What is the pH a 0.25 mol/L solution of methanoic acid? (Hint: include a reaction with water and the use of the ICE chart)
The chemical equation for the dissociation of methanoic acid (HCOOH) in water is:
HCOOH + H2O ⇌ H3O+ + HCOO-
Let x be the concentration of the H3O+ ions and the HCOO- ions in mol/L.
The initial concentration of HCOOH is 0.25 mol/L, and since it dissociates by losing one proton, the initial concentration of H3O+ and HCOO- is 0 mol/L.
Using the ICE table (Initial, Change, Equilibrium), we can set up the following table:
HCOOH + H2O ⇌ H3O+ + HCOO-
Initial: 0.25 0 0
Change: -x -x +x
Equilibrium: 0.25-x -x x
From the equation, we can conclude that the concentration of HCOOH at equilibrium is 0.25-x mol/L and the concentration of H3O+ and HCOO- at equilibrium is x mol/L.
Since methanoic acid is a weak acid, we can assume that x is negligible compared to 0.25. Therefore, we can approximate the equilibrium concentration of HCOOH as 0.25 mol/L, and the equilibrium concentration of H3O+ and HCOO- as x = 0 mol/L.
The pH of a solution is defined as the negative logarithm (base 10) of the H3O+ concentration. Since the equilibrium concentration of H3O+ is 0 mol/L, the pH is undefined or infinity (∞).