Buffer Solution: Definition, Equation, Formula, Questions and Examples

Buffer solutions, is vary important in building the pH stability in chemical and biological systems,This discovery was not done by ant single scientist at any particular time. The idea of buffers slowly grow with time and expand as per the need. The early work of the buffer solution were the idea of maintaining a stable pH can be go back to the 19th century. Buffer Solutions is a concept developed by Lawrence Joseph Henderson and Karl Albert Hasselbalch in the early 20th century.

This Story also Contains

1. Buffer Solution
2. ACTION OF BASIC BUFFER
3. Some Solved Examples
4. Summary

Henderson developed the Henderson-Hasselbalch equation in 1908, which depict the pH of a buffer solution in relation to the concentration of its acidic and basic components. Hasselbalch, in 1916, further defined his work, leading to the equation being named after both scientists. Why Buffer Solutions Are Important it is important because the buffers are important because they helps to maintain a relatively constant pH in a solution despite the addition of small amounts of acids or bases. This property is vital in many chemical processes, biological systems (such as blood), and industrial applications.

Buffer Solution

A solution whose pH does not change very much when H⁺(H₃O⁺) or OH^- are added to it is referred to as a buffer solution.
A buffer solution is prepared by mixing a weak and its salt having common anion(i.e HA + HB forms an acidic buffer) or a weak base and its salt having common cation(i.e BOH + BA forms a basic buffer).
It can be prepared to have a desired value of pH by controlling the amounts of acids and their salts in case of acidic buffer and of bases and their salts in basic buffer.

Acidic buffer:

$\quad \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$

Basic buffer :

$\quad \mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$


Consider an acidic buffer containing an acid HA and say common ions A^-. Now any H⁺ added to this solution within certain limits are neutralized by A^- ions as:
$\mathrm{H}^{+}+\mathrm{A}^{-} \rightleftharpoons \mathrm{HA}$
While the addition of OH^- ions externally (within certain limits) are neutralised by acid HA as:
$\mathrm{HA}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-}$
Hence in both the cases, effect of addition of H⁺ or OH^- is almost compensated for (i.e. pH almost remains constant).

Such a system (may be acidic or basic) finds enormous use not only in industrial processes but also most importantly in biological reactions. Like the pH of normal blood is 7.4 and for good health and even for the survival, it should not change below 7.1 or greater than 7.7, the body maintains it through a buffer system made of carbonate and bicarbonate ions and H₂PO₄^- and HPO₄^2-. Similarly, the pH of gastric juice is kept constant in order to operate good digestive functions.

Buffer solutions can be classified into three types:

(1) Acidic Buffer Solutions

Acidic buffer solutions are the solutions that are made from a weak acid and one of its salt with a strong base.

For example: Solution of $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COONa}$

It is to be noted that the pH of an acidic buffer may not be always less than 7. It depends upon the Ka values of the acid and also the concentration of the acid and the salt.

(2) Basic Buffer Solutions

Basic buffer solutions are the solutions that are made from a weak base and one of its salt with a strong acid.

For example: Solution of $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{NH}_4 \mathrm{Cl}$

It is to be noted that the pH of an basic buffer may not be more less than 7. It depends upon the Kb values of the base and also the concentration of the salt and base.

(3) Simple Buffer Solutions

Simple buffer solutions are the solutions that are made from the salt of a weak acid and weak base.

For example: Solution of $\mathrm{CH}_3 \mathrm{COONH}_4 \mid$

It is to be noted that the pH of simple buffer may be less than, greater than or equal to 7. It depends upon the Ka and Kb values of the acid and the base.

Buffer Action:

A buffer solution resists a change in its pH on addition of small amount of acid or base. This is because there is one component which can neutralise the acid and the other component can neutralise the base

e.g $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COONa}$

When small amount of base is added, then it is the acid which neutralises it

$\mathrm{OH}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{H}_2 \mathrm{O}+\mathrm{CH}_3 \mathrm{COO}^{-}$

When small amount of acid is added, then it is the acetate ion which neutralises it

$\mathrm{HCl}+\mathrm{CH}_3 \mathrm{COO}^{-} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}+\mathrm{Cl}^{-}$

as neutralisation occurs, the $\left[\mathrm{H}^{+}\right]$or $\left[\mathrm{OH}^{-}\right]$ does not alter much in the solution and pH change is almost negligible

Cases which are not a buffer solution

(1) Solutions of Strong Acid and its salt e.g. $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{KHSO}_4$

(2) Solutions of Strong Base and its salt e.g. NaOH and NaCl

For a solution to be classified as a buffer solution, there must be one weak acid or base and its respective conjugate base or acid

