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Calculate the pH at the equivalence point for the titration of 0.120 M methylamine with 0.120 M HCl (kb of methyalamine is 5.0 x 10^-4)

  • Chemistry -

    If we call methylamine, MeNH2, then
    the pH at the equivalence point is determined by the hydrolysis of the salt, MeNH3Cl formed at the equivalence point. This is Ka for MeNH3Cl
    .........MeNH3^+ + H2O -->MeNH2+ + H3O^+
    initial..0.06M..............0........0
    change.....-x................x.......x
    equil.....0.06-x.............x.......x

    Ka = (Kw/Kb) = (H3O^+)(MeNH2)/(MeNH3^+)
    Kw = 1E-14
    Kb = Kb for MeNH2
    Substitute into the Ka expression and solve for x, then convert to pH.
    (Note: the concn of the salt at the equivalence point is just 1/2 the concn initially. )

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