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

The maximum volume is 122.73 mL

Step-by-step 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
`(CH_3COO^-)/(CH_3COOH)`

= 4.76 +
`log (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
`(CH_3COO^-)/(CH_3COOH)`

log
`(CH_3COO^-)/(CH_3COOH)` = 4.76 - 3.76

log
`(CH_3COO^-)/(CH_3COOH)` = 1.00

`(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;

`([CH_3COO^-])/([CH_3COOH]) = 0.1`

`(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 =
`(moles)/(molarity )` =
`(36.82 \ mmol)/(0.300)`

= 122.73 mL