Posted by Justin on Tuesday, March 20, 2012 at 2:34pm.
............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.
Related Questions
Chemistry - I just did a lab on Acids and Bases and I am having trouble with the...
Chemistry (Acids and Bases) - I just did a lab on Acids and Bases and I am ...
Chemistry - IN the polystyrene beaker, mix 20mL of 0.1 M Acetic acid and 25mL of...
Chemistry - I did an experiment on Buffers: In a polystyrene beaker, mix 20 ml ...
chemistry - What is the pH of a sodium acetate solution prepared by adding 0....
Chemistry - I have a question about buffers. Part A So it starts with 20ml 0.1 ...
Chemistry - Hydrolysis of Salt: Using the universal indicator we had to ...
chemsitry - calculate the molarity of sodium in 12.4g of sodium acetate diluted...
chemistry - Okay let me write it this way. I mixed 30 mL of 0.1 M Acetic Acid ...
Chemistry(Please help) - For an acetic acid-sodium acetate buffer to have a pH ...
For Further Reading
