2007-04-14 06:27:14 · 4 answers · asked by W 1 in Science & Mathematics ➔ Chemistry
pH = 3 >> [H+] = 10^-3
For the equilibrium
CH3COOH <> CH3COO- + H+
at equilibrium
0.12-10^-3 . . . . .10^-3 . . . . .10^-3
Ka = (10^-3)(10^-3) / 0.12-10^-3 = 8.40 x10^-6
2007-04-14 06:36:09 · answer #1 · answered by Anonymous · 0⤊ 0⤋
CH3COOH --> CH3COO- + H+
I .12....................0......... . . .0
C X......................+X......... . . +X
E .12-X..................X............. . X
Since the products have an initial concentration of 0 they will both be x. Since we have the pH we could find out what the H+ concentration is using pH=-log[H+]
pH=-log[H+]
3=-log[H+]
10^-3 = [H+]
once you find the H+ concentration you find out what X is.
Ka= Products/Reactants
Ka= x^2/[.12-x]
Good Luck!
2007-04-14 06:39:45 · answer #2 · answered by specialk 2 · 0⤊ 0⤋
ka= -log(3.0)
take the negative log of the PH and you will have the ka value.
2007-04-14 06:33:28 · answer #3 · answered by dmblue 2 · 0⤊ 0⤋
pH = (pKa -logc)/2
pKa = 2pH +logc=6+log 0.12 =5.08
Ka = 10^-5.08 =810^-6
2007-04-14 07:14:39 · answer #4 · answered by maussy 7 · 0⤊ 0⤋