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Ch.16 - Acid-Base Equilibria
Chapter 16, Problem 112

The following observations are made about a diprotic acid H2A: (i) A 0.10 M solution of H2A has pH = 3.30. (ii) A 0.10 M solution of the salt NaHA is acidic. Which of the following could be the value of pKa2 for H2A: (i) 3.22, (ii) 5.30, (iii) 7.47, or (iv) 9.82?

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1
Determine the concentration of hydrogen ions [H^+] in the 0.10 M solution of H2A using the pH value: \([H^+] = 10^{-pH}\).
Use the concentration of [H^+] to find the ionization constant \(K_{a1}\) for the first ionization of H2A: \(H_2A \rightleftharpoons H^+ + HA^-\).
Recognize that the solution of NaHA is acidic, indicating that the second ionization \(HA^- \rightleftharpoons H^+ + A^{2-}\) is significant, and \(K_{a2}\) must be considered.
Compare the given pKa values to determine which one is consistent with the observed acidity of the NaHA solution, knowing that a lower pKa indicates a stronger acid.
Select the pKa value that is consistent with the acidic nature of the NaHA solution, considering that \(pKa_2\) should be higher than \(pKa_1\) but still low enough to contribute to the acidity.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Diprotic Acids

Diprotic acids are acids that can donate two protons (H⁺ ions) per molecule in a solution. They undergo two dissociation steps, each characterized by its own acid dissociation constant (Ka). Understanding the behavior of diprotic acids is crucial for predicting their pH and the resulting equilibrium in solution.
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pH and pKa Relationship

The pH of a solution is a measure of its acidity, defined as the negative logarithm of the hydrogen ion concentration. The pKa is the negative logarithm of the acid dissociation constant (Ka) and indicates the strength of an acid; lower pKa values correspond to stronger acids. The relationship between pH and pKa helps determine the degree of ionization of the acid in solution.
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Salt Hydrolysis

When a salt of a weak acid is dissolved in water, it can undergo hydrolysis, affecting the pH of the solution. In this case, NaHA, the salt of the weak acid H2A, can release H⁺ ions, making the solution acidic. Understanding how salts interact with water is essential for predicting the pH of solutions containing weak acids and their salts.
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Related Practice
Textbook Question

Butyric acid is responsible for the foul smell of rancid butter. The pKa of butyric acid is 4.84. (c) Calculate the pH of a 0.050 M solution of sodium butyrate.

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Textbook Question

Arrange the following 0.10 M solutions in order of increasing acidity: (i) NH4NO3, (ii) NaNO3, (iii) CH3COONH4, (iv) NaF, (v) CH3COONa.

Textbook Question
Ritalin is the trade name of a drug, methylphenidate, used to treat attention-deficit/hyperactivity disorder in young adults. The chemical structure of methylphenidate is

(a) Is Ritalin an acid or a base? An electrolyte or a nonelectrolyte?
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Open Question
Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that the pH in the stomach is 2.5, indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, Kb = 7 * 10-7; caffeine, Kb = 4 * 10-14; strychnine, Kb = 1 * 10-6; quinine, Kb = 1.1 * 10-6.
Open Question
The amino acid glycine H₂N¬CH₂¬COOH can participate in the following equilibria in water: H₂N¬CH₂¬COOH + H₂O ⇌ H₂N¬CH₂¬COO⁻ + H₃O⁺ with Ka = 4.3 × 10⁻³, and H₂N¬CH₂¬COOH + H₂O ⇌ +H₃N¬CH₂¬COOH + OH⁻ with Kb = 6.0 × 10⁻⁵. (c) What would be the predominant form of glycine in a solution with pH 13? With pH 1?
Textbook Question

The amino acid glycine (H2N–CH2–COOH) can participate in the following equilibria in water:

H2N–CH2–COOH + H2O ⇌ H2N–CH2–COO + H3O+ Ka = 4.3 × 10-3

H2N–CH2–COOH + H2O⇌ +H3N–CH2–COOH + OH- Kb = 6.0 × 10-5

(a) Use the values of Ka and Kb to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: H2N–CH2–COOH ⇌ +H3N–CH2–COO

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