The pH of a solution of 3-hydroxypropanoic acid is measured to be . Calculate the acid dissociation constant of 3-hydroxypropanoic acid. Be sure your answer has the correct number of significant digits.

3.1 × 10⁻⁵

Explanation:

The pH of a 0.53 M solution of 3-hydroxypropanoic acid (HC₃H₅O₃) is measured to be 2.39. Calculate the acid dissociation constant of 3-hydroxypropanoic acid. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Step 2: Write the acid dissociation reaction

3-hydroxypropanoic acid is a weak acid that dissociates according to the following equation.

HC₃H₅O₃(aq) ⇄ C₃H₅O₃⁻(aq) + H⁺(aq)

Step 3: Calculate the concentration of H⁺

We will use the definition of pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.39 = 4.07 × 10⁻³ M

Step 4: Calculate the acid dissociation constant (Ka) of 3-hydroxypropanoic acid

We will use the following expression.

Ka = [H⁺]² / Ca

Ka = (4.07 × 10⁻³)² / 0.53

Ka = 3.1 × 10⁻⁵

it's c. elements are placed in rows based on how may valencies electrons have.

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