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1

Try adding pure salt to aqueous CuSO4 and freeze out Glauber's salt (Na2SO4.10H2O): 2 NaCl + CuSO4 (aq) --Freeze--> Na2SO4.10H2O + CuCl2 (aq) This path avoids the formation of problematic CaSO4 to quote a source (http://homepages.see.leeds.ac.uk/~earlgb/Publications/Vam%20Driessche%20et%20al%20New%20Perspectives%20on%20Mineral%20Nucleation%20and%20Growth%...

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A traditional softener has a zeolite resin that "holds" positive ions. When it is charged the Na ions fill the resin. Then in service , the positive ions - Ca , Mg , etc , exchange places with the Na ions, putting Na in the water . In theory the back flush of the softener after charging, removes the Cl ions. So that in service the only Cl present would be ...

3

If the weak electrolyte is a weak acid, we have to reckon with the autoionization of water. With the autoionization, the solvated ions formed from the acid do not reach infinite dilution and therefore the dissociation is limited to below 100%. Let's look at acetic acid, for which (if I remember and can type correctly) $K_a=1.8×10^{-5}$. Thereby* $\dfrac{[... 3 For infinite dilution, water acts as a buffer with$\mathrm{pH}=7$. Therefore,$\alpha=f(\mathrm{p}K_\mathrm{a})$. For$\mathrm{p}K_\mathrm{a}=9$is$\alpha=\frac{1}{101}\overset{cca}=0.01$. For$\mathrm{p}K_\mathrm{a}=8$is$\alpha=\frac1{11}$. For$\mathrm{p}K_\mathrm{a}=7$is$\alpha=0.5$, as only a half of the acid is dissociated. For$\mathrm{p}K_\...

3

There are a number of ways to explain this. In words, the entropic gain with dissociation overwhelms any attraction between counterions as the electrolyte is diluted. Mathematically: for a m:n electrolyte undergoing dissociation according to the reaction $$\ce{A_mB_n -> mA^{\nu_A -} + nB^{\nu_B +}}$$ if the initial concentration of electrolyte \$\ce{...

1

I take for granted that what you call "electron gain enthalpy" is generally described as "electron affinity". This is the energy gained (or lost) when an electron is added to an atom, ion or molecule. In general the electron affinity is a measure for the tendency of elements to form anions. The trouble is that you may find tables giving these numerical ...

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