Textbook Question
A 0.148 M solution of a monoprotic acid has a percent ionization of 1.55%. Determine the acid ionization constant (Ka) for the acid.
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Ch.16 - Acids and Bases
Chapter 16, Problem 78
Find the pH and percent ionization of a 0.100 M solution of a weak monoprotic acid given the following Ka values: a. Ka = 1.0 * 10^-5, b. Ka = 1.0 * 10^-3, c. Ka = 1.0 * 10^-1.
Verified step by step guidance1
Identify the expression for the ionization of a weak monoprotic acid: HA ⇌ H⁺ + A⁻.
Write the expression for the acid dissociation constant (Ka): Ka = [H⁺][A⁻]/[HA].
Assume the initial concentration of HA is 0.100 M and the change in concentration due to ionization is x, so [H⁺] = [A⁻] = x and [HA] = 0.100 - x.
Substitute these values into the Ka expression: Ka = x^2 / (0.100 - x).
Solve for x (which represents [H⁺]) using the quadratic formula or by assuming x is small compared to 0.100 M, then calculate pH = -log[H⁺] and percent ionization = (x/0.100) * 100%.
Related Practice
Open Question
A 0.085 M solution of a monoprotic acid has a percent ionization of 0.59%. Determine the acid ionization constant (Ka) for the acid.
Open Question
Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8 * 10^-4.) a. 0.250 M HF b. 0.100 M HF c. 0.050 M HF.
Textbook Question
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Textbook Question
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