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natita [175]
2 years ago
6

Remember from the buffering capacity note, tell me which of the following acids(and their salts) can make the best buffered solu

tion to maintain the pH as close to 4.30 as possible.
a. Chloroacetic acid (Ka = 1.35 x 10^-3)
b. Propanoic acid (Ka = 1.30 x 10^-5)
c. Benzoic acid (Ka = 6.40 x 10^-5)
d. Hypochlorous acid (Ka = 3.50 x 10^-8)

Hint:
Calculate the required concentration of hyrogen ions from pH 4.30 (use antilog function of your calculator)
You, now, know [H+] and Ka values of each acid. Plug in the formula
Ka = [H+] [A-] / [HA] , and solve for the ratio [HA] / [A-]. The acid with the
ratio [HA] / [A-] closest to 1 is the best (why?)
Advanced Placement (AP)
1 answer:
lakkis [162]2 years ago
6 0

Answer:

The answer is Benzoic acid. Hence, (C) is the correct answer. The acid with the  ratio [HA] / [A-] closest to 1 is the best because a buffer is most effective when the amounts of acid and conjugate base are approximately equal and hence their ratio should be close to 1.

Explanation:

pH = - log₁₀[H⁺]

[H⁺] = 10∧-pH

Given,

pH = 4.30

Chloroacetic acid (Ka = 1.35 x 10^-3)

Propanoic acid (Ka = 1.30 x 10^-5)

Benzoic acid (Ka = 6.40 x 10^-5)

Hypochlorous acid (Ka = 3.50 x 10^-8)  

Using the Hints,

The required [H+] = 10⁻⁴°³  => 5.011 x 10⁻⁵

Using the formula,

Ka = [H+] [A-] / [HA] and finding [HA] / [A-] values for each of the given acids.

[HA] / [A-] = [H+] /Ka

Substituting the values,

For Chloroacetic acid, [HA] / [A-] = 5.011 x 10⁻⁵ / 1.35 x 10⁻³ => 0.037

For Propanoic acid,  [HA] / [A-] = 5.011 x 10⁻⁵ / 1.30 x 10⁻⁵ =>  3.85

For Benzoic acid,  [HA] / [A-] = 5.011 x 10⁻⁵ / 6.40 x 10⁻⁵ => 0.782

For Hypochlorous acid,  [HA] / [A-] = 5.011 x 10⁻⁵ / 3.50 x 10⁻⁸ => 1431.714

As Benzoic acid has  [HA] / [A-] value closest to 1, it can make the best buffered solution.

The acid with the

ratio [HA] / [A-] closest to 1 is the best because,

A buffer is most effective when the amounts of acid and conjugate base are approximately equal and hence their ratio should be close to 1.

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