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Calculationg molarity with diferent nH2O


flagstone

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Hello everybody!
I have a bit silly question maybe... I have to make solution with Na2HPO4.2H2O equal to 66,6mM (So I have to take 11,85g of it). OK, but I have Na2HPO4.12H2O. How to calculate it to get a solution with the same concentration of Na2HPO4.
My task is to make a Sorensen buffer with 66,6~,7mM KH2PO4 together with Na2HPO4.2H2O (66,6mM) = pH7.

Thank you!

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I know what is the molar mass of Na2HPO4.12H2O... The problem is that I have to get a 66,6mM solution of Na2HPO4.2H2O with molar mass 175.965 so I have to take 11.72g of it. However, I have Na2HPO4.12H2O with molar mass 346.04 so I have to take 23.046 of it which make sense because 23.046>11.72 at the expense of the nH2O. But is it right or wrong this logic?

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I know what is the molar mass of Na2HPO4.12H2O... The problem is that I have to get a 66,6mM solution of Na2HPO4.2H2O with molar mass 175.965 so I have to take 11.72g of it. However, I have Na2HPO4.12H2O with molar mass 346.04

Where did you get these data?

I would say rather that molar mass of Na2HPO4 . 2H2O is 177.99 g/mol (Na2HPO4 has 141.96 g/mol + 2 * 18.015 g/mol = ~177.99 g/mol)

and Na2HPO4 . 12H2O is 358.14 g/mol (Na2HPO4 has 141.96 g/mol + 12 * 18.015 g/mol = ~358.14 g/mol)

Edited by Sensei
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Where did you get these data?

I would say rather that molar mass of Na2HPO4 . 2H2O is 177.99 g/mol (Na2HPO4 has 141.96 g/mol + 2 * 18.015 g/mol = ~177.99 g/mol)

and Na2HPO4 . 12H2O is 358.14 g/mol (Na2HPO4 has 141.96 g/mol + 12 * 18.015 g/mol = ~358.14 g/mol)

Yeah, sorry - you are right. ;)

When you put the stuff in water it will dissociate into Na ions, HPO4 ions and water, whichever hydrate you start with.

And, once you make it up to the right volume with water you will end up with the same solution, whichever hydrate you start with.

 

 

As John has said, the water component doesn't actually matter here, you're really only concerned with the concentration of ions from the Na2HPO4.

Yes, I realized it a bit later... Thank you. :)

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