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Calculate the pH of each of following buffered solutions.?a. 0.10 M acetic acid/0.25 M sodium acetate b. 0.25 M acetic acid/0.10 M sodium acetate c. 0.080 M acetic acid/0.20 M sodium acetate d. 0.20 M acetic acid/0.080 M sodium acetate

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superman1987

Answer:

a. 5.10.

b. 4.35.

c. 5.10.

d. 4.35.

Explanation:

a. 0.10 M acetic acid/0.25 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.25)/(0.10)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

b. 0.25 M acetic acid/0.10 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.10)/(0.25)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

c. 0.080 M acetic acid/0.20 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.20)/(0.08)

∴ pH =  4.744 + 0.34 = 5.084 ≅ 5.10.

d. 0.20 M acetic acid/0.080 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.08)/(0.20)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

lheduardomarques

To perform calculations involving the pH of a solution, we can use the following logarithmic equation: pH = - log [H3O+] or pH = - log [H+]

pH of each of following buffered solutions

a. 0.10 M acetic acid/0.25 M sodium acetate = pH=3,22

b. 0.25 M acetic acid/0.10 M sodium acetate = pH=2,82

c. 0.080 M acetic acid/0.20 M sodium acetate pH=0,21

d. 0.20 M acetic acid/0.080 M sodium acetate pH=4,61

With this information, we can conclude that the behavior of weak acids, those that undergo partial hydrolysis in water, that is, are only partially deprotonated.

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