Final answer:
To calculate the pH of a 0.2M HCOOH solution at 25°C, we can use the equilibrium constant expression and solve for [H+]. The pH of the solution was found to be approximately 2.725.
Step-by-step explanation:
The Ka of HCOOH is given as 1.77 x 10^-4 M.
To calculate the pH of a 0.2M HCOOH solution at 25°C, we need to consider the dissociation of HCOOH into H+ and HCOO-. Using the equilibrium constant expression, Ka = [H+][HCOO-]/[HCOOH], we can solve for [H+].
Let x be the concentration of [H+] in the equilibrium:
Ka = x^2 / (0.2 - x)
1.77 x 10^-4 = x^2 / (0.2 - x)
Since x is small compared to 0.2, we can approximate 0.2 - x as 0.2:
1.77 x 10^-4 = x^2 / 0.2
Rearranging the equation and solving for x:
x^2 = 1.77 x 10^-4 * 0.2
x^2 = 3.54 x 10^-5
x = sqrt(3.54 x 10^-5)
x = 1.881 x 10^-3
Now, we can calculate the pH:
pH = -log[H+]
pH = -log(1.881 x 10^-3)
pH ≈ 2.725