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Chemistry II

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Calculate the pH at the equivalence point in the titration of 50ml of 0.20 M methylamine (Kb=4.3*10^-4) with a 0.40 M HCL solution.

  • Chemistry II - ,

    How many mL HCl is needed? That's
    50 x 0.2/0.4 = 25.00 mL
    Total volume = 75.00 mL
    Total mols = 0.050 x 0.2 =0.01
    (salt at equivalence point) = 0.01/0.075L = approximately 0.13
    If we call methylamine just BNH2 then the salt will be BNH3^+ and it will hydrolyze as follows:

    ......BNH3^+ + H2O ==> H3O^+ + BNH2
    I.....0.13..............0........0
    C.......-x..............x........x
    E...-0.13-x.............x........x

    Ka for BNH3^+ = (Kw/Kb for BNH2) = (x)(x)/(0.13-x)
    Solve for x = (H3O^+) and convert to pH.

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