Answer to Question #115857 in General Chemistry for J

Question

Calculate the pH of 0.0800 mol dm ^-3 C2H5COOH(aq)

Answer 1

C₂H₅COOH, or propionic acid, dissociates in water solution according to the equation:

C₂H₅COOH + H₂O $\Leftrightarrow$ H₃O⁺ + C₂H₅COO^-.

The acid dissociation constant expression involves the equilibrium concentrations of protonated, deprotonated forms and of hydronium cation:

$K_a = \frac{[H_3O^+][C_2H_5COO^-]}{[C_2H_5COOH]}$ .

The Ka of propionic acid is 10^-4.88. If the initial concentration of the acid was 0.0800 M and the equilibrium concentration of H₃O⁺ and C₂H₅COO^- are $x$, the K_a is:

$K_a = \frac{x^2}{0.0800 - x}$.

Then , the equilibrium concentration of hydronium ion is:

$x^2 + K_a·x - 0.0800·K_a = 0$

$x = 0.00102$

$[H_3O^+] = 0.00102$ M.

Finally, the pH of the solution is:

$pH = -\text{log}[H_3O^+]$

$pH = 2.99$.

Answer: the pH of 0.0800 mol dm^-3 C₂H₅COOH(aq) is 2.99.