Question

To 1.0 L of a 0.38 M solution of HClO2 is added 0.18 mol of NaClO. Calculate the [HClO2] at equilibrium.

Answer 1

Just to make sure, is that NaClO or NaClO2?

Answer 2

If the 0.18 mole is NaClO2, I get 0.25M HClO2 at equilibrium.

Note => In the process of calculating the HClO2 concentration using the ICE table analysis, Ka = 1.1E-2 which indicates by the simplification rule, one can not drop the 'x' in the equilibrium row when calculating the Hydronium ion concentration. Give a try... Post your work if ya get stuck.

Answer 3

If the salt is NaClO2 I would use the Henderson-Hasselbalch equation for buffer solutions. Try that and compare with the ICE method.

Answer 4

One must 'assume' a pH value with pKa from Ka = 1.1E-2. I used pH = 1.8 (pH = 1.5 - 2.0 is typical of chloric acid solutions in the 0.1 - 0.2M range) & pKa = -log(Ka) = -log(1.1E-2) = 1.96 and got [HOCl2] = 0.25M. Same as the ICE table analysis. Great suggestion DrBob!