Chemistry II

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For the titration of 50.00 mL of 0.106 M NH3 with 0.225 M HCl, calculate

(A) the initial pH and when the pH neutralization is
(B) 25.0% complete
(C) 50.0% complete
(D) 100.0% complete

  • Chemistry II -

    You need to be able to recognize where you are on the titration curve with these problems.
    50.00 x 0.106 = millimoles NH3
    x mL x 0.225M HCl = ? mmols HCl

    a. At the beginning you have a solution of NH3 and nothing else. So you know how to solve for the pH of a 0.106 M NH3 soln. Set up the ICE chart and go from there.
    d. At the equivalence point you have no acid and no base; the pH is determined by the hydrolysis of the salt. The salt is NH4Cl and it is the NH4^+ that is hydrolyzed. You will need to determine the (NH4Cl). You do that by mmoles NH3 at the start and divide by mL base(at start)+mL acid to get to the eq point). I'll call the (NH4^+) = 0.2 but that will not be right.
    .......NH4^+ + H2O ==> H3O^+ + NH3
    initial.0.2............0........0
    change..-x.............x.......x
    equil...0.2-x...........x......x

    Then Ka for NH4^+ = (Kw/Kb for NH3) = (H3O^+)(NH3)/(NH4^+)
    Substitute from the ICE chart and solve for x and convert that to pH.

    b and c.
    For all points between the beginning and the equivalence point, use the Henderson-Hasselbalch equation.

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