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 185 mL of a buffer that is 0.400 M in both acetic acid (CH3COOH) and its conjugate base (CH3COO−) . Calculate the maximum volume of 0.180 M HCl that can be added to the buffer before its buffering capacity is lost.

Answer 1

Step-by-step explanation:

It is given that 185 ml of buffer solution is 0.40 M
`CH_(3)COOH` and
`CH_(3)COO^(-)`.

Now, according to the Handerson equation,

pH =
`pK_(a) + log (CH_(3)COO^(-))/(CH_(3)COOH)`

It is known that
`pK_(a)` value of acetic acid is 4.76.

pH =
`pK_(a) + log (CH_(3)COO^(-))/(CH_(3)COOH)`

=
`4.76 + log (0.40)/(0.40)`

= 4.76

If pH of a buffer changes by 1 unit then it means the buffering capacity is lost.

Hence, when HCl is being added it reacts with
`CH_(3)COO^(-)` and gives
`CH_(3)COOH`. So, with increase in
`[CH_(3)COOH]` the log term gives a negative value. This means that new pH will be less than 4.76.

Therefore, calculate the concentration when pH = 3.6.

3.76 = 4.76 +
`log (CH_(3)COO^(-))/(CH_(3)COOH)`

`(CH_(3)COO^(-))/(CH_(3)COOH)` = 0.1 ....... (1)

Now, we assume that the moles of acid added or change in moles is x. Therefore, moles of acetic acid and conjugate base present are as follows.

No. of moles = Molarity × Volume

= 0.40 × 185 ml

= 74 mmol

Now, we put this value into equation (1) as follows.

`(74 - x)/(74 + x) = 0.1`

x = 60.5

This means that moles of acid added is 60.5 mmol.

As it is given that molarity is 0.180 M. Therefore, calculate the volume of acid as follows.

Volume of acid =
`(moles)/(molarity)`

=
`(60.5 mmol)/(0.180 M)`

= 336.1 ml

Thus, we can conclude that the maximum volume of 0.180 M HCl that can be added to the buffer before its buffering capacity is lost is 336.1 ml.