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..
A) 0.16M but i did it again and got 7.2M so i'm very stuck as to the very different answers i got.
B) 12.4% ionization.
If you missed A you probably missed B.
Call benzoic acid HB, then
................HB ==> H^+ + B^-
initial........0.15....0......0
change..........-x......x.....x.
equil........0.15-x.....x......x
K = (H^+)(B^-)/(HB)
Substitute and solve for x.
%ion = [(H^+)/0.15]*100 = ?
To calculate [H3O+] in a solution of benzonic acid, we can use the given equilibrium constant (K) and the initial concentration of the acid.
a) To start, we need to write the balanced equilibrium equation for the dissociation of benzonic acid:
HC7H5O2(aq) + H2O(l) ⇌ H3O+(aq) + C7H5O2-(aq)
The equation tells us that one molecule of benzonic acid (HC7H5O2) reacts with one molecule of water (H2O) to produce one hydronium ion (H3O+) and one benzoate ion (C7H5O2-).
Next, we need to set up an equilibrium expression using the equilibrium constant (K):
K = [H3O+][C7H5O2-] / [HC7H5O2]
Given that K = 6.4x10^-5, we can substitute this value into the equilibrium expression:
6.4x10^-5 = [H3O+][C7H5O2-] / 0.15
Before we can solve for [H3O+], we need to determine the concentration of the benzoate ion ([C7H5O2-]) in the solution. Since benzonic acid is a weak acid, we can assume that it dissociates very little, so the amount of benzoate ion formed is negligible compared to the initial concentration of benzonic acid. Therefore, we can approximate the concentration of the benzoate ion as zero for this calculation.
6.4x10^-5 = [H3O+](0) / 0.15
Simplifying the equation, we find:
[H3O+] = 0
So, the concentration of hydronium ions ([H3O+]) in the solution is zero.
b) To calculate the percent ionization of HC7H5O2, we need to compare the concentration of the ionized form (H3O+) to the initial concentration of the acid (HC7H5O2).
Percent Ionization = ([H3O+] / [HC7H5O2]) * 100
Since we found that [H3O+] is zero, the percent ionization will also be zero.
Therefore, the percent ionization of HC7H5O2 is 0%.
It's important to note that in this case, the value of K is very small (6.4x10^-5), indicating that the acid does not dissociate significantly in water, resulting in a low percent ionization.