ap chemistry

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what volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to acheive a PH of 8.00?

  • ap chemistry -

    1000 mL x 0.05M = 50 millimols.
    x mL x 0.1M = ? mmols.
    -----------------
    ............HOCl + OH^- ==> OCl^- + H2O
    initial.....50......0........0........0
    add.................x................
    change......-x....-x.........x........x
    equil.......50-x....0.........x........x
    I used 3E-8 for Ka for HOCl
    pH = pKa + log(base)/(acid)
    8.00 = 7.52 + log(x)/(50-x)
    Solve for x = millimols OH^- added. Then mL = mmols/M. I get approximately 380 mL

  • ap chemistry -

    I believe you made a slight error in the last calculation from mmol --> mol. The answer should be around 3.8L.

  • ap chemistry -

    Ka=3.5X10^-8
    pH+pOH=14
    14-8.00=pOH
    pOH=-log[OH-], so [OH-]=10^(-6.00)=1x10^-6
    HOCl + OH--------H20 + OCl-
    .05M X 0
    -X -X +X
    .05-X 0 X
    THEN USE HENDERSON-HASSELBACK EQUATION
    pH=pKa+LOG[A-/HA]
    8.00=-LOG(3.5X^-8)+ LOG [X/ (.05-X)]
    8.00=7.46+LOG [X/(.05-X)]
    .544=LOG[X/(.05-X]
    TO GET RID OF LOG, RAISE 10^.544
    3.5=X/(.05-X)
    3.5(.05-X)=X
    .175-3.5X=X
    .175=3.5X+X
    .175=X(3.5+1)
    .175=4.5X
    X=.0388 mol
    .0388mol x (1 L/ .100 mol)=3.9 L

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