Chemical Systems and Equilibrium Question? (Calculating ionization constant)?
Methanoic acid is also called formic acid. It has the chemical formula HCOOH(l). It is a colourless fuming liquid that is mainly used as a preservative. It exhibits the following equilibrium in water:
HCOOH(aq) + H2O(l) → HCOO–(aq) + H3O+(aq)
If the ionization constant for the above acid is 1.8 x 10–4, the pH of methanoic acid in a 0.35 mol/L solution will be ________.
- hcbiochemLv 74 years agoFavorite Answer
While the equation used in Jan's answer is a legitimate equation and gives the correct answer, I prefer that students use the expression for Ka in these calculations.
For the ionization of formic acid (HFa + H2O <--> H3O+ + Fa-), you can write the expression of the ionization constant:
Ka = [Fa-][H3O+]/[HFa] = 1.8X10^-4
In the solution, [H3O+] = [Fa-] = x, and [HFa] will be approximately 0.35 M. So,
1.8X10^-4 = x^2 / 0.35
x = 7.94X10^-3 = [H3O+]
pH = -log (7.94X10^-3) = 2.10
- JanLv 74 years ago
pH = 0.5(pKa - lg(0.35)) = 2.1