In a solution prepared by dissolving 0.100 mol of propionic acid in enough water to make 1.00 L of solution, the pH is observed to be 1.35. What is K_a for propionic acid (HC₃H₅O₂)? A) 4.5 * 10^-² B) 2.0 * 10^-² C) 5.0 * 10^-¹² D) 3.6 * 10^-² E) none of these

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First, we need to use the pH value to find the concentration of H+ ions in the solution. pH=-log[H+] 10^-pH=[H+] 10^-1.35=[H+] [H+]= 0.000047943 Next, we can use the equation for Ka of propionic acid to solve for the value of Ka. Ka=[H+][C2H5COO-]/[C2H5COOH] Since we know that propionic acid completely dissociates in water, we can assume that [C2H5COOH]=[H+]. Ka=[H+]^2/[C2H5COOH] Ka=(0.000047943)^2/0.1 Ka= 1.35x10^-5 Finally, we can use the equation for Kb and Kw to solve for the value of Kb and then use it to calculate pKb. Ka x Kb = Kw Kb = Kw/Ka Kb = (1.0x10^-14)/(1.35x10^-5) Kb = 7.41x10^-10 pKb = -log(Kb) = 9.13 Therefore, K_a for propionic acid is 3.6x10^-5 (from Ka) and pKb is 9.13 (from Kb). The answer is D.

In a Solution Prepared by Dissolving 0