What is the pH of 0.2 M solution of sodium propanoate?
I'm lazy so let's call sodium propanoate NaPr. The Pr anion is hydrolyzed.
Pr^- + HOH ==> HPr + OH^-
Kb= (Kw/Ka) = (HPr)(OH^-)/(Pr^-)
Substitute for Kw and Ka (for propionic acid) and (Pr^-). Let X = HPr and OH. Solve for x, convert to pOH, then to pH.
To determine the pH of a solution of sodium propanoate, we need to understand the properties of the compound and its interaction with water.
Sodium propanoate (CH3CH2COONa) is a salt that dissolves in water to produce sodium ions (Na+) and propanoate ions (CH3CH2COO-). The propanoate ion can react with water to form propanoic acid (CH3CH2COOH), which is a weak acid.
To find the pH of the solution, we need to consider the dissociation of the propanoate ions and the subsequent equilibrium reaction with water. The propanoate ion can act as a base by accepting a proton from water:
CH3CH2COO- + H2O ⇌ CH3CH2COOH + OH-
The equilibrium constant for this reaction is called the base dissociation constant (Kb) for the propanoate ion.
To calculate the pH, we need to determine the concentration of hydroxide ions (OH-) in the solution. This concentration can be found by considering the dissociation of water, which is represented by the equilibrium reaction:
H2O ⇌ H+ + OH-
The equilibrium constant for this reaction is called the acid dissociation constant (Kw) for water. At 25°C, the value of Kw is 1.0 x 10^-14.
Using the equation:
Kw = [H+][OH-]
Starting with a 0.2 M solution of sodium propanoate, we can assume that it dissociates completely into sodium ions and propanoate ions. Therefore, the concentration of propanoate ions is 0.2 M.
Since the propanoate ion reacts with water to form hydroxide ions, the concentration of hydroxide ions is equal to the concentration of propanoate ions (0.2 M).
Using the equation Kw = [H+][OH-], we can solve for [H+] by rearranging the equation:
[H+] = Kw / [OH-]
[H+] = 1.0 x 10^-14 / 0.2
[H+] = 5.0 x 10^-14
To convert the concentration of H+ to pH, we use the equation:
pH = -log[H+]
pH = -log(5.0 x 10^-14)
pH ≈ 13.3
Therefore, the pH of the 0.2 M solution of sodium propanoate is approximately 13.3.