Respuesta :
Explanation:
It is given that 185 ml of buffer solution is 0.40 M [tex]CH_{3}COOH[/tex] and [tex]CH_{3}COO^{-}[/tex].
Now, according to the Handerson equation,
pH = [tex]pK_{a} + log \frac{CH_{3}COO^{-}}{CH_{3}COOH}[/tex]
It is known that [tex]pK_{a}[/tex] value of acetic acid is 4.76.
pH = [tex]pK_{a} + log \frac{CH_{3}COO^{-}}{CH_{3}COOH}[/tex]
= [tex]4.76 + log \frac{0.40}{0.40}[/tex]
= 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 [tex]CH_{3}COO^{-}[/tex] and gives [tex]CH_{3}COOH[/tex]. So, with increase in [tex][CH_{3}COOH][/tex] 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 + [tex]log \frac{CH_{3}COO^{-}}{CH_{3}COOH}[/tex]
[tex]\frac{CH_{3}COO^{-}}{CH_{3}COOH}[/tex] = 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.
[tex]\frac{74 - x}{74 + x} = 0.1[/tex]
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 = [tex]\frac{moles}{molarity}[/tex]
= [tex]\frac{60.5 mmol}{0.180 M}[/tex]
= 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.
