For propanoic acid (HC3H5O2, Ka = 1.3 10-5), determine the concentration of all species present, the pH, and the percent dissociation of a 0.29 M solution.
[H+]
[C3H5O2-]
[OH -]
[HC3H5O2]
pH
percent dissociation
Propanoic acid = HPr
HPr ==> H^+ + Pr^-
Set up an ICE chart and substitute into the below equation.
Ka = (H^+)Pr^-)/(HPr)
Solve for (H^+) and convert to pH.
(Pr^-) = (H^+)
You can find OH^- from (H^+)(OH^-) = 1E-14
%diss = (H^+)/(HPr)*100 = ??
To determine the concentration of all species present, the pH, and the percent dissociation of a 0.29 M solution of propanoic acid (HC3H5O2), we need to consider its dissociation reaction:
HC3H5O2 ⇌ H+ + C3H5O2-
Given that the Ka of propanoic acid is 1.3 × 10^-5, we can set up an ice-table to determine the concentrations of all species at equilibrium.
Let's assume that x Moles/L of HC3H5O2 dissociates to form x Moles/L of H+ and C3H5O2-.
Using the initial concentration and the equilibrium concentrations, we can write:
Initial: HC3H5O2 = 0.29 M
Equilibrium: H+ = x M
C3H5O2- = x M
Using the equilibrium expression for Ka, we have:
Ka = [H+] [C3H5O2-] / [HC3H5O2]
Substituting the given values:
1.3 × 10^-5 = x * x / (0.29 - x)
Since x is small compared to 0.29, we can assume that 0.29 - x ≈ 0.29.
1.3 × 10^-5 = x^2 / 0.29
Rearranging the equation, we get:
x^2 = 1.3 × 10^-5 * 0.29
x^2 = 3.77 × 10^-6
Taking the square root of both sides, we find:
x ≈ 6.14 × 10^-3
Now that we have the concentration of H+ and C3H5O2- at equilibrium, we can calculate the concentration of [OH-].
Since water is always present in solution, we can use the equation:
[H+] [OH-] = Kw = 1 × 10^-14
Let's assume [OH-] = y M.
With [H+] ≈ 6.14 × 10^-3 M, we have:
(6.14 × 10^-3) (y) = 1 × 10^-14
Solving for y, we get:
y = (1 × 10^-14) / (6.14 × 10^-3)
y ≈ 1.63 × 10^-12 M
Now, we can calculate the concentration of HC3H5O2:
[HC3H5O2] = Initial concentration - x
[HC3H5O2] = 0.29 M - 6.14 × 10^-3 M
[HC3H5O2] ≈ 0.29 M
Finally, we can calculate the pH using the equation:
pH = -log[H+]
pH = -log(6.14 × 10^-3)
pH ≈ 2.21
To find the percent dissociation, we use the equation:
% dissociation = (amount dissociated / initial amount) × 100
The amount of HC3H5O2 dissociated is x, and the initial amount is 0.29 M. Therefore:
% dissociation = (6.14 × 10^-3 / 0.29) × 100
% dissociation ≈ 2.12%
So, the concentrations of the species present in the solution are:
[H+] ≈ 6.14 × 10^-3 M
[C3H5O2-] ≈ 6.14 × 10^-3 M
[OH-] ≈ 1.63 × 10^-12 M
[HC3H5O2] ≈ 0.29 M
The pH of the solution is approximately 2.21, and the percent dissociation is approximately 2.12%.
To determine the concentration of all species present, the pH, and the percent dissociation of a 0.29 M solution of propanoic acid, we can use the principles of acid dissociation and equilibrium.
1. First, let's determine the initial concentration of propanoic acid ([HC3H5O2]initial):
Given: [HC3H5O2]initial = 0.29 M
2. Next, let's set up an ICE table (Initial, Change, Equilibrium) to track the changes in concentration:
HC3H5O2 (aq) ⇌ H+ (aq) + C3H5O2- (aq)
Initial: [HC3H5O2]initial 0 0
Change: -x +x +x
Equilibrium: [HC3H5O2]initial - x x x
3. Now, we can equate the expression for the acid dissociation constant (Ka) to the equilibrium concentrations of the species:
Ka = [H+][C3H5O2-]/[HC3H5O2]initial - x
Since the concentration of water is constant, we can consider the initial concentration of water as the initial concentration of propanoic acid. Therefore, we can rewrite the equation as:
Ka = [H+][C3H5O2-]/[HC3H5O2]initial
Rearranging the equation gives:
[H+][C3H5O2-] = Ka * [HC3H5O2]initial
Plugging in the given values:
[H+][C3H5O2-] = (1.3 * 10^-5)*(0.29)
4. Since the concentration of [H+] and [C3H5O2-] are equal at equilibrium, we can calculate their concentration using this equation:
[H+] = [C3H5O2-] = √(Ka * [HC3H5O2]initial)
Plugging in the values:
[H+] = [C3H5O2-] = √((1.3 * 10^-5)*(0.29))
5. Now, we can calculate the hydroxide ion concentration ([OH-]) using the equation:
[OH-] = Kw/[H+]
Given that Kw = 1.0 x 10^-14 at 25°C, we can plug in the values to calculate [OH-].
6. Finally, we can use the equation pH = -log[H+] to calculate the pH.
7. To calculate the percent dissociation, we can use the equation:
Percent dissociation = ([H+]/[HC3H5O2]initial) * 100
Now, with the given information and using the explained steps, you can calculate the concentration of all species, the pH, and the percent dissociation for the propanoic acid solution.