Answer :
Answer: The pH change of the buffer is 0.30
Explanation:
To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})[/tex]
[tex]pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})[/tex] .....(1)
We are given:
[tex]pK_b[/tex] = negative logarithm of base dissociation constant of aniline = 9.13
[tex][C_6H_5NH_3^+]=0.306M[/tex]
[tex][C_6H_5NH_2]=0.418M[/tex]
pOH = ?
Putting values in equation 1, we get:
[tex]pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH_{initial}=14-8.99=5.01[/tex]
To calculate the molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Moles hydrochloric acid solution = 0.124 mol
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M[/tex]
The chemical reaction for aniline and HCl follows the equation:
[tex]C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-[/tex]
Initial: 0.418 0.124 0.306
Final: 0.294 - 0.430
Calculating the pOH by using using equation 1:
[tex]pK_b[/tex] = negative logarithm of base dissociation constant of aniline = 9.13
[tex][C_6H_5NH_3^+]=0.430M[/tex]
[tex][C_6H_5NH_2]=0.294M[/tex]
pOH = ?
Putting values in equation 1, we get:
[tex]pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH_{final}=14-9.29=4.71[/tex]
Calculating the pH change of the solution:
[tex]\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30[/tex]
Hence, the pH change of the buffer is 0.30