To find the pH of a 0.35 M solution of sodium formate, we first need to understand that sodium formate is the sodium salt of formic acid. In solution, it dissociates into sodium ions (Na+) and formate ions (HCOO–), which can act as a weak base.
Since we know the Ka of formic acid (HCOOH) is 1.8 x 10-4, we can find the Kb of the formate ion using the relationship:
Kw = Ka * Kb
Where Kw is the ion product of water (1.0 x 10-14 at 25°C). Rearranging gives us:
Kb = Kw / Ka
Substituting the values, we get:
Kb = (1.0 x 10-14) / (1.8 x 10-4) ≈ 5.56 x 10-11
Next, we apply the Kb to find the concentration of hydroxide ions (OH–) produced in the solution of sodium formate. Assuming x is the concentration of OH–:
Kb = [OH–]2 / [HCOO–]
Where [HCOO–] can be assumed to be approximately 0.35 M because the change due to dissociation is negligible. Thus, we have:
5.56 x 10-11 = x2 / 0.35
Solving for x:
x2 = 5.56 x 10-11 * 0.35
x2 ≈ 1.946 x 10-11
x ≈ √(1.946 x 10-11) ≈ 4.41 x 10-6
This x value represents the [OH–] concentration. Now we can calculate the pOH:
pOH = -log[OH–] = -log(4.41 x 10-6) ≈ 5.35
Finally, we can find the pH using the relationship:
pH + pOH = 14
Thus,:
pH = 14 – pOH ≈ 14 – 5.35 = 8.65
Therefore, the pH of a 0.35 M solution of sodium formate is approximately 8.65.