i of freezing point depression/melting point depression

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cc609

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The formula for freezing point depression is change in Temperature = i* Kfreezing*molal

For determining i, I know that NaCl for example would be 2, and H2So4 would be 3.
However, do we disregard the i value if there is a strong acid? For example, if there was say, a strong acid such as HCl, technically with an i of 2, and a Na2So4, which has i of 3....the i value is higher for the latter, but HCl is a strong acid so it would technically dissociate more would it not? Would someone help me with this one :'(

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The formula for freezing point depression is change in Temperature = i* Kfreezing*molal

For determining i, I know that NaCl for example would be 2, and H2So4 would be 3.
However, do we disregard the i value if there is a strong acid? For example, if there was say, a strong acid such as HCl, technically with an i of 2, and a Na2So4, which has i of 3....the i value is higher for the latter, but HCl is a strong acid so it would technically dissociate more would it not? Would someone help me with this one :'(

Yes,,,,,The i value for HCl would be two particles..... usually we are dissolving solutes into solution and not strong acids.

Dr. Romano
 
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They would have to specify "completely dissociates" or give the Ka Value for HSO4- and then you could calculate the ions generated by: [Ka*[HSO4-]]^1/2 then multiply that number by 2 to figure out how many i values H & SO4-2 are generated. If you say the SO4- concentration is x then I think the i value would be:

(1) for first Hydrogen
(1 - x ) for HSO4
(x) for 2nd Hydrogen
(x) for SO4-2

If you sum this all together, i would equal 2+x

depending on the Ka value this could be either really close to 2 or close to 3 or somewhere in the middle....
 
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