calculate the PH of a mixture containing 30 mL, 0.10M HC2H3O2? (Ka=1.8x10^-8 of HC2H3O2)
3 answers
If this is a mixture you didn't post all of the problem.
yeah i forgot to put
+ 30 mL, 0.10 M of NaOH
+ 30 mL, 0.10 M of NaOH
You also listed Ka wrong. That's 1.8E-5.
acetic acid is HAc = HC2H3O2
millimols HAc = mL x M = 30 x 0.1 = 3
millimols NaOH = 3
........HAc + NaOH ==> NaAc + H2O
I.......3.......0........0.....0
add.............3.............
C......-3......-3........3......3
E.......0.......0........3......3
So you have at equilibrium a solution of sodium acetate, NaAc, and the molarity is 3 millimols/60 mL = 0.05M
The acetate ion hydrolyzes in water as follows:
..........Ac^- + HOH ==> HAc + OH^-
I.......0.05..............0.....0
C.........-x..............x.....x
E......0.05-x.............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x)
Solve for x = (OH-) and convert to pH.
acetic acid is HAc = HC2H3O2
millimols HAc = mL x M = 30 x 0.1 = 3
millimols NaOH = 3
........HAc + NaOH ==> NaAc + H2O
I.......3.......0........0.....0
add.............3.............
C......-3......-3........3......3
E.......0.......0........3......3
So you have at equilibrium a solution of sodium acetate, NaAc, and the molarity is 3 millimols/60 mL = 0.05M
The acetate ion hydrolyzes in water as follows:
..........Ac^- + HOH ==> HAc + OH^-
I.......0.05..............0.....0
C.........-x..............x.....x
E......0.05-x.............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x)
Solve for x = (OH-) and convert to pH.
2 answers