CALCULATION OF PH OF ACIDIC BUFFER SOLUTION

When a solution contains CH₃COOH and CH₃COONa, then the following equilibrium will be established:$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$

The equilibrium equation for the given system can be calculated using the following equation:

$
\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{[\mathrm{Salt}]\left[\mathrm{H}^{+}\right]}{[\mathrm{Acid}]}
$

$\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]$is the concentration of salt $\left[\mathrm{CH}_3 \mathrm{COOH}\right]$ is the initial concentration of acid

Rearranging the above equation, we get

$\begin{aligned} & {\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \frac{[\text { Acid }]}{[\text { Salt }]}} \\ & -\log _{10}\left[\mathrm{H}^{+}\right]=-\log _{10} \mathrm{~K}_{\mathrm{a}}-\log _{10}[\text { Acid }]+\log _{10}[\text { Salt }] \\ & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{\text { Acid }}\end{aligned}$

This equation is also known as the Henderson-Hasselbalch equation.

Some examples

• Find the pH of a solution having 0.1M CH₃COOH(K_a = 10^-5) and 0.2M CH₃COONa.
  
  We know that pH of a solution is given as:
  
  $\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{\text { Acid }} \\ & \text { Thus, } \mathrm{pH}=-\log _{10} \mathrm{~K}_{\mathrm{a}}+\log _{10} \frac{[0.2]}{[0.1]} \\ & \Rightarrow \mathrm{pH}=-\log _{10} 10^{-5}+\log _{10} 2 \\ & \Rightarrow \mathrm{pH}=5+0.30=5.30\end{aligned}$
• Find the pH of a solution containing 0.25 moles of HCN(K_a = 10^-5) and 0.10 moles of NaCN present in 1 litre solution.
  
  We know that pH of a solution is given as:

$\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{\text { Acid }} \\ & \text { Thus, } \mathrm{pH}=-\log _{10} \mathrm{~K}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]} \\ & \Rightarrow \mathrm{pH}=-\log _{10} 10^{-5}+\log _{10} \frac{0.10}{0.25} \\ & \Rightarrow \mathrm{pH}=5+\log _{10} \frac{2}{5} \\ & \Rightarrow \mathrm{pH}=5-0.39=4.6\end{aligned}$

BASIC BUFFER

Basic buffer solution contains a weak base and its salt with strong acid. Some examples of basic buffers are:

• NH₄OH + NH₄Cl
• NH₄OH + (NH₄)₂SO₄
• CH₃-NH₂ + [CH₃-NH₃⁺]Cl^-

The pH of the basic buffer is given as:

$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}$

We already know that pH = 14 - pOH. Thus can be calculated using this equation.

For example: basic buffer we have:

$\begin{aligned} & \mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{Cl}^{-} \\ & \mathrm{NH}_4 \mathrm{Cl} \rightarrow \mathrm{NH}_4^{+} \mathrm{Cl}^{-} \quad(\text { Strong electrolyte }) \\ & \text { Thus, } \mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}\end{aligned}$

In this system:

• [NH₄OH]: Initial concentration of [NH₄OH] is taken as at equilibrium negligible dissociation of NH₄OH is there because of common-ion effect.
• [NH₄⁺]: The concentration of NH₄OH is mostly from 100% dissociation of NH₄Cl.

Again, as we know:

$
\mathrm{K}_{\mathrm{b}}=\frac{[\mathrm{Salt}]\left[\mathrm{OH}^{-}\right]}{\text {Base }}
$

Thus, $\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{b}} \frac{[\text { Base }]}{[\text { Salt }]}$
Using $\log$ on both sides, we get :

$
\begin{aligned}
& -\log _{10}\left[\mathrm{OH}^{-}\right]=-\log _{10} \mathrm{~K}_{\mathrm{b}}+\log _{10}[\text { Salt }]-\log _{10}[\text { Base }] \\
& \text { Hence, } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}
\end{aligned}
$

ACTION OF BASIC BUFFER

Basic buffer solution contains equimolar quantities of a weak base and its salt with strong acid. For example: ammonium hydroxide i.e. NH₄OH and ammonium chloride i.e NH₄Cl.

On Adding Acid: H⁺ release and combines with OH^- of base.

On Adding Base: OH^- releases and combines with NH₄⁺ of salt.

