Chemistry(Please help)

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What is the pH of a 5.06E-2 M aqueous solution of sodium acetate?

I know that I would set this up the same way I did for the previous question that I posted. I am just confused on how to find the Ka. I know it is kw/kb but for what compound is what I do not know.

  • Chemistry(Please help) -

    You want to find Kb for acetate. That is Kw/Ka and Ka is for acetic acid.

  • Chemistry(Please help) -

    ok thank you!

  • Chemistry(Please help) -

    so for Kb

  • Chemistry(Please help) -

    So for Kb I did kw/ka = 1.0e-14/1.8e-5 = 5.55e-10 . Now I would multiply this by 5.06e-6 and take the square root and then covert to pH by taking the -log. Is this correct?

  • Chemistry(Please help) -

    Yes, however the number is 5.06E-2 and not -6. Note, too, that a quadratic may be necessary. You will substitute 0.0506-x for (acetate) in the equation and the x may (or may not) be negligible.

  • Chemistry(Please help) -

    I set this up as 5.55e-10 = x^2/0.0506-x

    After converting to pH I got 5.27 but it said that this was incorrect so maybe I do need to use a quadratic, but I am not sure how to go about this.

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