Question

Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 125 mL of a buffer that is 0.360 M in both acetic acid ( CH 3 COOH ) and its conjugate base ( CH 3 COO − ) . Calculate the maximum volume of 0.300 M HCl that can be added to the buffer before its buffering capacity is lost.

Answer 1

Answer:

The maximum volume is 122.73 mL

Explanation:

125 mL of butter that is 0.360 M in CH₃COOH and CH₃COO⁻.

pKa of acetic acid = 4.76

Using Henderson Hasselbalch equation.

pH =pKa + log $\frac{CH_3COO^-}{CH_3COOH}$

= 4.76 + $log \frac{0.3600}{0.3600}$

= 4.76

Assuming the pH of the buffer changes by a unit of 1 , then it will lose its' buffering capacity.

When HCl is added , it reacts with  CH₃COO⁻ to give CH₃COOH.

However, [CH₃COOH] increases and the log term  results in a negative value.

Let  assume, the new pH is less than 4.76

Let say 3.76; calculating the concentration when the pH is 3.76; we have

3.76 = 4.76 + log $\frac{CH_3COO^-}{CH_3COOH}$

log $\frac{CH_3COO^-}{CH_3COOH}$ = 4.76 - 3.76

log $\frac{CH_3COO^-}{CH_3COOH}$ = 1.00

$\frac{CH_3COO^-}{CH_3COOH}$ = 0.1

Let number of  moles of acid be x   (i.e change in moles be x);  &

moles of acetic acid and conjugate base present be = Molarity × Volume

= 0.360 M × 125 mL

= 45 mmol

replacing initial concentrations and change in the above expression; we have;

$\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.1$

$\frac{45-x}{45+x} =0.1$

0.1(45 + x) = 45 - x

4.5 + 0.1 x = 45 -x

0.1 x + x = 45 -4.5

1.1 x = 40.5

x = 40.5/1.1

x = 36.82

So, moles of acid added = 36.82 mmol

Molarity = 0.300 M

So, volume of acid = $\frac{moles}{molarity }$ =  $\frac{36.82 \ mmol}{0.300}$

= 122.73 mL