A weak acid, HA, has a pKapKa of 4.3574.357 . If a solution of this acid has a pH of 4.0054.005 , what percentage of the acid is not ionized? Assume all H+H+ in the solution came from the ionization of HA.

Answers

Answer 1

Answer:

Answer:

69.3%

Explanation:

The question should read as follows:

A weak acid, HA, has a pKa of 4.357. If a solution of this acid has a pH of 4.005, what percentage of the acid is not ionized? Assume all H⁺ in the solution came from the ionization of HA.

The Henderson-Hasselbalch equation relates the pKa and pH of a solution to the ratio of ionized (A⁻) and unionized (HA) forms of a weak acid:

pH = pKa + log ([A⁻]/[HA])

Substituting and solving for [A⁻]/[HA]:

4.005 = 4.3574 = log([A⁻]/[HA])

-0.3524 = log([A⁻]/[HA])

[A⁻]/[HA] = 0.444/1

The percentage of acid that is not ionized (i.e. the percentage of acid in the HA form) is calculated:

[HA]/([A⁻] + [HA]) x 100% = 1/(1+0.444) x 100% = 69.3%

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To find more about atomic nucleus, refer the link below:

brainly.com/question/10658589

#SPJ2

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Answers

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Answers

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Explanation:

Answer:

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Explanation:

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Answers

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