The dissociation for a weak base BOH in water found to be 1.25 x 10^-6. What is the concentration of H+ in a 3.2M soluton of BOH.

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asked by LT
  1. Do you mean the dissociation for a weak base is 1.25 x 10^-5 or the dissociation constant is 1.25 x 10^-5?

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  2. constant
    sorry :(

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    posted by LT
  3. BOH <==> B^+ + OH^-

    Kb = (B^+)(OH^-)/(BOH) = 1.25 x 10^-6

    Have you been introduced to the ICE chart?
    Initial = I
    change = C
    equilibrium = E

    Initial concns (before any dissociation):
    (BOH) = 3.2 M
    (B^+) = 0
    (OH^-) = 0

    change in concns (after dissociation):
    (B^+) = y
    (OH^-) = y
    (BOH) = -y

    equilibrium concns:
    (B^+) = 0 + y = y
    (OH^-) = 0 + y = y
    (BOH) = 3.2 - y = 3.2-y

    Plug the equilibrium values into the Kb expression and solve for y. You will get a quadratic which you can solve with the quadratic formula OR you can make the simplifying assumption and see if that will work. At any rate, y = (OH^-) so you take the - log to get pOH, then subtract from 14 to get pH, then convert to (H^+).

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  4. I got the pH how do i turn that into (H+)

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    posted by LT
  5. pH = -log(H^+)
    Say pH = 4.3.
    Then 4.3 = -log(H^+)
    -4.3 = log(H^+)
    Now enter -4.3 into your calculator, then hit the 10x button and up will pop 5.01187 x 10^-5. Of course you would round that to 5.01 x 10^-5 M

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