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March 28, 2017

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A reaction mixture is in a 5.4 L flask at 25 oC initially contains only 7.64 grams of acetic acid. (CH3COOH). Set up your I.C.E. table and use it to answer the following questions. HINT: You will need to look up the Ka for this equilibrium reaction.
CH3COOH(aq) ¡ê CH3COO-(aq) + H+(aq)


a. How many moles of Acetic acid are in the solution at equilibrium?
b. What is the Kb for this reaction?
c. What is the pH of the equilibrium solution?
d. What is the pOH of the equilibrium solution?

  • chemistry - ,

    I get tired of typing, CH3COOH we will call HAc. That is 7.64 g/molar mass HAc = 7.64/60 = 0.127 moles. You should do it more accurately than that since I estimated the molar mass of HAc.
    It ionizes as
    .................HAc -->H^+ + Ac^-
    begin(moles)...0.127...0.....0
    change............-x......+x.....+x
    equilibrium...0.127-x.....x......x

    Ka = (H^+)(Ac^-)/((HAc)

    a. See above.
    b. Did you make a typo here? If you mean Ka, look that up in your text or on-line. If you REALLY mean Kb, it is Kw/Ka.
    c. Substitute the ICE values into the Ka expression and solve for x. That will be moles H^+. Convert that to M by (H^+)= moles/L [Note: I must point out here that the problem does NOT tell you the volume of the solution. It tells you that the acetic acid is in a 5.4L flask BUT it doesn't say what else is there NOR its volume. I think the intent of the problem is that the flask CONTAINS 5.4 L of solution]. After moles H^+ are converted to M, then pH =-log(H^+).
    d. pH + pOH = pKw = 14. Solve for pOH.

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