Question regarding diprotic acids.

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Watergrizzly

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I think I am getting slightly confused regarding diprotic acids. I recall during one of Chad's videos that the second hydrogen doesn't fully dissociate. However, when working through problems (normality, molarity, and such) it seems that it is always treated as if it does? Is there an instance I am forgetting where what Chad mentioned comes into effect? Thank you!

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Yes if you put diporitc acids in water, the second H does not dissociate 100%.
However when you react it with a base... All H will react with OH of the base
 
I think I am getting slightly confused regarding diprotic acids. I recall during one of Chad's videos that the second hydrogen doesn't fully dissociate. However, when working through problems (normality, molarity, and such) it seems that it is always treated as if it does? Is there an instance I am forgetting where what Chad mentioned comes into effect? Thank you!
It depends on which diprotic acid you are referring to. If given H2SO4, which is most likely seen on the DAT, assume as a good approximation that BOTH hydrogens come off. Thus a 0.1 M H2SO4 would be 0.2 M in Hydrogen ions when doing the pH calculation. If the acid is weak, H2CO3 for example, we would only use the first ionization step to calculate the pH. Why ? You will notice that Ka values would be given. Ka1 is approximately 4 x10 exp -7 and Ka2 is 5 x10 exp -11..... Since Ka1 is much larger than Ka2, we use only the first ionization step. It is unlikely that you will see this on the DAT. However, any General Chem text will illustrate this for you. Chang and Zumbdahl do a nice job. For the DAT, H2SO4 is the more likely question, and all you need to do is to assume BOTH hydrogens ionize.

Hope this helps.
 
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