The dissociation constant (\(K_a\)) for propionic acid (\(C_2H_5COOH\)) is approximately \(1.85 \times 10^{-5}\). Calculated using the given pH value (3.13) and the ICE table approach, the equilibrium expression was solved using the quadratic formula.
To calculate the dissociation constant (\(K_a\)) for propionic acid (\(C_2H_5COOH\)), we can use the ICE table approach and the given pH value.
The equilibrium expression for the dissociation of propionic acid is:
\[ C_2H_5COOH(aq) + H_2O \rightleftharpoons H_3O^+(aq) + C_2H_5COO^-(aq) \]
The ICE table is as follows:
\[ \begin{array}{cccc} & C_2H_5COOH & \rightleftharpoons & H_3O^+ & + & C_2H_5COO^- \\ \hline \text{Initial (M)} & & & & \\ \text{Change (M)} & -x & & +x & +x \\ \text{Equilibrium (M)} & (3.16 \, \text{g})/(74.08 \, \text{g/mol} \times 2 \, \text{L}) & & x & x \end{array} \]
Since the pH is given as 3.13, we can find the concentration of \(H_3O^+\) (\(x\)):
\[ \text{pH} = -\log[H_3O^+] \implies [H_3O^+] = 10^{-\text{pH}} \]
Now, substitute the values into the equilibrium expression:
\[ K_a = \frac{[H_3O^+][C_2H_5COO^-]}{[C_2H_5COOH]} \]
\[ K_a = \frac{x^2}{(3.16 \, \text{g})/(74.08 \, \text{g/mol} \times 2 \, \text{L})} \]
Solve for \(x\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
After finding \(x\), substitute it back into the equilibrium expression to find \(K_a\).
The calculated dissociation constant (\(K_a\)) for propionic acid is approximately \(1.85 \times 10^{-5}\).