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Chem

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In a 1.0× 10–2 M solution of CH3NH3Br(aq), identify the relative molar amounts from highest to lowest for :
-Br-
-OH-
-CH3NH3+
-HBr
-CH3NH2
-H3O+
-H2O

  • Chem -

    ........CH3NH3Br ==> CH3NH3^+ + Br^-
    I......0.01...........0..........0
    C.....-0.01..........0.01.......0.01
    E........0...........0.01.......0.01

    Then the hydrolysis of the cation.
    ......CH3NH3^+ + H2O ==> H3O^+ + CH3NH2
    I.....0.01................0........0
    C......-x.................x........x
    E.....0.01-x..............x........x

    Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.01-x)
    Solve for x which gives you H3O^+ and CH3NH2 as well as CH3NH3^+.

    Finally, (H^+)(OH^-) = Kw = 1E-14 which allows you to calculate OH^-

    Now you can arrange them in any order you wish.
    Post your work if you get stuck.

  • Chem -

    I can't find the value x by calculation. It ends up to give me a quadratic equation and the value of x is negative! please help me ASAP...

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