To determine the pH of a solution of methanoic acid (HCOOH) with a concentration of 1.15 M, we need to consider the dissociation of methanoic acid and the equilibrium expression for its acid dissociation constant (Ka).
The dissociation of methanoic acid can be represented by the equation:
HCOOH ⇌ H⁺ + COO⁻
The acid dissociation constant (Ka) is given as 1.8 x 10^-4.
The equation for Ka is:
Ka = [H⁺][COO⁻] / [HCOOH]
Assuming that the concentration of H⁺ and COO⁻ are equal, we can simplify the equation to:
Ka = [H⁺]² / [HCOOH]
Since the initial concentration of methanoic acid is 1.15 M, we can assume that the concentration of H⁺ formed from the dissociation is x. Therefore, the concentration of methanoic acid remaining would be 1.15 - x.
Plugging in the values into the simplified equilibrium expression, we get:
1.8 x 10^-4 = (x)² / (1.15 - x)
Solving this equation for x will give us the concentration of H⁺ ions in the solution. However, this calculation involves solving a quadratic equation. After solving the quadratic equation, we find that x ≈ 0.012 M.
Now, we can calculate the pH using the formula:
pH = -log[H⁺]
pH = -log(0.012) ≈ 1.92
Therefore, the pH of a 1.15 M solution of methanoic acid is approximately 1.92.