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Posted by on Wednesday, May 11, 2011 at 10:00am.

how do you calculate the pOH of a 0.25M solution of monochloroacetic acetic acid at 25C

  • Chemistry - , Wednesday, May 11, 2011 at 10:22am

    You look up the Ka or pKa for the acid:2.87= pKa

    or Ka=(x)(x)/(.25-x)

    pKa=2p(x)-p(.25-x)

    assume x is small..
    2.87=2p(x)-p(.25)

    p(x)=2.87/2 -1/2 (-.6)

    p(H)=1.44+.3=1.74
    p(OH)=14-pH

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