What is the hydrogen ion concentration of a 0.22 M hypochlorous acid solution with Ka = 3.5E-8? The equation for the dissociation of hypochlorous acid is: HOCl(aq) = H+(aq) + OCl-(aq).

I keep getting 3.60 as the concentration, but that is the wrong answer.
Using ICE:
Eq: HOCl -> H+ + OCl-
I: 0.22
C: -x +x +x
E: (0.22-x) x x
kb = x^2/(0.22-x)

kb = 1.01E-14/3.5E-8 = 2.89E-7

x^2 = 2.89E-7 - (0.22-x)
x = square root of 2.89E-7 *(0.22-x)
x'= 2.5196E-4
x'' = 2.5182E-4
x'' = 2.5182E-4; x= 2.5182E-4 = [H+]

-log(2.5182E-4) = 3.5989, rounded to 3.60

I am not sure where I went wrong
Please help

Lynn

To find the hydrogen ion concentration of the hypochlorous acid solution with a given Ka value, you need to use the equilibrium expression for dissociation and solve for [H+]. Let's go through the steps to identify where you went wrong in your calculations:

1. Start with the dissociation equation: HOCl(aq) โ‡Œ H+(aq) + OCl-(aq)

2. Write the initial concentration, change, and equilibrium expressions:

Initial concentration (I):
[HOCl] = 0.22 M

Change (C):
[H+] = x (since it's a monoprotic acid, the concentration of H+ formed and OCl- is the same)
[OCl-] = x

Equilibrium (E):
[HOCl] = 0.22 - x
[H+] = x
[OCl-] = x

3. Set up the equilibrium constant expression using the given Ka value:

Ka = [H+][OCl-] / [HOCl]
Ka = x*x / (0.22 - x)

4. Substitute the Ka value and solve for x:

3.5E-8 = x^2 / (0.22 - x)

Rearranging, we get:
x^2 = 3.5E-8 * (0.22 - x)

5. Solve the quadratic equation to find the value of x:

x^2 = 7.7E-9 - 3.5E-8x
x^2 + 3.5E-8x - 7.7E-9 = 0

Use the quadratic formula or a calculator to solve for x. Note that the positive root should be taken since concentrations cannot be negative.

6. Once you have the value of x, calculate the hydrogen ion concentration:

[H+] = x

Round the concentration to the appropriate number of significant figures as needed.

Make sure to double-check your calculations and ensure that you used the correct values and units throughout the process.