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amm1812
3 years ago
5

Suppose that you have 115 mL of a buffer that is 0.460 M in both benzoic acid ( C 6 H 5 COOH ) and its conjugate base ( C 6 H 5

COO − ) . Calculate the maximum volume of 0.390 M HCl that can be added to the buffer before its buffering capacity is lost.
Chemistry
1 answer:
77julia77 [94]3 years ago
7 0

<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]})

pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})        .....(1)

We are given:

pK_a = negative logarithm of acid dissociation constant of benzoic acid = 4.2

[C_6H_5COOH]=0.460M

[C_6H_5COO^-]=0.460M

pH = ?

Putting values in equation 1, we get:

pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2

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:

pK_a = 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

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}        ......(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

The chemical reaction for aniline and HCl follows the equation:

C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-

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

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

Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

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