An acid solution with a concentration of 0.500 M has a pH = 3.21. What is the Ka of the acid? (LO 16.8) (a) 1.2 * 10-5 (b) 1.7 * 10-6 (c) 7.6 * 10-7 (d) 5.4 * 10-3

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Calculate the concentration of hydrogen ions \([H^+]\) using the pH value. Use the formula \([H^+] = 10^{-\text{pH}}\).

Substitute the given pH value into the formula to find \([H^+]\).

Write the expression for the acid dissociation constant \(K_a\) for the weak acid, which is \(K_a = \frac{[H^+][A^-]}{[HA]}\).

Assume that the initial concentration of the acid \([HA]\) is 0.500 M and that the change in concentration due to dissociation is negligible, so \([HA] \approx 0.500\).

Substitute the values of \([H^+]\) and \([HA]\) into the \(K_a\) expression to solve for \(K_a\).>

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Key Concepts

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

pH and pKa

pH is a measure of the hydrogen ion concentration in a solution, defined as pH = -log[H+]. The pKa is the negative logarithm of the acid dissociation constant (Ka) and provides insight into the strength of an acid. A lower pKa indicates a stronger acid, while a higher pKa indicates a weaker acid. Understanding the relationship between pH and pKa is essential for calculating the dissociation constant of an acid.

Acid Dissociation Constant (Ka)

The acid dissociation constant (Ka) quantifies the strength of an acid in solution. It is defined by the equilibrium expression for the dissociation of an acid (HA) into its conjugate base (A-) and hydrogen ions (H+). The larger the Ka value, the stronger the acid, as it indicates a greater tendency to donate protons. Calculating Ka from pH involves using the concentration of the acid and the concentration of the dissociated ions at equilibrium.

Equilibrium Concentrations

In a chemical reaction, equilibrium concentrations refer to the concentrations of reactants and products when the reaction has reached a state where the rates of the forward and reverse reactions are equal. For weak acids, this involves determining the concentrations of the undissociated acid and the ions produced at equilibrium. These concentrations are crucial for calculating the Ka value, as they directly influence the equilibrium expression used in the calculation.

Related Practice

Textbook Question

What is the concentration of hydroxide ions 3OH-4 in a glass of wine with pH = 3.64? (LO 16.5, 16.6) (a) 2.3 * 10-4 M (b) 6.4 * 10-3 M (c) 6.8 * 10-9 M (d) 4.4 * 10-11 M

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Textbook Question

Consider the conjugate bases, 1X-, Y-, Z-2 in Problem 2. If you mix equal concentrations of reactants and products, which of the following reactions will proceed to the left? (LO 16.3) (a) HX + Y- HY + X- (b) HX + Z- HZ + X- (c) HY + X- HX + Y- (d) HZ + Y- HY + Z-

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Textbook Question

Consider the reaction: SO2 + OH- S HSO3-. Which reaction scheme shows the correct use of the curved arrow notation representing thedonation of an electron pair and the correct labeling of the Lewis acid and Lewis base? (LO 16.14)(a) (b) (c) (d)

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