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I am following a protocol that requires me to prepare $10~ \mathrm{mL}$ $\mathrm{1~M}~ \ce{Ca(NO3)2}$. However, there is only $\ce{Ca(NO3)2*4H2O}$ in the lab.

Originally I thought that all I have to do to figure how much $\ce{Ca(NO3)2}$ to weigh is enter image description here

and then just add $10~ \mathrm{mL}$ water. However, that might dilute the solution and make the concentration less than $\mathrm{1~M}$ as there is already water in the hydrate.

So then I found a protocol and made new calculations as followed: enter image description here

How can I verify that I got $\mathrm{1~M}$ $\ce{Ca(NO3)2}$ as opposed to $\mathrm{1~M}$ $\ce{Ca(NO3)2*4H2O}$?

Which protocol is correct?

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  • $\begingroup$ The hydro-molecular equation is used to determine the volume of water contained within any given mass of a hydrated compound. This volume is then subtracted from the final volume of the solution. $\endgroup$
    – Holms Peterson
    Jul 23, 2020 at 8:16

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Originally I thought that all I have to do to figure how much $\ce{Ca(NO3)2}$ to weigh is enter image description here

and then just add $10~ \mathrm{mL}$ water.

To properly make an aqueous solution of a specified concentration you should dissolve the desired amount of material in about $80\%$ ($8~ \mathrm{mL}$) the desired volume of water. Then dilute to the final solution volume (add water until the solution volume is $10~ \mathrm{mL}$).

How can I verify that I got $\mathrm{1~M}$ $\ce{Ca(NO3)2}$ as opposed to $\mathrm{1~M}$ $\ce{Ca(NO3)2*4H2O}$?

Hydrates are for the solid state. You do not have "hydrates" in the aqueous phase. Therefore the two solutions are identical and indistinguishable.

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  • $\begingroup$ But the volumes of water added are different. How can they be the same solutions? I want it such that the concentration of Ca(NO3)2 is 1 M, no more, no less. $\endgroup$
    – wswr
    Jan 14, 2016 at 1:16
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    $\begingroup$ As AK says dilute solution TO 10ml total volume, not WITH 10ml. $\endgroup$
    – MaxW
    Jan 14, 2016 at 3:01

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