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.
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 = ?
Okay so i got
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.
To solve part (a) of the problem, you need to use the equilibrium expression for the dissociation of benzonic acid, which is given by:
HC7H5O2 ⇌ H+ + C7H5O2-
The equilibrium constant expression for this dissociation reaction is:
K = [H+][C7H5O2-] / [HC7H5O2]
Given that the value of K is 6.4x10^-5 and the concentration of the benzonic acid solution is 0.15 M, you can set up the equilibrium expression as follows:
6.4x10^-5 = [H+][C7H5O2-] / 0.15
Since benzonic acid is a weak acid, you can assume that the concentration of [C7H5O2-] is approximately equal to the concentration of [H+]. Let's denote both concentrations as x:
6.4x10^-5 = x*x / 0.15
Now, you can solve for x. Rearranging the equation:
0.15 * 6.4x10^-5 = x^2
9.6x10^-6 = x^2
Taking the square root of both sides:
x = √(9.6x10^-6)
x ≈ 3.1x10^-3
Therefore, the concentration of [H3O+] in the 0.15 M solution of benzonic acid is approximately 3.1x10^-3 M.
For part (b), the percent ionization of HC7H5O2 can be calculated using the formula:
% ionization = (concentration of H+ produced / initial concentration of HC7H5O2) x 100
In this case, the concentration of H+ produced is approximately equal to the concentration of [H3O+], which we found to be 3.1x10^-3 M. The initial concentration of HC7H5O2 is given as 0.15 M.
% ionization = (3.1x10^-3 / 0.15) x 100
% ionization ≈ 2.07%
Hence, the percent ionization of HC7H5O2 is approximately 2.07%.