Chemistry, 07.10.2019 23:00 crazymadhatter0

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.

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Answer from: bgallman153p71edg

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 = $pK_{a} + log \frac{CH_{3}COO^{-}}{CH_{3}COOH}$

It is known that $pK_{a}$ value of acetic acid is 4.76.

pH = $pK_{a} + log \frac{CH_{3}COO^{-}}{CH_{3}COOH}$

= $4.76 + log \frac{0.40}{0.40}$

= 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 + $log \frac{CH_{3}COO^{-}}{CH_{3}COOH}$

$\frac{CH_{3}COO^{-}}{CH_{3}COOH}$ = 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.

$\frac{74 - x}{74 + x} = 0.1$

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 = $\frac{moles}{molarity}$

= $\frac{60.5 mmol}{0.180 M}$

= 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.

Answer from: Quest

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Answer from: Quest

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