Final answer:
To calculate the pH of the aqueous solution of CH₂ClCOOH, we need to use its Ka value to find the concentration of H₃O⁺ ions. After setting up the equation for Ka and solving for x, we find that the concentration of H₃O⁺ ions is approximately 0.043 M. Using the equation pH = -log[H₃O⁺], we can calculate the pH of the solution to be approximately 1.37.
Step-by-step explanation:
To calculate the pH of the aqueous solution of CH₂ClCOOH, we need to find the concentration of H₃O⁺ ions. Since CH₂ClCOOH is a weak acid, we can use its Ka value to calculate the concentration of H₃O⁺ ions. The equation for the dissociation of CH₂ClCOOH is CH₂ClCOOH(aq) + H₂O(l) ⟶ CH₂ClCOO⁻(aq) + H₃O⁺(aq). The expression for Ka is Ka = [CH₂ClCOO⁻][H₃O⁺] / [CH₂ClCOOH]. Given that the Ka of CH₂ClCOOH is 0.0014 and its initial concentration is 0.80 M, we can use the equation for Ka to find the concentration of H₃O⁺ ions and then calculate the pH using the equation pH = -log[H₃O⁺].
First, we can assume that the change in CH₂ClCOOH concentration is negligible compared to its initial concentration because it is a weak acid. Let x be the concentration of H₃O⁺ ions. The change in concentration of CH₂ClCOOH is then 0.80 - x. The change in concentration of CH₂ClCOO⁻ is x, and the change in concentration of H₃O⁺ is also x.
Using the equation for Ka, we can write the expression: 0.0014 = (x)(x) / (0.80 - x). Solving for x gives x = 0.043 M. Then, we can calculate the pH using the equation pH = -log(0.043), which is approximately 1.37. Therefore, the correct answer is not provided in the options given. The correct pH value for the 0.80 M CH₂ClCOOH(aq) solution is approximately 1.37.