What is the percent ionization of 0.025 M chlorous acid, HClO2 solution?
Ka HClO2 = 1.1*10^(-2)
Answer: 48%
What I did:
HClO2(aq) + H2O(l) -><-
ClO2(aq) + H3O(aq)
initial-change-end table results:
x^(2)/(.025-x)=1.1*10^(-2) x=1.6*10^(-2)
% ionization = (1.1*10^(-2))/(1.6*10^(-2))*100 = 69%
Two problems with what you did.
1. %ion = [(H3O^+)/0.025]*100 = ??
2. Your equation is set up correctly; i.e., x^2/(0.025-x) = Ka.
However, with Ka so large and (HClO2) so small, you MUST solve the quadratic equation. That is, you may not assume 0.025-x = 0.025.
3. I get 47.9% which rounds to 48% if you solve the quadratic and substitute into #1 above correctly.
thank you; I was pretty sure I couldn't assume x negligible for Ka but I thought that the test for negligibility was the same as the test for % ionization.
To find the percent ionization of a 0.025 M chlorous acid (HClO2) solution, you can use the equilibrium constant (Ka) and the initial concentration of the acid.
The equation representing the dissociation of HClO2 is:
HClO2(aq) + H2O(l) ⇌ ClO2-(aq) + H3O+(aq)
Using an initial-change-end table, we can set up the following:
Initial: HClO2 = 0.025 M
Change: -x M (dissociation of HClO2)
: x M (formation of ClO2- and H3O+)
End: HClO2 = (0.025 - x) M
ClO2- = x M
H3O+ = x M
Now, we can write the expression for the equilibrium constant (Ka):
Ka = [ClO2-][H3O+]/[HClO2]
Substituting the values from the table, we get:
Ka = (x)(x)/(0.025 - x)
Given that Ka = 1.1*10^(-2), we can rearrange the equation:
1.1*10^(-2) = (x^2)/(0.025 - x)
Solving this equation will give us the value of x, which corresponds to the concentration (in M) of ClO2- and H3O+ ions at equilibrium.
By solving this equation, we find that x = 1.6*10^(-2) M.
Now, to calculate the percent ionization, we divide the concentration of the dissociated species (x) by the initial concentration of the acid (0.025 M), and multiply by 100:
% ionization = (1.6*10^(-2))/(0.025)*100 ≈ 64%
Therefore, the percent ionization of the 0.025 M chlorous acid solution is approximately 64%.
To calculate the percent ionization of a weak acid like chlorous acid (HClO2), you can use the formula for the equilibrium constant (Ka) and the initial concentration of the acid.
First, write the balanced equation for the dissociation of HClO2:
HClO2(aq) + H2O(l) -> H3O+(aq) + ClO2-(aq)
Next, setup an initial-change-end table to determine the equilibrium concentration of the species involved. Let x represent the amount of H3O+ and ClO2- formed at equilibrium.
HClO2(aq) + H2O(l) -> H3O+(aq) + ClO2-(aq)
Initial: 0.025M 0M 0M 0M
Change: -x -x +x +x
Equilibrium: 0.025-x 0-x x x
Now, use the equation for the equilibrium constant (Ka) to set up an equation:
Ka = [H3O+] * [ClO2-] / [HClO2]
Since the equilibrium concentration of HClO2 is 0.025 - x, you can substitute the expressions into the Ka equation:
1.1x10^(-2) = x * x / (0.025 - x)
Next, solve for x by rearranging the equation:
1.1x10^(-2) = x^2 / (0.025 - x)
Multiply both sides by (0.025 - x) to eliminate the denominator:
(0.025 - x) * 1.1x10^(-2) = x^2
Now, simplify the equation:
0.000275 - 0.011x + x^2 = 0
Rearrange the equation:
x^2 - 0.011x + 0.000275 = 0
Next, use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = -0.011, and c = 0.000275. Plugging in these values, you can solve for x:
x = (0.011 ± √((-0.011)^2 - 4(1)(0.000275))) / (2(1))
x = (0.011 ± √(0.000121 - 0.0011)) / 2
x = (0.011 ± √(-0.000979)) / 2
Since the square root of a negative number is undefined, this means that there is no real solution for x. This indicates that the percent ionization of HClO2 in the initial solution is less than 100%.
However, if you were to approximate x as 0.016 (since it is small compared to the initial concentration of 0.025), you would calculate:
% ionization = (Ka / x) * 100
= (1.1x10^(-2) / 0.016) * 100
= 0.69 * 100
= 69%
So, the approximate percent ionization of the 0.025 M chlorous acid solution would be 69%. However, this is only an approximation and the exact value would be slightly different due to the quadratic equation.