I am considering the solubility product of sodium citrate. I need this value, I do not have it. However, I do know that the solubility product of calcium citrate is (7±2)×10^−14.

Is it safe to assume that sodium citrate will have a very similar Ksp?

I am concerned about whether or not the addition of sodium citrate will force borate ions out of the water as sodium tetraborate.

Furthermore, could I hypothesise on the solubility based on the likely strength of the hydrogen bonds? I am able to view the charges of each individual atom within a molecule, and the citrate's oxygen anions are far more charged than those in the borate ions. The sodium citrate's hydrogen bonds should therefore be stronger, and result in a higher solubility?

Thank you.


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  • $\begingroup$ No, because sodium is singly charged and calcium is doubly charged, so there will be a different number of terms in your solubility product formula, and you should expect quite a large difference in the $K_\text{sp}$ between the two salts. $\endgroup$ – a-cyclohexane-molecule Jan 25 '18 at 1:53

No. The reason you can't find the $K_{\rm sp}$ of sodium citrate is most likely because it is highly soluble. Except for a few exotic cases, sodium salts are very soluble in water.

Adding sodium citrate to the solution does reduce the solubility of borax, however this is not going to be a problem unless you are using solution with really high concentration. As a reference, the solubility of sodium citrate and borax are $\pu{770 g/L}$ at $\pu{25^\circ\!C}$ and $\pu{51 g/L}$ at $\pu{20^\circ\!C}$, respectively.

The solubility of a solid in water depends on many factors. In this case where the two solids share the same cation, we can say that thermodynamically it depends on the extent that water stabilizes the anion compared to the crystal lattice. Your reasoning that the citrate anion's hydrogen bonding with water is stronger than borate anion's is valid and I think it's fine to say that sodium citrate is more soluble than borax based on this. Just that you need to keep in mind that the crystal energy also plays a role here as well, and need to be examined with care if the oxygens are not "far more" charged in one species than in the other.

  • $\begingroup$ @GunaPrashant Updated $\endgroup$ – Weijun Zhou Jan 25 '18 at 1:03

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