For propanoic acid HC3H5O2, Ka = 1.3 ✕ 10−5 and 0.102 M solution.
[HC3H5O2] ____M
[C3H5O2− ] ____M
percent dissociation ____percent
If it helps, I have:
[H+] = 0.00115
[OH-] = 8.696e-12
pH = 2.94
These are verified to be correct.
So you have the answers and they are verified. What do you need from us?
[H⁺] = [C₃H₅O₂ˉ] = √Ka[HC₃H₅O₂]) = √(1.3 x 10ˉ⁵)(0.102) M = 0.00115M
[HC₃H₅O₂] = 0.102M
%Dissociation = ([H⁺] or [C₃H₅O₂ˉ])]/[HC₃H₅O₂])100% = [(0.00115M)/(0.102M)]100% = 1.13% dissociated
To find the concentration of [HC3H5O2] and [C3H5O2-], we can start by assuming that x is the amount of propanoic acid (HC3H5O2) that dissociates into the propanoate ion (C3H5O2-).
Using the given Ka value and the equilibrium expression for the dissociation of propanoic acid, we have:
Ka = [C3H5O2-][H3O+] / [HC3H5O2]
Since propanoic acid is a weak acid, we can assume that the concentration of [H3O+] is equal to the concentration of [C3H5O2-]. Therefore, we let [H3O+] = [C3H5O2-] = x.
Substituting these values into the equilibrium expression and rearranging, we have:
Ka = x^2 / (0.102 - x)
Solving this quadratic equation will give us the value of x, which represents the concentration of [C3H5O2-] as well as the concentration of [H3O+].
Let's solve the equation:
1.3 x 10^-5 = x^2 / (0.102 - x)
Simplifying further, we get:
1.3 x 10^-5 (0.102 - x) = x^2
0.001326 - 1.3 x 10^-5x = x^2
Rearranging and setting the equation to zero:
x^2 + 1.3 x 10^-5x - 0.001326 = 0
This is a quadratic equation. We can use the quadratic formula to solve for x:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = 1, b = 1.3 x 10^-5, and c = -0.001326.
After substituting the values into the quadratic formula, we get two possible values for x: x1 and x2.
The concentration of [C3H5O2-] is equal to x1 or x2, and the concentration of [HC3H5O2] is equal to 0.102 - x1 or 0.102 - x2.
Finally, to find the percent dissociation of propanoic acid, we can use the equation:
% dissociation = (x1 or x2) / (0.102) * 100
This will give us the percent dissociation of propanoic acid.