5% rule and % ionization

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5% rule and % ionization

Postby charlotte_jacobs_4I » Wed Jan 23, 2019 11:50 am

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

Rian Montagh 2K
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Re: 5% rule and % ionization

Postby Rian Montagh 2K » Wed Jan 23, 2019 12:03 pm

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.

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Re: 5% rule and % ionization

Postby JacobHershenhouse3G » Wed Jan 23, 2019 1:41 pm

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!

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