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chemistry

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Consider the titration of 100.0 mL of 0.500 M NH3 (Kb = 1.8 x 10-5) with 0.500 M HCl. At the
stoichiometric point of this titration, the [H+] is:

  • chemistry -

    NH3 + HCl ==> NH4Cl
    (NH4Cl) at eq. point = 0.500 x (100/200) = 0.25M
    The pH is determined by the hydrolysis of the salt, NH4Cl.
    .......NH4^+ + H2O ==> H3O^+ + OH^-
    initial.0.25M..........0........0
    change...-x.............x.......x
    equil....0.25-x.........x........x

    Ka for NH4^+ = (Kw/Kb for NH3) = (H3O^+)(OH^-)/(NH4^+)
    Substitute into the equil equation from the equil line in the ICE chart and solve for x = (OH^-) and convert to pH.

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