Asked by Justin
What is initial concentration of a sodium acetate if the pH (at epuilibrium) is measured to be 9.30?
Answers
Answered by
DrBob222
............Ac^- + HOH ==> HAc + OH^-
initial.....y........0.......0.....0
change.....-x...............x......x
equil......y-x...............x......x
y = initial concn Ac^- which is our unknown.
pH = 9.30; convert to H^+ which is 5.01E-10
Therefore OH^- = Kw/(H^+) = 1E-14/5E-10 = 2E-5 but you should be more accurate.
Kb for acetate = (Kw/Ka for acetic acid) = (HAc)(OH^-)/(Ac^-)
(Kw/Ka) = (x)(x)/(y-x)
For x you insert 2E-5; solve for y.
initial.....y........0.......0.....0
change.....-x...............x......x
equil......y-x...............x......x
y = initial concn Ac^- which is our unknown.
pH = 9.30; convert to H^+ which is 5.01E-10
Therefore OH^- = Kw/(H^+) = 1E-14/5E-10 = 2E-5 but you should be more accurate.
Kb for acetate = (Kw/Ka for acetic acid) = (HAc)(OH^-)/(Ac^-)
(Kw/Ka) = (x)(x)/(y-x)
For x you insert 2E-5; solve for y.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.