when CuSO4.5H2O is dissolved in water, what physical and chemical effects happens to its water of crystallization? While solving a stoichiometric problem the ques was that 1litre of CuSO4.5H2O was dissolved in 1 liter water and the concentration of Cu2+ ions were asked. When calculating molarity the volume which those 5 water of crystallization were also taken along with the solvent water. I had the doubt that if the 5H2O is attached to CuSO4.5H2O then how can we include that also while calculating volume of solvent (for denominator of molarity). Those 5 water molecules are not free, they are bonded then why do we consider it?
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1$\begingroup$ The waters of crystallisation simply mix with the rest of the water in the solution. The $\mathrm{ Cu^{2+}}$ is, however, solvated and forms $\mathrm{Cu(H_2O)_6^{2+}}$ with two of the water ligands further away than the others. One litre of CuSO4.5H2O does not makes sense as it is a solid. $\endgroup$– porphyrinCommented Nov 19, 2020 at 17:34
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2$\begingroup$ "blue vitriol"? Are you learning from a century old book? $\endgroup$– MithoronCommented Nov 19, 2020 at 20:38
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When a sample of $\ce{CuSO4⋅5H2O}$ is dissolved in water, the $5$ molecules water are mixed with the water used as solvant, and cannot be distinguished any more from the other molecules of water. It does not really matter, as if you want to calculate the molarity, you never dissolve your sample of $\ce{CuSO4·5H2O}$ in exactly $1$ litre water. No ! You dissolve it in a volume of water big enough to get one liter solution, whatever the exact amount of water used for this dissolution. It should be a little less than one liter water. The only important parameter is the final volume of the solution, namely one liter.
