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Alex_Xolod [135]
3 years ago
14

Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that yo

u 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.
Chemistry
1 answer:
Artist 52 [7]3 years ago
6 0

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.

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