Answer:
The pH of a solution of a monoprotic acid can be related to the acid dissociation constant (Ka) of the acid by the following equation:
pH = -log[H3O+]
where [H3O+] is the hydronium ion concentration in the solution. For a monoprotic acid HA, the equilibrium equation for its dissociation in water is:
HA + H2O ⇌ H3O+ + A-
where A- is the conjugate base of the acid.
The equilibrium constant expression for this reaction is:
Ka = [H3O+][A-]/[HA]
Given that the acid is monoprotic, the concentration of [HA] that dissociates is equal to the initial concentration of the acid. Let [HA] = 1.82 M, and let x be the concentration of [H3O+] and [A-] formed when the acid dissociates. Then, we can set up an ICE table as follows:
| | HA | H3O+ | A- |
|-------------|------------|--------------|-------------|
| Initial | 1.82 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 1.82 - x | x | x |
Substituting these concentrations into the expression for Ka, we get:
Ka = (x)(x) / (1.82 - x)
We also know that the pH of the solution is 2.97, so:
pH = -log[H3O+]
2.97 = -log(x)
x = 10^(-2.97) = 7.02 x 10^(-3) M
Substituting this value of x into the expression for Ka, we get:
Ka = (7.02 x 10^(-3))(7.02 x 10^(-3)) / (1.82 - 7.02 x 10^(-3))
Ka = 1.77 x 10^(-4)
Therefore, the value of Ka for the acid is 1.77 x 10^(-4).