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%
B is wrong pH is .00001953 =2.0x10^-5
2.0x10^-5/.0027 = .74%
6 1/2 yrs and nobody saw this?
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.
To find the percent ionization for each of the acids given, we can use the formula:
% ionization = ([H+] formed / [HA] initial) x 100
a. For the 0.022 M HClO2 solution with a pH of 3.88, we need to first calculate the concentration of H+ ions using the pH value:
pH = -log[H+]
3.88 = -log[H+]
Taking the inverse log of both sides, we get:
[H+] = 10^(-3.88)
[H+] = 1.32 x 10^-4
Now we can calculate the percent ionization:
% ionization = (1.32 x 10^-4 / 0.022) x 100
% ionization = 0.60%
b. For the 0.0027 M HClO2 solution with a pH of 4.70:
pH = -log[H+]
4.70 = -log[H+]
Taking the inverse log of both sides:
[H+] = 10^(-4.70)
[H+] = 1.20 x 10^-5
% ionization = (1.20 x 10^-5 / 0.0027) x 100
% ionization = 0.44%
So the percent ionization for the given acids are:
a. 0.60%
b. 0.44%
To find the percent ionization of an acid, you first need to determine the concentration of hydrogen ions (H+) in the solution. The concentration of H+ ions can be calculated using the pH of the solution.
The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration. So, to calculate the concentration of H+ ions, you can use the equation:
[H+] = 10^(-pH)
Once you have the concentration of H+ ions, you can calculate the percent ionization by dividing the concentration of H+ ions formed by the original acid concentration (MHA) and multiplying by 100.
Let's apply this to the given examples:
a. For the 0.022 M HClO2 solution with a pH of 3.88:
The concentration of H+ ions can be calculated as:
[H+] = 10^(-3.88) = 1.32 x 10^-4
The percent ionization is then calculated as:
% ionization = (1.32 x 10^-4) / 0.022 x 100 = 0.60%
b. For the 0.0027 M HClO2 solution with a pH of 4.70:
The concentration of H+ ions can be calculated as:
[H+] = 10^(-4.70) = 1.20 x 10^-5
The percent ionization is then calculated as:
% ionization = (1.20 x 10^-5) / 0.0027 x 100 = 0.44%