chemistry

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What would be the pH of a solution of hy-
poiodous acid (HOI) prepared by dissolving
144 grams of the acid in 200 mL of pure water
(H2O)? The Ka of hypoiodous acid is 2¡¿10−11

1. 7
2. 10
3. 13
4. 5
5. 1

  • chemistry -

    HOI ==> H^+ + OI^-

    Ka = (H^+)(OI^-)/(HOI)
    At equilibrium,
    (H^+) = x
    (OI^-) = x
    (HOI) = moles/L. moles = 144 g/molar mass HOI and L = 0.2. So at equilibrium (HOI) = M-x
    Solve for x and convert to pH.

  • chemistry -

    the freakin' answer is #4. the pH is 5. don't post up answers if they won't f-ing help, smarta**.

  • chemistry -

    said it twice for added effect......

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