If I dissolve $\ce{K_2HPO_4}$ and $\ce{CaHPO_4}$ in water (two different reactions), the latter shows better solubility. Is it because $\ce{Ca^{2+}}$ has ha higher valence of $2$ than $\ce{K^+}$, which has only $1$? Or has it something to do with the lattice enthalpy?


1 Answer 1


The charge and size of an ion influences both lattice stability and stability of the hydrated ion.

  1. The less stable the lattice the more soluble the ion (less energy required to disrupt the lattice). The more highly charged and smaller the ion is, the larger are the electrostatic repulsions within the lattice which will act to destabilize the lattice.
  2. The more stable the solvated ion, the more soluble the ion. The higher the charge on the ion and the smaller its radius, the more stable the solvated ion.

The size of the potassium ion is around 152 pm, the size of the calcium ion is around 114 pm. So the calcium ion is both the smaller and the more highly charged of the two. Therefor we would expect the calcium ion to have the less stable lattice and be more highly solvated (stabilized) when in solution. Both of these factors would lead us to expect $\ce{CaHPO4}$ to be more soluble than $\ce{K2HPO4}$


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