<u>Answer:</u> The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL
<u>Explanation:</u>
We are given:
Concentration of buffer = 0.460 M
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
.....(1)
We are given:
= negative logarithm of acid dissociation constant of benzoic acid = 4.2
![[C_6H_5COOH]=0.460M](https://tex.z-dn.net/?f=%5BC_6H_5COOH%5D%3D0.460M)
![[C_6H_5COO^-]=0.460M](https://tex.z-dn.net/?f=%5BC_6H_5COO%5E-%5D%3D0.460M)
pH = ?
Putting values in equation 1, we get:
![pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2](https://tex.z-dn.net/?f=pH%3D4.2%2B%5Clog%28%5Cfrac%7B0.460%7D%7B0.460%7D%29%5C%5C%5C%5CpH%3D4.2)
When the pH of the buffer changes by 1 unit, the buffering capacity is said to be lost.
pH change for loosing buffer capacity = [4.2 - 1] = 3.2
Calculating the ratio of conjugate base and its acid by using equation 1:
= 4.2
pH = 3.2
Putting values in equation 1, we get:
![3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1](https://tex.z-dn.net/?f=3.2%3D4.2%2B%5Clog%28%5Cfrac%7B%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D%29%5C%5C%5C%5C%5Cfrac%7B%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D%3D0.1)
- To calculate the number of moles for given molarity, we use the equation:
......(2)
For benzoic acid and its conjugate base:
Molarity of benzoic acid and its conjugate base = 0.460 M
Volume of solution = 115 mL
Putting values in equation 2, we get:
![0.460=\frac{\text{Moles of benzoic acid and its conjugate base}\times 1000}{115mL}\\\\\text{Moles of benzoic acid and its conjugate base}=\frac{0.460\times 115}{1000}=0.053mol](https://tex.z-dn.net/?f=0.460%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20benzoic%20acid%20and%20its%20conjugate%20base%7D%5Ctimes%201000%7D%7B115mL%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20benzoic%20acid%20and%20its%20conjugate%20base%7D%3D%5Cfrac%7B0.460%5Ctimes%20115%7D%7B1000%7D%3D0.053mol)
The chemical reaction for aniline and HCl follows the equation:
![C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-](https://tex.z-dn.net/?f=C_6H_5COO%5E-%2BHCl%5Crightarrow%20C_6H_5COOH%2BCl%5E-)
Let the moles of acid added to carry out the change is 'x' moles
- Calculating the moles of acid added:
![\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434](https://tex.z-dn.net/?f=%5Cfrac%7B%5BC_6H_5COO%5E-%5D-x%7D%7B%5BC_6H_5COOH%5D%2Bx%7D%3D0.1%5C%5C%5C%5C%5Cfrac%7B0.053-x%7D%7B0.053%2Bx%7D%3D0.1%5C%5C%5C%5Cx%3D0.0434)
Calculating the volume of acid added by using equation 2, we get:
Moles of acid added = 0.0434 moles
Molarity of solution = 0.390 M
Putting values in equation 2, we get:
![0.390M=\frac{0.0434\times 1000}{\text{Volume of acid}}\\\\\text{Volume of acid}=\frac{0.0434\times 1000}{0.390}=111.3mL](https://tex.z-dn.net/?f=0.390M%3D%5Cfrac%7B0.0434%5Ctimes%201000%7D%7B%5Ctext%7BVolume%20of%20acid%7D%7D%5C%5C%5C%5C%5Ctext%7BVolume%20of%20acid%7D%3D%5Cfrac%7B0.0434%5Ctimes%201000%7D%7B0.390%7D%3D111.3mL)
Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL