Chemistry

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.240 M HClO(aq) with 0.240 M KOH(aq). The ionization constant for HClO can be found here.

asked by Judith
  1. Divide the titration curve into 4 parts.
    a. beginning
    b. between beginning and eq pt.
    c. eq pt
    d. after eq pt.

    First, determine mL to eq pt.
    mLacid x M aci = mL base x M base and that determination tells you how to divide into the four part.
    a.......... HClO ==> H^+ + ClO^-
    I..........0.240M.....0.....0
    C............-x.......x......x
    E.........0.240-x.....x......x

    Substitute the E line into the Ka expression for HClO and solve for x = (H^+), then convert to pH.

    b. mmols to start = mL x M = 50*0.240 = about 12.
    mmols KOH added. You don't give a value but lets say 25 mL. So 25 x 0.240 = about 6. Therefore, you have 6 acid left and you formed 6 of the salt from
    .......HClO + KOH ==> KClO + H2O
    I.......12.....0........0......0
    add............6.............
    C.......-6.....-6........6......6
    E........6.......0.......6.......6

    Use the Henderson-Hasselbalch equation to solve for the pH.
    pH = pKa for HClO + log (6/6)
    ?

    c. The pH is determined by the hydrolysis of the salt.
    ........ClO^- + HOH ==>HClO + OH^-
    I......0.12............0......0
    C........-x............x......x
    E........-0.12-x........x.....x

    Kb for ClO^- = (Kw/Ka for HClO) = (x*x)/(0.12-x).
    Substitute and solve for x = OH^- and convert to pH.
    NOTE: The concn of the salt at the eq.pt is just 1/2 of 0.240. Since they are the same molarity you have added 50 mL KOH so the M salt is just 1/2 of what acid or base started.

    d. After the eq. pt is just excess KOH.

    posted by DrBob222

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