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
    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 -

    pOH=-log[OH-], so [OH-]=10^(-6.00)=1x10^-6
    HOCl + OH--------H20 + OCl-
    .05M X 0
    -X -X +X
    .05-X 0 X
    8.00=-LOG(3.5X^-8)+ LOG [X/ (.05-X)]
    8.00=7.46+LOG [X/(.05-X)]
    TO GET RID OF LOG, RAISE 10^.544
    X=.0388 mol
    .0388mol x (1 L/ .100 mol)=3.9 L

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