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### autoprotolysis

Posted: Wed Jan 23, 2019 12:49 pm
will we be expected to derive 14 = pOH + pH from 10^-14=[H3O+][OH-] on the test?

### Re: autoprotolysis

Posted: Wed Jan 23, 2019 1:02 pm
While I can't confirm what is on the test, you just take the negative log of both sides and then use the formula log(ab) = loga + logb to derive the final equation.

### Re: autoprotolysis

Posted: Wed Jan 23, 2019 1:08 pm
No, because 10^-4= Kw at 298K is an experimentally determined value.

### Re: autoprotolysis

Posted: Wed Jan 23, 2019 4:13 pm
Whether or not it is on the test, deriving it is simple because you only have to take the negative log of the concentrations to get the respective pH and pOH. It is important in general to remember the relationships like those such as Kw because they will come in handy on the midterm and final as well.

### Re: autoprotolysis

Posted: Wed Jan 23, 2019 4:32 pm
While I am not sure about this equation specifically, on his syllabus it says we should know how to derive pKw=pH+pOH so we might also have to know how to derive this.

### Re: autoprotolysis

Posted: Wed Jan 23, 2019 4:43 pm
i don't think we'll need to know how to derive it but just know the relationship. but it's pretty simple to derive it so i would know it to just help you out later on