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pH Problem !!


Sarahisme

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Thats a bit hard to do without acidity constants eh? Do they give you those?

 

Edit: Now that i think about it if you can't calculate I think it's experiment 3 because for that to be true it would have had to dissociate 100% and i don't think anything dissocitates perfectly.

 

~Scott

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Thats a bit hard to do without acidity constants eh? Do they give you those?

 

Edit: Now that i think about it if you can't calculate I think it's experiment 3 because for that to be true it would have had to dissociate 100% and i don't think anything dissocitates perfectly.

 

~Scott

 

 

The answer in most probability shud b the 4th one. Since H2SO4 dissociates in 2 steps as:

H(+) + HSO4(-) AND

H(+) + SO4(2-)

Hence the complete dissociation of H2SO4 does not occur due to common ion effect(H here).

Hence the ionsization will be incomplete and hence higher pH(less acidic)

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Get out, thats diprotic and seeing as you aren't given acidic constants you would have to assume that theoreticly the second dissociation could be 99%?? Or do you mean that the pH would be >2?

 

~Scott

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No, the acid is definitely diprotic but there is 2nd dissociation which is hindered by the strong 1st dissociation of H2SO4[due to common ion effect H(+)]

Hence the higher pH. BenSon the higher pH means high value and weaker acid thus the actual pH shud theoretically lie between 1.99 and 2.2 under STP and in water.

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It doesn't matter how dissociated the H2SO4 is, the maximum [H+] it can give is 0.02 M, which corresponds to a pH of 1.7. To reach pH 1.4, [H+] must be 0.04 M - impossible with 0.01 M H2SO4

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Hey sarah, seeing as your a regular why don't you start one thread in the chemistry forums that you post all your questions in? It will save some room and probably be more responsive then the homework help section. Just a thought.

 

~Scott

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