁺] [NH₃] [NH₄⁺] [OH^-]/[NH₄⁺][NH₃]

K_a × K_b = K_w

If we take (–log) of both sides, we will get:

pK_a  + pK_b = pK_w =14

A very important conclusion can be drawn from the above equation. If pK_a of an acid is lower then its conjugate base must have higher pK_b and vise versa, which means strong acid has a weak conjugate base.

We know that smaller the pK_a, the stronger the acid. Very strong acids have pK_a less than 1, moderately strong acids have pK_ain between 1 to 5 and weak acids have pK_a in between 5 to 14.

What is the difference between pH and pK_a?

Always remember that there is an important difference between pH and pK_a, we use pH scale to measure the acidity and pK_avalue indicates the strength of an acid. The pH is the characteristic of a solution, it means we can get solutions of different pH by dissolving the same acid in different quantities, like 1×10^-2 M solution of HCl has pH 2 and 1×10^-4 M solution of HCl has pH 4(HCl is a strong acid which dissociates completely i.e. its α is 1). On the other hand, pK_a is the characteristic of the particular compound, for example, pK_a of HCl is -7, HF is 3.5×10^-4 and pK_a of HCN is 4.9×10^-10. It tells us how readily the compound gives up a proton H⁺. By pK_a value you can also calculate the K_c

Relation Between equilibrium constant and pka

NH_3(aq) + H₂O_(l) ⇌ NH₄⁺_(aq)  + OH^- _(aq)

K_c = [NH₄⁺][OH^-]/[NH₃][H₂O]               -----------(3)

If we write equation for reactant acid H₂O:

H₂O_(l) ⟶ H⁺_(aq) + OH^- _(aq)

K_{a (Reactant acid)} = [H⁺][OH^-]/[ H₂O]            -----------(4)

If we write equation for product acid NH₄⁺

NH₄⁺_(aq)  + H₂O_(l) ⟶ H⁺_(aq) + NH_3(aq)            

K_{a (Product acid)} = [H⁺] [NH₃]/ [NH₄⁺]          -----------(5)

When we compare equation 3, 4 and 5, we can infer that:

K_c = K_{a (Reactant acid)} / K_{a (Product acid)}

Now you are able to measure the strength of an acid, but what are the factors which make an acid strong or weak? Is it something which is hidden in its structure? In the next post we will try to reveal its secret.​

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