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what is the approximate pH of a solution labeled 0.050 M HClO? I realized my mistake on my last post but this isnt a strong acid nor a strong base so how do i solve this?

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3 answers
  1. HClO ==> H^+ + ClO^-

    Set up ICE chart, substitute into Ka expression, and solve for (H^+). Convert to pH.

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  2. i still don't get it.
    (H+)=x/0.050M

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  3. Absolutely not. Where's the ICE chart?
    HClO ==> H^+ + ClO^-

    initial:
    HClO = 0.05 M
    H^+ = 0
    ClO = 0

    change:
    HClO = -x
    H^+ = +x
    ClO^- = +x

    equilibrium:
    HClO = 0.05-x
    H^+ = x
    ClO^- = x

    Ka = (H^+)(ClO^-)/(HClO)
    Ka = 3.0 x 10^-8 in my book. Yours may be different.
    (x)^2/(0.05 = 3.0 x 10^-8
    x^2 = 1.5 x 10^-9
    x = 3.87 x 10^-5 for pH = 4.41

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