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chemistry 12

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Calculate [H3O+] and [OH-] in 0.18M hydrobromic acid, HOBr(aq), having a K of 2.0x10^-9?

I got 9.0x10^-8

  • chemistry 12 -

    I don't think your one (1) answer is right for either H^+ or OH^-
    Post your work and I'll look for the error.

  • chemistry 12 -

    i got

    K= (HOBr2)(OH)/(HOBr)
    2.0x10^-9 = (x)(x)/0.18M

    x= 11x10^-6
    (OH-) =11x10^-6
    (H3O)= 1.0x10^-14/11x10^-6 =9.0x10^-8

  • chemistry 12 -

    #1. Hydobromic acid is HBr. That is a strong acid. HOBr, I assume, is correct since that is a weak acid but it is called hypobromous acid.
    HOBr + H2O ==> H3O^+ + OBr^-

    Ka = 2.0E-9 = (H3O^+)(OBr^-)/(HOBr)
    (H3O^+) = x
    (OBr^-) = x
    (HOBr) = 0.18-x which can be called 0.18 without changing the result but it makes the math easier.
    Solve for x = (H3O^+)

    After you know (H3O^+), then
    (H3O^+)(OH^-) = 1E-14 and solve for (OH^-).
    I obtained approximately 2E-5 for H^+ which makes OH^- about 5E-10. You should recalculate more accurately than I did but my answers are close.

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