Determine the pH of a 0.020 M solution of HClO₂ (Ka = 1.2 × 10⁻²).

Question

I keep getting a pH of 1.81, but the answer is pH=1.97.

Answer 1

Hello Alizee,

Make sure you always first start by writing down the reaction:

HClO2 <---> ClO2- + H+

Now what this means is that if I drop some HClO2 (in this case .02 M worth) I will end up with some H+ ions, ClO2- ions, and HClO2 that does not dissolve. So how do we determine how much? we use our equilibrium constant equation relating the products to reactants.

Ka = [H+] [ClO2-] / [HClO2]

So now that we have our Equilibrium constant equation we go ahead and run an ICE Chart (In General Chemistry these are done in molarity typically; so that's how I will do here):

[HClO2] [H+] [ClO2-]

Initial .02 0 0

Change -x +x +x

Equilibrium .02-x x x

So what this says is I start the instant I put .02 M HClO2 in solution, I only have HClO2. But......some trivial amount of time later we will have H+ and ClO2- as well and this dissolving will reach some equilibrium and result in "x" [H+] being made. So let's plug in these variables into our equilbrium constant equation:

Ka = [H+] [ClO2-] / [HClO2]

1.2 x 10^-2 = x * x / (.02-x)

Using the quadratic equation, you should get x = .0106132. This means I have .0106132 [H+] in solution becuase by our definition in the ICE chart we said at equilibrium x = [H+]. So using our pH equation:

pH = -log[H+]

pH = -log(.0106132)

pH = 1.97

Hope this helps!