a) Calculate [H3O^+] in a 0.15M solution of benzonic acid, HC7H5O2(aq) having K of 6.4x10^-5.
b) Calculate the per cent ionization of the HC7H5O2.
I got..
Kb = [HC7H5O3][OH]/HC7H5O2
6.3x10^-5 x Kb =1.0x10^-14
Kb = 1.0x10^-14/6.3x10^-4 = 1.5x10^-10
1.5x10^-10 = (x)(x)/0.15
i'm not sure where i got from there.
4 answers
Where i go from there **
You start over. Benzoic acid is an acid, not a base. Let's call benzoic acid HB.
..........HB ==> H^+ + B^-
init.....0.15.....0.....0
change....-x......x.....x
equil....0.15-x...x.....x
Ka + (H^+)(B^-)/(HB)
Substitute the equilibrium line above into the Ka expression and solve for x = (H^+) = (H3O^+).
b. %ion = [(H^+)/0.15]*100 = ?
..........HB ==> H^+ + B^-
init.....0.15.....0.....0
change....-x......x.....x
equil....0.15-x...x.....x
Ka + (H^+)(B^-)/(HB)
Substitute the equilibrium line above into the Ka expression and solve for x = (H^+) = (H3O^+).
b. %ion = [(H^+)/0.15]*100 = ?
Okay so i got
Ka = (x)(x)/0.15 = 6.5x10^-5
Ka = (x)(x)/0.15 = 6.5x10^-5
OK. Ka = 6.5E-5 = (x*x)/0.15; now solve for x which is (H^+) (H3O^+ to be more exact) which is what the question asked. Then use the x value to solve for percent ionization as I showed you in part b.
1 answer