What is the pH of a 0.115 mol/L solution of sodium acetate, CH3COONa, at 25oC? (pKa=4.74 at 25oC for CH3COOH.)
The pH is determined by the hydrolysis of the CH3COONa, which I will call NaAc.
.........Ac^- + HOH --> HAc + OH^-
I.......0.115............0.....0
C.......-x...............x.....x
E.....0.115-x............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.115-x)
Substitute and solve for x = OH^- and convert to pH.
find Kb from pKa
Find CH3COO- concentration
Use ICE table to find concertation of OH-
Find pH from there
To find the pH of a solution of sodium acetate, we need to consider the dissociation of sodium acetate into its respective ions in water. Sodium acetate (CH3COONa) dissociates into sodium ions (Na+) and acetate ions (CH3COO-).
At equilibrium, some of the acetate ions react with water to form acetic acid (CH3COOH) and hydroxide ions (OH-). This reaction is described by the equation:
CH3COO- + H2O ⇌ CH3COOH + OH-
To determine the pH, we need to determine the concentration of hydronium ions (H3O+) or hydroxide ions (OH-) in the solution.
Since sodium acetate is a salt of a weak acid (acetic acid), and its conjugate base (acetate ion) is derived from a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log ([A-]/[HA])
Where:
pH = the acidity or basicity of a solution
pKa = the logarithmic acid dissociation constant of the weak acid (CH3COOH)
[A-] = the concentration of the conjugate base (acetate ion, CH3COO-)
[HA] = the concentration of the weak acid (acetic acid, CH3COOH)
In this case, the acetate ion acts as a base (A-) and acetic acid acts as an acid (HA). Given that the molar concentration of sodium acetate is 0.115 mol/L, the concentrations of acetate ion and acetic acid are also 0.115 mol/L.
Now, we substitute the values into the Henderson-Hasselbalch equation:
pH = 4.74 + log (0.115/0.115)
Since the logarithm of 1 is 0, the equation simplifies to:
pH = 4.74 + 0
Therefore, the pH of the 0.115 mol/L solution of sodium acetate is 4.74 at 25°C.