Problem 58 - Chemistry: An Atoms-Focused Approach

Question

Poisonous Plant Gifblaar is a small South African shrub and one of the most poisonous plants known because it contains fluoroacetic acid. If a \(0.480 M\) solution of fluoroacetic acid has a pH of \(1.44,\) what is the \(K_{\mathrm{a}}\) of the acid?

Step-by-Step Solution

Answer

Question: Given the molarity of fluoroacetic acid is 0.480 M and its pH is 1.44, calculate the acid dissociation constant (\(K_a\)) of fluoroacetic acid. Answer: The acid dissociation constant (\(K_a\)) of fluoroacetic acid is approximately \(3.98 \times 10^{-3}\).

Step 1: Find the hydrogen ion concentration (\(H^+\)) from the given pH

Use the pH formula to find the hydrogen ion concentration. The formula for pH is: pH = \(-\log_{10} [H^+]\) Given the pH, we can find the hydrogen ion concentration: \(1.44 = -\log_{10} [H^+]\) Now, solve for \([H^+]\): \([H^+] = 10^{-1.44}\)

Step 2: Write the acid dissociation equilibrium equation

Fluoroacetic acid (HFA) dissociates as follows in water: HFA \(\rightleftharpoons\) H\(^+\) + FA\(^-\) Now, let's write down the equilibrium expression for the dissociation of fluoroacetic acid: \(K_a = \frac{[H^+][FA^-]}{[HFA]}\)

Step 3: Calculate the equilibrium concentrations

To find the equilibrium concentrations, we will create a table representing the initial concentrations, changes, and equilibrium concentrations: | | HFA | H\(^+\) | FA\(^-\) | |---|-----|-------|-------| | Initial | 0.480 | 0 | 0 | | Change | -x | +x | +x | | Equilibrium | 0.480-x | x | x | Since we know the equilibrium concentration of \(H^+\) from Step 1: x = \(10^{-1.44}\) This means that the equilibrium concentrations of the species are: [HFA] = 0.480 - \(10^{-1.44}\) [H\(^+\)] = \(10^{-1.44}\) [FA\(^-\)] = \(10^{-1.44}\)

Step 4: Calculate the \(K_a\) of fluoroacetic acid

Use the equilibrium concentrations of HFA, H\(^+\), and FA\(^-\) calculated in Step 3 and plug them into the equilibrium expression we found in Step 2: \(K_a = \frac{(10^{-1.44})(10^{-1.44})}{(0.480 - 10^{-1.44})}\) Now, solve for \(K_a\) to find the acid dissociation constant of fluoroacetic acid: \(K_a \approx 3.98 \times 10^{-3}\)