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What would be the pH of a 0.1 M aqueous solution of the phenolate ion, C6H5O–? (Ka for phenol, C6H5OH, is 1.3 x 10–10) I got the answer 5.44 but I think it's wrong. Do I have to subtract it from 14?

  • Chemistry -

    My answer was slightly different for pOH than yours (5.11) but yes you need to subtract from 14 since I suspect you solved for OH^-.
    Call phenolate ion P^-, then
    .......P^- + HOH ==> HP + OH^-

    Kb for phenolate = (Kw/Ka for phenol) = (x)(x)/(0.1-x) and solve for x = (OH^-) then convert to pOH and pH.

  • Chemistry -

    I reworked the problem and realized I didn't find Kb at first and plugged in Ka instead. Thanks!

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