• On adding acid to the basic buffer, its H⁺ ions react with OH^- ions of the base and forms H₂O. Thus, in this case, solution feels that its [OH^-] has decreased, thus to neutralize this effect, NH₄OH dissociate in small amounts and gives [OH^-] so as to restore concentration of [OH^-]
• On adding base to the basic buffer, its [OH^-] ions react with NH₄⁺ ions and forms NH₄OH. In this case, the solution feels that its NH₄OH concentration is increased. Thus, in this case, the reaction will not proceed forward because of common ion effect.

Recommended topic video on (Buffer Solution )

Some Solved Examples

Example.1

1.In some solutions, the concentration of H₃O⁺ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as :

1) Ideal solutions

2) Colloidal solutions

3) True solutions

4) (correct) Buffer solutions

Solution

The solution which resists the change in pH on dilution or with the addition of a small amount of acid or base is called a buffer solution.

Hence, the answer is the option (4).

Example.2

2.Fear and excitement generally cause one to breathe rapidly and it results in the decrease of CO₂ concentration in blood. In what way will it change the pH of blood?

1)pH will increase

2)pH will decrease

3)pH will adjust to 7

4) (correct)No change

Solution

pH of blood remains same because it has a buffer solution of $\mathrm{H}_2 \mathrm{CO}_3 / \mathrm{HCO}_3^{-}$
Hence, the answer is the option(4).

Example.3

3.The pK_a of a weak acid HA is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is

1)7.0

2)4.5

3)2.5

4) (correct)9.5

Solution

Let us consider the dissociation equilibrium of the acid HA

$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$

Writing the expression for the equilibrium constant

$\mathrm{k}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$

When the acid is 50% dissociated, $\left[\mathrm{A}^{-}\right]=[\mathrm{HA}]$

$\begin{aligned} & \therefore\left[\mathrm{H}^{+}\right]=\mathrm{k}_{\mathrm{a}} \Rightarrow \mathrm{pH}=\mathrm{pk}_{\mathrm{a}} \\ & \text { Given } \mathrm{pK}_{\mathrm{a}}=4.5 \quad \therefore \mathrm{pH}=4.5 \\ & \therefore \mathrm{pOH}=14-4.5=9.5\end{aligned}$

Hence, the answer is the option (4).

Example.4

4.What volume of 0.1 M sodium formate solution should be added to 50 ml of 0.05 M formic acid to produce a buffer solution of pH = 4.0; pK_a of formic acid is 3.80?

1) (correct)39.62 mL

2)39.62 L

3)396.2 mL

4)396.3 L

Solution

Suppose, Vml of 0.1 M HCOONa is mixed to 50 ml of 0.05 M HCOOH

$
[\text { Molarity }]=\frac{\text { Total millimole }}{\text { Total volume }}
$

In mixture $[\mathrm{HCOONa}]=\frac{0.1 \times \mathrm{V}}{(\mathrm{V}+50)}$

$
\begin{aligned}
& {[\mathrm{HCOOH}]=\frac{50 \times 0.05}{\mathrm{~V}+50}} \\
& \mathrm{pH}=-\log \mathrm{Ka}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 4.0=3.80+\log _{10} \frac{(0.1 \times \mathrm{V})}{(\mathrm{V}+50) / 2.5 /(\mathrm{V}+50)} \\
& \mathrm{V}=39.62 \mathrm{ml}
\end{aligned}
$

Hence, the answer is the option (1).

Example.5

5. An acidic buffer is obtained by mixing:

1)100 mL of 0.1 M CH₃COOH and 100 mL of 0.1 M NaOH

2) (correct)100 mL of 0.1 M HCl and 200 mL of 0.1 M CH₃COONa

3)100 mL of 0.1 M CH₃COOH and 200 mL of 0.1 M NaOH

4)100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl

Solution

An acidic buffer contains a weak acid and its salt.

Here, Given the weak acid is CH₃COOH and its salt will be CH₃COONa

So,100 mL of 0.1 M HCl = 10 meq HCl

200 mL of 0.1 M CH₃COONa = 20 meq of CH₃COONa

After mixing 10 meq of HCl react with 10 meq of CH₃COONa then 10meq of CH₃COOH form.

But from other options, the same amount of weak acid and its salt cannot be obtained.

Finally, 10 meq of CH₃COOH and 10 meq of CH₃COONa will be present.

So upon mixing 100ml of 0.1M HCl and 200ml of 0.1M CH₃COONa, a buffer solution will be obtained.

Hence, the answer is the option (2).

Summary

The main preference of a buffer solution is to take control the pH level reasonably stable even when modest amounts of acids or bases are added. Since many chemical and biological processes are sensitive to pH variations, this is significant. The weak acid and its conjugate base, or the weak base and its conjugate acid, cooperate in a buffer solution to neutralize additional acids or bases and reduce pH variations.