Question

Calculate the pH of $0.5{{ }}L$ of a $0.2M\;N{H_4}Cl - \;0.2M\;N{H_3}$ ​ buffer before and after addition of (a) $0.05$mole of $NaOH$and (b) $0.05$ mole of $HCI$. Assume that the volume remains constant.
[Given: $pKb{{ }}of{{ }}N{H_3} = 4.74\;$ ]

Answer 1

a buffer solution is known for its ability to resist a change in its own $pH$even after addition of an acid or base. Using the values of $pKb$ and the concentrations of the buffer solution we can find the $pH$ of the solution through the formula,
$pOH = pKb + \log \left[ {\dfrac{{salt}}{{base}}} \right]$
where $pOH$ and $pKb$ is the measure of the basicity of solution
$\left[ {\dfrac{{salt}}{{base}}} \right]$ is the concentration of both salt and base.
$pH = pKa + \log \left[ {\dfrac{{salt}}{{acid}}} \right]$
$pH$ and $pKa$ is the measure of acidity

Formula used: Where the concentration of salt and base and acid is required.
$pOH = pKb + \log \left[ {\dfrac{{salt}}{{base}}} \right]$

Complete step by step answer:
Buffer solution is an important mixture that can remain at the same level of pH despite the addition of an acid or base in small quantities. There are two types of buffer solutions:
The acidic buffer solution which consists of a weak acid and the salt of this weak acid. It has a pH below $7$ . It has a very important role in blood in the regulation of the blood.
The basic buffer consists of a weak base and the salt of this weak base. It has a $pH$above $7$ .
The above question has three parts. It will be helpful to solve it one by one systematically:
The first part is to find the $pH$of the solution before addition of any external base or acid.
This can be found using the following steps:
$pOH = pKb + \log \left( {\dfrac{{salt}}{{base}}} \right)$
It is given that :
$pKb{{ }}of{{ }}N{H_3} = 4.74\;$
The concentration of salt ammonium chloride = $0.2M$
Concentration of base, $N{H_3}$ = $0.2M$
On substituting the values , we get

$\Rightarrow 4.74 + \log \left[ {\dfrac{{0.2}}{{0.2}}} \right]$
Dividing the numerator and denominator, we get,
$\Rightarrow 4.74 + \log \left[ 1 \right]$
$pOH = 4.74$
Now that we have found pOH we can find pH using the steps below:
$pH + pOH = 14$
$pH = 14 - pOH$
Plugging in the value of pOH, we get,
$pH = 14 - 4.74$
$pH = 9.26$
Therefore, the answer to the first part is $9.26$.
This step involves addition of base $NaOH$. The pOH is obtained first using the steps mentioned below. On finding pOH we can find the pH later.
the concentration of base, $NaOH$ = $0.05M$
concentration of salt, $N{H_4}Cl$ = $0.2M$
substituting these values, we get,
$pOH = 4.74 + \log \left[ {\dfrac{{0.2}}{{0.05}}} \right]$
plugging in the values of concentration of salt and base, we get,
$\Rightarrow 4.74 + \log \left[ 4 \right]$
$\Rightarrow 4.74 + 0.63$
$\Rightarrow 5.37$
To find the pH, we need to follow the steps below,
$pH + pOH = 14$
$pH = 14 - 5.37$
The answer we get is,
$pH = 8.63$
The answer to the question will be $8.63$on addition of base.
The next step is adding acid $HCl$. The pH can be derived as follows:
$pH = pKb + \log \left[ {\dfrac{{salt}}{{acid}}} \right]$
concentration of salt, = $0.2M$
concentration of acid $HCl$ = $0.05M$
plugging these values we get,
$pH = 4.74 + \log \left[ {\dfrac{{0.2}}{{0.05}}} \right]$
After dividing the two concentrations of salt and acid, we get
$pH = 4.74 + \log 4$
$pH = 4.74 + 0.63$
Therefore, the answer to the question will be,
$pH = 5.37$
Therefore, the answer to the second part of the question is $5.37$.

Note: Remember that if the buffer is primarily basic then we have to use the $pKb$ to find the $pOH$and only then can we find the $pH$ . It is important to distinguish whether the buffer is a basic buffer or an acidic buffer.