Chemistry

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Calculate the pOH, pH, and percent protonation of solute in the following aqueous solution.
0.059 M quinine, given that the pKa of its conjugate acid is 8.52.

Please show me how to do it and not just the answers. Thanks!

  • Chemistry -

    Quinine has a tertiary amine group so acts as base
    QN + H2O -----> QNH+ + OH-
    given pka is 8.52
    -log ka=8.52
    ka= 3*10^-09,
    for conjugate acid base pair at 25degrees Celsius kw= ka*kb
    therefore Kb= kw/ka
    = 10^-14/ 3*10^-09 = 3.4 *10^-06
    from equation kb=[QNH+][OH-]/[QN]
    QN + H2O -----> QNH+ + OH-
    initial 0.018
    equlli 0.018-x x x
    since quinine is a weak base 0.018-x can be written as 0.018
    therefore kb= x^2/0.018
    x= squre root of ( 3.4*10^-6*0.018)
    =2.47 *10^-04
    POH= -log[OH-]= -logx= -log 2.47*10^-4
    =3.61
    PH= 14-POH =10.39
    percent protonation= [QNH+]/Qn
    =x/0.018 *100
    =1.4%

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