## 5% rule and % ionization

charlotte_jacobs_4I
Posts: 63
Joined: Fri Sep 28, 2018 12:29 am

### 5% rule and % ionization

What are the steps for solving for a percent ionized/deprotonized question and how does that relate to the 5% rule?

Rian Montagh 2K
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

### Re: 5% rule and % ionization

To find the % ionization you use an ICE table to find the concentrations of the initial concentration of the unionized substance and the concentration of the ionized version of the substance at equilibrium. Then do [ionized]/[initial unionized] * 100. This is the percent.

This relates to the 5% rule because usually [ionized] is x, and if x is less than 5% of the initial then you can use the approximation of x. So, if the percent x/initial * 100 (the same calculation as the first one written) is less than 5%, you confirm that the approximation is ok to use.

JacobHershenhouse3G
Posts: 32
Joined: Thu Jan 10, 2019 12:17 am

### Re: 5% rule and % ionization

Simply use the equation %deprot= [H3O+]/initial[reacting acid] or %prot [OH-]/initial[reacting base]. The 5 % rule is a check on if the change (-x) to the initial concentration of the given acid or base is significant or negligible (if negligible, then the simplification of our k equation is alright). So, if the %(de)protonation is less than 5%, we say that the change is negligible (like the millionaire example) and we are able to simplify and approximate our final value. Hope this helps!