Calculationg molarity with diferent nH2O

[flagstone]

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 Na₂HPO_4.
My task is to make a Sorensen buffer with 66,6~,7mM KH₂PO₄ together with Na2HPO4.2H2O (66,6mM) = pH7.

Thank you!

[hypervalent_iodine]

Are you familiar with the equation, number of moles = mass / molar mass? Mass is the value you're after, and you have the number of moles you want as per your OP. So, what's the molar mass of Na₂HPO_4.12H₂O?

[flagstone]

I know what is the molar mass of Na₂HPO_4.12H₂O... 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 Na₂HPO_4.12H₂O 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?

[John Cuthber]

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.

[Sensei]

I know what is the molar mass of Na₂HPO_4.12H₂O... 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 Na₂HPO_4.12H₂O 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 February 23, 2016 by Sensei

[hypervalent_iodine]

As John has said, the water component doesn't actually matter here, you're really only concerned with the concentration of ions from the Na₂HPO_4.

[flagstone]

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 Na₂HPO_4.

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