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Math Help - Equilibrium Equation?

  1. #1
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    Equilibrium Equation?

    A saturated solution of hydrogen sulfide (0,1 M concentration) dissociates into cation [H+] and anion [HS-], where [H+] = [HS-]. When this solution dissociates, the following equilibrium equation is established: . Find [H+]. (If [H+] = x, then [HHS] = 0.1 - x.)


    I found this problem in my algebra book and I'm not entirely sure how to do it. I've took chemistry before but it's been quite a while, so..

    I was wondering if i can seek some assistance, please?

    I've had already tried: x^2/0.1-x = 1.0 X 10^-7

    To: x^2 = 1.0 X 10^-7/0.1-x

    Then: x^2 = 0.1-x/1.0 X 10^7

    I'm unsure what to do from there.
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  2. #2
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    Quote Originally Posted by Ponderous View Post
    A saturated solution of hydrogen sulfide (0,1 M concentration) dissociates into cation [H+] and anion [HS-], where [H+] = [HS-]. When this solution dissociates, the following equilibrium equation is established: . Find [H+]. (If [H+] = x, then [HHS] = 0.1 - x.)


    I found this problem in my algebra book and I'm not entirely sure how to do it. I've took chemistry before but it's been quite a while, so..

    I was wondering if i can seek some assistance, please?

    I've had already tried: x^2/0.1-x = 1.0 X 10^-7

    To: x^2 = 1.0 X 10^-7/0.1-x

    Then: x^2 = 0.1-x/1.0 X 10^7

    I'm unsure what to do from there.
    \frac{x^2}{0.1-x} = 10^{-7}

    x^2 = 10^{-8} - x \times 10^{-7}

    x^2+x \times 10^{-7} - 10^{-8} = 0

    You can solve this just like a normal quadratic

    Just to point out that since (10^{-7})^2 << 4 \times 1 \times 10^{-8} you may take b^2-4ac = -4ac
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