chemistry question: acids and bases

[toph99]

Ok i have a question. I did a titration in chem class, and found the molar mass of the solid unknown acid that we used, but there is a question asking what the molar mass would be if the acid was diprotic(i believe that means it gives off two H+ ions per molecule) so to find the molar mass of a diprotic acid would i multiply the old mass by two, or divide it in half? thanks

 

[LethalWolfe]

Originally posted by: toph99
Ok i have a question. I did a titration in chem class, and found the molar mass of the solid unknown acid that we used, but there is a question asking what the molar mass would be if the acid was diprotic(i believe that means it gives off two H+ ions per molecule) so to find the molar mass of a diprotic acid would i multiply the old mass by two, or divide it in half? thanks

If my gf was here I'd ask her (bio & chem degree from IU), but I don't know s**t so I'm stepping back now...


Lethal

 

[McPhreak]

Neither. Just add one to the original molar mass since it's giving off one extra proton.

 

[FeathersMcGraw]

Originally posted by: toph99
Ok i have a question. I did a titration in chem class, and found the molar mass of the solid unknown acid that we used, but there is a question asking what the molar mass would be if the acid was diprotic(i believe that means it gives off two H+ ions per molecule) so to find the molar mass of a diprotic acid would i multiply the old mass by two, or divide it in half? thanks

If the solute gives off twice as many ions, you need half as much to achieve the same concentration.

Edit: better disregard this answer. I think I misunderstood the question.

 

[miniMUNCH]

Basically what Mcphreak said.

But also you have take the H+'s into consideration for your mole calculations.

[austin powers]...we're not supposed to talk about the mole but there it is!! Molely Molely Molely Molely Molely!!![\austin powers]

 

[MichaelD]

If you drop acid while listen to bass music, you'll see all the possible chemical combinations...quite clearly, I might add.