what is the pH of a 0.025 M aqueous solution of sodium propionate, NaC3H5O2? What is the concentration of propionic acid in the solution?
To determine the pH of the aqueous solution of sodium propionate (NaC3H5O2), we need to consider the dissociation of the compound in water. Sodium propionate is a salt, which means it dissociates into its respective ions in water.
The sodium ion (Na+) does not have any effect on the pH of the solution since it is the conjugate base of a strong acid (NaOH). However, the propionate ion (C3H5O2-) is the conjugate base of the weak acid propionic acid (HC3H5O2), and it can hydrolyze in solution.
To find the pH, we need to calculate the concentration of hydroxide ions (OH-) in the solution. Since the propionate ion can hydrolyze, we can consider it as the source of hydroxide ions. The hydrolysis reaction is as follows:
C3H5O2- + H2O ⇌ HC3H5O2 + OH-
The concentration of HC3H5O2 (propionic acid) can be determined from the initial concentration of NaC3H5O2 using the stoichiometry of the reaction. For every mole of NaC3H5O2, one mole of HC3H5O2 is formed.
Therefore, the concentration of propionic acid (HC3H5O2) is also 0.025 M.
Now, to find the pH, we need to determine the concentration of hydroxide ions (OH-) using the concentration of propionic acid. Since propionic acid is a weak acid, we can assume that it does not fully dissociate, and consider it in equilibrium with its conjugate base and hydrogen ions:
HC3H5O2 ⇌ C3H5O2- + H+
The equilibrium constant for this reaction, known as the acid dissociation constant (Ka), can be used to find the concentration of hydroxide ions (OH-) from the concentration of propionic acid (HC3H5O2).
pH = -log10 [H+]
Now, using the Ka value for propionic acid (Ka = 1.3 x 10^-5), we can set up an equilibrium expression:
Ka = [C3H5O2-][H+] / [HC3H5O2]
Since the concentration of C3H5O2- is equal to the concentration of HC3H5O2, we can simplify the equation to:
Ka = [H+]^2 / [HC3H5O2]
Rearranging the equation, we get:
[H+]^2 = Ka * [HC3H5O2]
Taking the square root of both sides:
[H+] = √(Ka * [HC3H5O2])
Substituting the values, we get:
[H+] = √(1.3 x 10^-5 * 0.025)
[H+] ≈ 9.53 x 10^-4 M
Finally, we can calculate the pH:
pH ≈ -log10 [H+]
pH ≈ -log10 (9.53 x 10^-4)
pH ≈ 3.02
Therefore, the pH of the 0.025 M aqueous solution of sodium propionate is approximately 3.02.
Additionally, since the concentration of propionic acid (HC3H5O2) in the solution is also 0.025 M, we can say that the concentration of propionic acid in the solution is 0.025 M.
This is a hydrolysis problem. The sodium propionate ion (which I'll call Pr^-) hydrolyzes.
Pr^- + HOH ==> HPr + OH^-
Set up an ICE chart and substitute into the Kb expression for Pr^-. (HPr is propionic acid).
Kb = (Kw/Ka) = (HPr)(OH^-)/(Pr^-).
(HPr_ = x = (OH^-). (Pr^-) = 0.025M from the problem. Solve for x and convert to pH. Post your work if you get stuck.