## 6D.3 7th ed, calculating Ka and pKa

JulieAljamal1E
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

### 6D.3 7th ed, calculating Ka and pKa

I have two questions about 6D.3. It asks to find the Ka and pKa values given the information that you have 0.10 M HClO2 with pH 1.2. In the solutions it shows that the concentration of H3O is 0.06 mol/L which is solved by taking 10^-pH in this case 10^-1.2. I understand that, but then they say that [H3O]=[ClO2], why is that? Secondly, after knowing these concentrations, to calculate Ka you should set up the equation [H3O][ClO2]/[HClO2]. Plug in the appropriate values so (0.06x0.06)/(0.10)... that's what I thought to do because the given concentration of HClO2 is 0.10. But the solutions show that you must do (0.06x0.06)/(0.10-0.06). Why do we subtract .06 from 0.10? Thank you :)

Sean_Rodriguez_1J
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

### Re: 6D.3 7th ed, calculating Ka and pKa

It says that [H3O]=[ClO2] because for every ClO2- ion that is made, a H3O+ ion must be made as well (They have a molar ratio of 1:1). We subtract .06 from 0.10 based on the ICE chart for this model. It looks like this (ignore water because it's the solvent of the reaction):
HClO2(aq) + H2O(l) ------> ClO2- + H3O+
I 0.10 0 0
C -0.06 +0.06 +0.06
E 0.10 - 0.06 0.06 0.06
For the equilibrium concentration we must account for the small amount of HClO2 that reacted.