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

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 :

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

Types of Buffer Solution

Buffer solutions can be classified into three types:

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.

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.

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 K_a and K_b 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

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

1. Find the pH of a solution having 0.1M CH₃COOH(K_a = 10^-5) and 0.2M CH₃COONa.

Solution: We know that pH of a solution is given as:

$\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{\text { Acid }} \\ & \text { Thus, } \mathrm{pH}=-\log _{10} \mathrm{~K}_{\mathrm{a}}+\log _{10} \frac{[0.2]}{[0.1]} \\ & \Rightarrow \mathrm{pH}=-\log _{10} 10^{-5}+\log _{10} 2 \\ & \Rightarrow \mathrm{pH}=5+0.30=5.30\end{aligned}$

2. 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.

Solution: We know that pH of a solution is given as:

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

Calculation Of pH Of Basic Buffer Solution

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

• NH₄OH + NH₄Cl
• $\mathrm{NH}_4 \mathrm{OH}+\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$
• $\mathrm{CH}_3-\mathrm{NH}_2+\left[\mathrm{CH}_3-\mathrm{NH}_3^{+}\right] \mathrm{Cl}^{-}$

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 :

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

Some Solved Examples

Question 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 correct answer is option (4).

Question 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 correct answer is option (4).

Question 3: The pKa 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 correct answer is option (4).

Question 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; pKa 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 correct answer is option (1).

Question 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 correct answer is option (2).