In a solution prepared by dissolving 0.100 mole of propanoic acid in enough water to make 1.00 L of solution,the pH is observed to be 2.832.The K_a for propanoic acid (HC₃H₅O₂)is: A)1.47 $\times$ 10^-³ B)2.20 $\times$10^-⁵ C)1.49 $\times$10^-² D)4.55 $\times$ 10^-¹⁰ E)none of these

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Quizplus · Accepted Answer

The pH of 2.832 corresponds to an H⁺ concentration of 1.47 x 10⁻³ M. Using the formula for Ka, Ka = [H⁺][A⁻]/[HA], and assuming [A⁻] ≈ [H⁺] and [HA] ≈ initial concentration - [H⁺], the calculated Ka is approximately 2.20 x 10⁻⁵.

In a Solution Prepared by Dissolving 0 ×\times× 10-3 B)2 ×\times× 10-5 C)1

In a Solution Prepared by Dissolving 0 $\times$ 10^-³
B)2 $\times$ 10^-⁵
C)1