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31 December, 16:32

A solution is made by dissolving 0.0150 mol of HF in enough water to make 1.00 L of solution. At 26 °C, the osmotic pressure of the solution is 0.449 atm. What is the percent ionization of this acid?

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  1. 31 December, 20:14
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    Given:

    M = 0.0150 mol/L HF solution

    T = 26°C = 299.15 K

    π = 0.449 atm

    Required:

    percent ionization

    Solution:

    First, we get the van't Hoff factor using this equation:

    π = i MRT

    0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)

    i = 1.219367

    Next, calculate the concentration of the ions and the acid.

    We let x = [H+] = [F-]

    [HF] = 0.0150 - x

    Adding all the concentration and equating to iM

    x + x + 0.0150 - x = 1.219367 (0.0150)

    x = 3.2905 x 10^-3

    percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%

    Also,

    percent dissociation = (i - 1) (100) = (1.219367 * 1) (100) = 21.94%
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