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.

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

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.

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

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.

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

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.​

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

Ionization of weak acids and bases

Weak acids and bases are miser kind of species, because even though they have H⁺ or OH^- ions, they don’t give them quickly and when they do give them, they only give a part of them. When weak acids/bases are dissolved in water they partly dissociate or ionize. To know how much H⁺ or OH^- ions they will release in water, we need to study their ionization reactions.

Let’s take an example of a weak acid HX and study its ionization in water:

HX_(aq) + H₂O_(l) ⇌ H₃O⁺_(aq)  + X^-_(aq)

If we have taken c mol/lit of HX initially at time t=0, when the concentration of H₃O⁺ and X^- were 0 and only a fraction of moles (α) underwent dissociation.

Suppose out of 1 mole of acid, α mole of acid undergoes dissociation.

Then from c moles of acid (c.α) will be dissociated.

At time t, (cα) moles of HX dissociate and produce (cα) moles of H₃O⁺ and (cα) moles of X^-. So at the time of equilibrium HX is left with (c-cα) moles/lit and H₃O⁺ and X^- each has (cα) moles/lit.

Dissociation Constant and Degree of Dissociation

Relation between Degree of Dissociation α and Dissociation Constant K_a:

Now we will calculate the equilibrium constant K_a  as it is the ionization reaction of an acid:

K_a = (cα) (cα)/ (c-cα)

K_a = cα² / (1- α)

K_a is the ionization or dissociation constant of acid HX and α is the degree of dissociation or the extent of ionization.

If we write equation of K_a in terms of molar concentration, we will get:

K_a = [H₃O⁺] [X^-]/ [HX]

Or

K_a = [H⁺] [X^-]/ [HX]

As you can see here that K_a is directly proportional to the H⁺ concentration, which means acids which have higher K_a value are stronger.

Similarly you can calculate the equilibrium constant K_b for a weak base. Let’s take an example of a weak base MOH and study its ionization in water:

MOH _(aq) ⇌ M⁺_(aq)  + OH^-_(aq)

If the initial concentration of MOH is c mole/lit and degree of dissociation is α, then at equilibrium MOH is left with (c-cα) moles/lit and M⁺ and OH^- each has (cα) moles/lit. So the K_b will be:

K_b = [M⁺] [OH^-]/ [MOH]

Or

K_b = cα² / (1- α)

K_b is directly proportional to the OH^- concentration, which means bases which have higher K_b value are stronger.

In the next post we will see if there is any relation between K_a and K_b.

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.