Answer :
Answer: The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL
Explanation:
We are given:
Concentration of buffer = 0.460 M
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})[/tex] .....(1)
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of benzoic acid = 4.2
[tex][C_6H_5COOH]=0.460M[/tex]
[tex][C_6H_5COO^-]=0.460M[/tex]
pH = ?
Putting values in equation 1, we get:
[tex]pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2[/tex]
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:
[tex]pK_a[/tex] = 4.2
pH = 3.2
Putting values in equation 1, we get:
[tex]3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1[/tex]
- To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] ......(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:
[tex]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[/tex]
The chemical reaction for aniline and HCl follows the equation:
[tex]C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-[/tex]
Let the moles of acid added to carry out the change is 'x' moles
- Calculating the moles of acid added:
[tex]\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434[/tex]
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:
[tex]0.390M=\frac{0.0434\times 1000}{\text{Volume of acid}}\\\\\text{Volume of acid}=\frac{0.0434\times 1000}{0.390}=111.3mL[/tex]
Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL