Question

Find the initial concentration of a solution of HClO with pH=4.28 and Ka= 3.0 x 10^-8. ( include ICE chart if possible)

Answer 1

-X a-x Ka[H+][clo-]/[Hclo] = x^2/(a - x) (5.248 x 10^ (-5))^2/(a 5.248 x 10^ (-5)) 3.0 x 10^ (-8) - Initial concentration of HC1。= a 0.0919 M (approximately 0.092 M)

Answer 2

In a solution of this acid in water, you know that [H+]= [ClO-]. From the pH = 4.48, you can calculate [H+] = 10^-4.48 = 3.31 X 10^-5 = [ClO-]

Now, using the expression for Ka, you have:

Ka = 3.0 X 10^-8 = [H+][ClO-] / [HClO] Plugging in and solving for [HClO] gives:

3.0 X 10^-8 = (3.31 X 10^-5)^2/[HClO]

[HClO] = 0.0365 M so 3.7 X 10^-2 M is the bet answer.