Final answer:
The acid ionization constant (Ka) for a monoprotic acid with an initial concentration of 8.0×10⁻² M and 0.60% dissociation is calculated to be 2.88×10⁻⁸.
Step-by-step explanation:
To calculate the acid ionization constant (Ka) for the monoprotic acid with a given percent dissociation, we should recognize that the percent dissociation expresses how much of the original acid has ionized into its ions at equilibrium. Given that the acid has a 0.60% dissociation, this means that 0.60% of the initial concentration has become ions.
Given a 8.0×10⁻² M solution and a 0.60% dissociation:
- Calculate the concentration of ionized acid, which will be 0.60% of 8.0×10⁻² M.
- Ionized acid concentration: (0.60/100) x 8.0×10⁻² M = 4.8×10⁻⁵ M.
- For the reaction HA → H+ + A-, the [H+] and [A-] at equilibrium is 4.8×10⁻⁵ M, and un-ionized [HA] = 8.0×10⁻² M - 4.8×10⁻⁵ M.
- The Ka expression is Ka = [H+] [A-] / [HA]. We plug the equilibrium concentrations into this expression to determine Ka.
Ka calculation:
Ka = ((4.8×10⁻⁵ M) ^2) / (8.0×10⁻² M - 4.8×10⁻⁵ M)
Approximate [HA] as 8.0×10⁻² M since the ionized amount is very small compared to the initial concentration:
Ka = ((4.8×10⁻⁵) ^2) / 8.0×10⁻²
Ka = (2.304×10⁻⁹) / 8.0×10⁻²
Ka = 2.88×10⁻⁸
Thus, the acid ionization constant (Ka) for this acid is 2.88×10⁻⁸.