What is initial concentration of a sodium acetate if the pH (at epuilibrium) is measured to be 9.30?
............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.