What is the percent ionization for each of the following acids?
a. 0.022 M HClO2 solution of pH= 3.88
Answer obtained: .60%
pH = 10^(-3.88) = 1.32 x 10^-4
% ionization= [H+] formed divided by
MHA (original acid concent) x 100
= (1.32x10^-4)/0.022 x 100
= .60%
b. 0.0027 M HClO2 solution of pH=4.70
Answer obtained from solving: .44%
pH = 10^(-4.70) = 1.20 x 10^-5
% ionization
= (1.20x1-^-5)/0.0027 x 100
= .44%
2 answers
I can't check your answers because you didn't provide the Ka for HClO2. I don't have it listed in any of my references.
B is wrong pH is .00001953 =2.0x10^-5
2.0x10^-5/.0027 = .74%
6 1/2 yrs and nobody saw this?
2.0x10^-5/.0027 = .74%
6 1/2 yrs and nobody saw this?
2 answers