A 0.15 m solution of chloroacetic acid has a ph of 1.86. what is the value of ka for this acid?

General
1

a 0.15 m solution of chloroacetic acid has a ph of 1.86. what is the value of ka for this acid?

2

LectureNotes said a 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of Ka for this acid?

Answer:
To find the Ka (acid dissociation constant) value for chloroacetic acid from the given pH value, we need to follow these steps:

  1. First, we need to determine the concentration of H+ ions in the solution. The pH of a solution is the negative base-10 logarithm of the concentration of H+ ions.

    pH = -log[H+]

    Given that the pH is 1.86, we can calculate the concentration of H+ ions:

    [H+] = 10^(-pH)
    = 10^(-1.86)
    = 1.23 x 10^(-2) M

  2. Chloroacetic acid (CH2ClCOOH) can be represented by the following equilibrium equation when it dissociates in water:

    CH2ClCOOH ⇌ CH2ClCOO- + H+

    We can establish an ICE table where:

    • [CH2ClCOOH] = initial concentration of chloroacetic acid (0.15 M)
    • [H+] = change in concentration
    • [CH2ClCOO-] and [H+] = equilibrium concentrations

    Using the ICE table, we find:

    [CH2ClCOOH] = 0.15 M - x
    [CH2ClCOO-] = x
    [H+] = x

  3. The Ka expression for chloroacetic acid is given by:

    Ka = [CH2ClCOO-][H+] / [CH2ClCOOH]

    Substitute the equilibrium concentrations into the Ka expression:

    Ka = (x * x) / (0.15 - x)

  4. Since Ka value is related to x, we can make an assumption that x is very small compared to 0.15 (initial concentration) leading to the simplification:

    0.15 ≈ 0.15 - x

    This simplifies to:

    x ≈ [H+] = 1.23 x 10^(-2) M

  5. Finally, calculate the Ka value using the approximate x value:

    Ka = (1.23 x 10^(-2)) * (1.23 x 10^(-2)) / (0.15 - 1.23 x 10^(-2))

    Ka ≈ 9.80 x 10^(-5)

Therefore, the calculated Ka value for chloroacetic acid is approximately 9.80 x 10^(-5).