<u>Answer:</u> The pH change of the buffer is 0.30
<u>Explanation:</u>
To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:
![pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})](https://tex.z-dn.net/?f=pOH%3DpK_b%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjugate%20acid%7D%5D%7D%7B%5B%5Ctext%7Bbase%7D%5D%7D%29)
.....(1)
We are given:
= negative logarithm of base dissociation constant of aniline = 9.13
![[C_6H_5NH_3^+]=0.306M](https://tex.z-dn.net/?f=%5BC_6H_5NH_3%5E%2B%5D%3D0.306M)
![[C_6H_5NH_2]=0.418M](https://tex.z-dn.net/?f=%5BC_6H_5NH_2%5D%3D0.418M)
pOH = ?
Putting values in equation 1, we get:

To calculate pH of the solution, we use the equation:

To calculate the molarity, we use the equation:

Moles hydrochloric acid solution = 0.124 mol
Volume of solution = 1 L
Putting values in above equation, we get:

The chemical reaction for aniline and HCl follows the equation:

<u>Initial:</u> 0.418 0.124 0.306
<u>Final:</u> 0.294 - 0.430
Calculating the pOH by using using equation 1:
= negative logarithm of base dissociation constant of aniline = 9.13
![[C_6H_5NH_3^+]=0.430M](https://tex.z-dn.net/?f=%5BC_6H_5NH_3%5E%2B%5D%3D0.430M)
![[C_6H_5NH_2]=0.294M](https://tex.z-dn.net/?f=%5BC_6H_5NH_2%5D%3D0.294M)
pOH = ?
Putting values in equation 1, we get:

To calculate pH of the solution, we use the equation:

Calculating the pH change of the solution:

Hence, the pH change of the buffer is 0.30