The Ka of HClO is 3.0x 10^-8 at 25°c. What is the percent ionization of HClO in a 0.015M aqueous solution of HCLO at 25°c?
0.14
percent ionization=concentration of ionized acid/initial concentration of acid*100
[H3O+]equil/HA]init*100%
To find the percent ionization of HClO in a 0.015M aqueous solution, we can use the information given about its Ka value. The Ka value is an equilibrium constant for the dissociation of HClO into H+ and ClO-. The equation for this dissociation reaction is:
HClO ⇌ H+ + ClO-
The Ka expression for this reaction is:
Ka = [H+][ClO-] / [HClO]
Given that the Ka value for HClO is 3.0x10^-8, we can set up the equation:
3.0x10^-8 = [H+][ClO-] / [HClO]
Since the initial concentration of HClO is 0.015M, we substitute [HClO] with 0.015M:
3.0x10^-8 = [H+][ClO-] / 0.015
Next, we need to solve for [H+] to find the concentration of the ion H+ resulting from the dissociation of HClO. Rearranging the equation, we have:
[H+][ClO-] = 3.0x10^-8 x 0.015
[H+][ClO-] = 4.5x10^-10
Since HClO is a weak acid, we can assume that the concentration of [H+] formed from the dissociation is much smaller compared to the initial concentration of HClO. Therefore, we can approximate the concentration of H+ as 0. We make this approximation because the ionization of weak acids is usually less than 5% and ensures accurate results.
Using this approximation, we have:
[H+] ≈ 0
Thus, the percent ionization of HClO is approximately 0%.
Note: This approximation is valid for weak acids with a small Ka value like HClO. For stronger acids or acids with larger Ka values, this approximation may not be accurate, and a more detailed calculation may be required.
............HClO ==> H^+ + ClO^-
I..........0.015.....0......0
C...........-x.......x......x
E.........0.015-x....x......x
Ka = (H^+)(ClO^-)/(HClO)
Plug in the E line into Ka expession, solve for x, then %ionization = [(x)/0.015]*100 = ?