chemistry

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Calculate the pH and percentage protonation of the solute in an aqueous
solution of 0.39 M quinine if the pKa of its
conjugate acid is 8.52.

  • chemistry -

    Let's call quinine, BN
    .........BN + HOH ==> BNH^+ + OH^-
    initial.0.39M.........0........0
    change...-x...........x.........x
    equil...0.39-x........x........x

    Kb = (BNH^+)(OH^-)/(BN)
    You have pKa of the conjugate acid, convert to pKb for the base, substitute into the Kb expression and solve for (OH^-) then convert that to pH.
    %protonation = [(OH^-)/(BN)]*100 = ?

  • chemistry -

    -log ka=8.52
    ka= 3*10^-09
    Kb= kw/ka
    = 10^-14/ 3*10^-09 = 3.4 *10^-06
    kb= x^2/0.39
    x= squre root of ( 3.4*10^-6*0.39)
    =1.15*10^-3
    POH= -log[OH-]= -logx= -log 1.15*10^-3
    =2.939
    PH= 14 - 2.939
    =11.06

    IS THIS RIGHT?

  • chemistry -

    nvm, it IS right, ty xD

  • chemistry -

    I calculate 3.31E-6 for Kb which runs through the rest of the problem and ends with 11.055 which I would round to 11.06.

  • chemistry -

    for the %protonation = [(OH^-)/(BN)]*100
    i tried doing 1.15*10^-3 / 0.39 * 100
    = .295, but the answer is incorrect.
    can you tell me what i did wrong?

  • chemistry -

    I think it goes back to the incorrect conversion of pKa to pKb to Kb. Then for OH^- I obtained 1.136E-3.
    0.001136/0.39 = 0.00291 and that times 100 = 0.291% which rounds to 0.29 to 2 significant figures.
    Your problem may very well be that you are reporting too many s.f. I'll bet 0.29% will work. These data bases are "good" about that and catch a lot of trouble for it. Let me know.

  • chemistry -

    YES! it was the OH^- that messed up my calculation. The answer you provided me was correct, thanks for all the help!

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