Shown below is the Lewis Structure for propanoic acid.
i. Is propanoic acid a monoprotic, diprotic, or triprotic acid?
ii. Write the acid dissociation of propanoic acid.
iii. Determine the pH of a 0.015 M propanoic acid solution (Ka = 1.35×10-5).
iv. Determine the [OH-1] of the solution from iii.
The Correct Answer and Explanation is :
Based on the Lewis structure for propanoic acid depicted in the provided image, let’s address each of the questions:
i. Propanoic Acid as a Monoprotic, Diprotic, or Triprotic Acid
Propanoic acid, which has the chemical formula C2H5COOH, is a monoprotic acid. This means it can donate only one proton (H+) per molecule during the dissociation process. The presence of one carboxylic acid group (-COOH) in its structure is responsible for its ability to release one hydrogen ion in solution.
ii. Acid Dissociation of Propanoic Acid
The dissociation of propanoic acid in water can be written as:
[ \text{CH}_3\text{CH}_2\text{COOH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ ]
This reaction represents the release of a hydrogen ion from the carboxylic acid group, leaving behind the propanoate ion.
iii. Determination of the pH of a 0.015 M Propanoic Acid Solution
To find the pH, we first use the acid dissociation constant (Ka) and the initial concentration of propanoic acid to set up the expression based on the equilibrium constant:
[ \text{Ka} = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]} ]
Given that Ka = (1.35 \times 10^{-5}) and assuming that the change in concentration of H+ at equilibrium is x, the equation becomes:
[ 1.35 \times 10^{-5} = \frac{x^2}{0.015 – x} ]
Approximating that (0.015 – x \approx 0.015) (since x will be small), the equation simplifies to:
[ x^2 = 1.35 \times 10^{-5} \times 0.015 ]
[ x^2 = 2.025 \times 10^{-7} ]
[ x = \sqrt{2.025 \times 10^{-7}} ]
[ x \approx 1.42 \times 10^{-4} ]
Thus, the pH of the solution is:
[ \text{pH} = -\log[\text{H}^+] ]
[ \text{pH} = -\log[1.42 \times 10^{-4}] ]
[ \text{pH} \approx 3.85 ]
iv. Determination of the ([OH^-]) of the Solution
The relationship between the concentration of hydrogen ions ([\text{H}^+]) and hydroxide ions ([\text{OH}^-]) in a solution is governed by the water ion-product, (K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}) at 25°C. Using the concentration of H+ calculated:
[ [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{1.42 \times 10^{-4}} ]
[ [\text{OH}^-] = 7.04 \times 10^{-11} ]
Therefore, the concentration of hydroxide ions in the 0.015 M propanoic acid solution is approximately (7.04 \times 10^{-11}) M.