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Find the Ph:

of 7.01 x 10-5 M Ba(OH)2


  • Chemistry -

    Ba(OH)2 ==> Ba^+2 + 2OH^-
    [Ba(OH)2] = 7.01E-5
    [OH^-] = 2*7.01E-5
    pOH = -log(OH^-)
    pH + pOH = pKw = 14.

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