A 0.490 M solution of an unknown monoprotic weak acid, HA, is 2.80% ionized. What is the value of Ka for HA?
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
You know HA is 0.490. If it is ionized 2.80%, then H^+ and A^- must be 0.490 x 0.0280 = ??. Substitute into Ka expression and solve.
To find the value of Ka for the weak acid HA, we need to use the percent ionization and the initial concentration of the acid.
The percent ionization is given as 2.80%. This means that 2.80% of the initial concentration of HA has dissociated into H+ ions and A- ions.
The initial concentration of HA is given as 0.490 M.
So, we can calculate the concentration of H+ ions and A- ions:
- The concentration of H+ ions (or [H+]) is 2.80% of 0.490 M.
- The concentration of A- ions (or [A-]) is also 2.80% of 0.490 M.
Now, let's assume that the concentration of undissociated HA is x M. Since HA dissociates into H+ and A-, the equilibrium concentrations would be (0.490 - x) M for HA, and (2.80% of 0.490) M for both H+ and A- ions.
Since HA is a monoprotic weak acid, the equation representing the ionization of the acid is:
HA ⇌ H+ + A-
The equilibrium constant expression (Ka) for this reaction can be written as:
Ka = [H+][A-] / [HA]
Now we substitute the concentrations and calculate Ka:
Ka = [(0.028 × 0.490) / (0.490 - x)]
Since the percent ionization is given as 2.80%, it implies that 2.80% of HA has dissociated, which means (0.028 × 0.490) M of H+ and A- ions are formed. So, we can substitute these values in the equation:
Ka = (0.028 × 0.490) / (0.490 - x)
Now, we have an equation with one unknown (x), which represents the concentration of undissociated HA. To find the value of x, we would need additional information, such as the pH of the solution or the value of Ka itself. Without this information, we can only express the Ka value in terms of x.
So, the value of Ka for HA cannot be determined exactly without additional information.