Probably not. I could tell better if you had shown your work. I think you solve for x but failed to realize that x = OH^-. Remember
Ac^- + HOH ==> HAc + OH^-
So your 5.27 probably is pOH. Subtract from 14 to get pH.
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) - DrBob222, Saturday, April 7, 2012 at 10:58pm
You want to find Kb for acetate. That is Kw/Ka and Ka is for acetic acid.
Chemistry(Please help) - Hannah, Saturday, April 7, 2012 at 11:03pm
ok thank you!
Chemistry(Please help) - Hannah, Saturday, April 7, 2012 at 11:10pm
so for Kb
Chemistry(Please help) - Hannah, Saturday, April 7, 2012 at 11:12pm
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) - DrBob222, Saturday, April 7, 2012 at 11:26pm
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) - Hannah, Sunday, April 8, 2012 at 4:30am
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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