Is Mg(OH)2 more insoluble than MgCO3?

Question

In removing temporary hardness by boiling, $\ce{Mg(HCO3)}$ is converted to $\ce{Mg(OH)2}$ but not $\ce{Mg(CO3)}$. The answer that I found in our coursebook was that it has a higher solubility product, but if its solubility product is higher, then it should be more soluble than $\ce{Mg(CO3)}$. Can you tell me the correct reason?

Answer 1

I finally found the answer.

In temporary hardness, we have to remove $\ce{Mg^2+}$ ions by precipitating it. In $\ce{Mg(OH)2}$, $\ce{Mg^2+}$ ions in water are present in the concentration that is cube root of the solubility product, but in $\ce{Mg(CO3)}$ it is the square root of the solubility product. So, despite $\mathrm{K_{sp}}$ of $\ce{Mg(OH)2}$ being higher than $\ce{Mg(CO3)}$, $\ce{Mg^2+}$ ions in $\ce{Mg(OH)2}$ are comparatively less than $\ce{Mg^2+}$ ions in $\ce{Mg(CO3)}$.

Answer 2

The solubility data for $MgCO_3$ and $Mg(OH)_2$ are these: for $MgCO_3$ 0.01 g/100 mL cold H2O; for $Mg(OH)_2$ 0.0009 g/100 mL H2O @ 18C, 0.004 g/100 mL H2O @ 100C (CRC Handbook).

The published Ksp data are all over the place: for MgCO3, I have found $10^{-5}$, $6.8× 10^{-6}$, and $ 3.5 × 10^{-8} $(twice). For Mg(OH)2, in the same set of 4 documents, I have $ 1.8 ×10{-11}$ (twice), $1.6 × 10^{-12} $ and $5.6 × 10^{-12}$

Calculating Ksp from the solubility data gives: for Mg(OH)2 @ 18C, $1.5 × 10^{-11}$ and @ 100C, $1.31 × 10^{-9}$. For MgCO3, Ksp calculates to $1.4 × 10^{-6}$.

From the solubility data, Mg(OH)2 is by far less soluble than MgCO3. From the Ksp data, any way you look at them, the solubility product of Mg(OH)2 is smaller than for MgCO3, indicating a lower solubility. So, no conflict.

There is a complexity in comparing a solubility product of MgCO3 (which multiplies 2 concentrations) vs Mg(OH)2, which multiplies 3 concentrations. Not to mention the fact that solubilities and Ksp are given for hydrates of MgCO3, which I did not deal with. And you might consider that it is not we who are removing $Mg^{+2}$ ions from the temporary hard water, but that heating removes CO2 from the water, leading to production of Mg(OH)2 in solution and then precipitation. The loss of CO2 minimizes the possible precipitation of MgCO3, especially since it is more soluble at high temperatures, where the transformation is occurring.