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### 5%

Posted: **Sun Jan 20, 2019 10:31 pm**

by **SydBenedict2H**

How exactly does one actually test the 5% rule (if x less than 5% of initial concentration then the approximation is valid)? He always says it in lecture but hasn't actually done the math and I need to see things like that before i fully grasp them regardless of how basic i know it is.

### Re: 5%

Posted: **Sun Jan 20, 2019 10:36 pm**

by **melissa_dis4K**

In the 1/16/19 lecture, he explained towards the end that because x was less than 5% of the initial value it was okay, since in that calculation it was 1.3%. He got the 1.3% from calculating the % ionization of aceitic acid. I was also confused on what he meant but I think that he means that if the % ionization is less than 5% it is valid. But I am not entirely sure, that is what I a understanding. If anyone else knows please let me know if I am misunderstanding. Thank you!

### Re: 5%

Posted: **Sun Jan 20, 2019 10:39 pm**

by **404975170**

You compare the disassociation of the conjugate acid to the conjugate base or vice verse by putting the concentration of the product for example conjugate acid over the concentration of its reactant concentration base times 100%.

The picture shows how it is applied for the problem I did at the bottom.

### Re: 5%

Posted: **Sun Jan 20, 2019 11:06 pm**

by **Angel Chen 2k**

The 5% rule is essentially a technique that allows us to simply calculations ( do not have to use the quadratic formula) when we are able to recognize that the change in composition ( x) is less than 5 % of the initial value.

### Re: 5%

Posted: **Sun Jan 20, 2019 11:16 pm**

by **Hadji Yono-Cruz 2L**

5% refers to the x value and if it is less than 5% then x will be insignificant to calculations, so the quadratic formula would not have to be used and you can simply plug in the values that are given.

### Re: 5%

Posted: **Mon Jan 21, 2019 9:48 am**

by **Jordan Lo 2A**

if the x (change in concentration) is less than 5% of the initial concentration, you don't have to include "-x" in the denominator when solving the ICE table