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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?

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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 $\ce{Mg(CO3)}$.

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The solubility data for MgCO3 and Mg(OH)2 are these: for MgCO3 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 10e-5, 6.8 x 10e-6, and 3.5 x 10e-8 (twice). For Mg(OH)2, in the same set of 4 documents, I have 1.8 x 10e-11 (twice), 1.6 x 10e-12 and 5.6 x 10e-12.

Calculating Ksp from the solubility data gives: for Mg(OH)2 @ 18C, 1.5 x 10e-11 and @ 100C, 0.33 x 10e-9. For MgCO3, Ksp calculates to 1.4 x 10e-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++ 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.

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  • $\begingroup$ I think that the removal of $\ce{Mg^{++}}$ions from the temporary hard water is by precipitating them in the form of $\ce{MgCO3_\mathrm{(s)}} $ and not in the form of $\ce{Mg(OH)2_\mathrm{(s)}}$as in the following equation:$$\ce{Mg(HCO3)2_\mathrm{(aq)}\overset{\Delta}{ ->} MgCO3_\mathrm{(s)} + H2O_\mathrm{(l)} + CO2_\mathrm{(g)}}$$ $\endgroup$ – Adnan AL-Amleh Aug 28 at 23:31
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    $\begingroup$ Ca(HCO3)2 can be a component of temporary hard water, and will deposit CaCO3 when heated. CaCO3 is much less soluble than Ca(OH)2. The reverse is true for magnesium: the hydroxide is much less soluble than the carbonate. As Mg(HCO3)2 is heated, and CO2 is evolved, the pH rises and the OH- concentration increases. Finally, at pH ~10.5, the OH- concentration is high enough to ppt Mg(OH)2 but with [CO3--] low enough to not ppt MgCO3. What all our calculations do not show is the possibility of mixed ppts! Aldrich offers (MgCO3)4.Mg(OH)2.5H2O. Very complex!! $\endgroup$ – James Gaidis Aug 30 at 3:03
  • $\begingroup$ thanks for the interest.Can we represent the removal of $\ce{Mg++}$ ions from the temporary hard water with the following equations? $$\ce{Mg(HCO3)2_\mathrm{(aq)}->Mg^{++}_\mathrm{(aq)} + HCO3^-_\mathrm{(aq)}}$$ $$\ce{HCO3^-_\mathrm{(aq)} + H2O_\mathrm{(l)}-> H2CO3 + OH-_\mathrm{(aq)}}$$ $$\ce{H2CO3 \overset{\Delta}{->}CO2_\mathrm{(g)} +H2O_\mathrm{(l)} }$$ $$\ce{Mg^{++}_\mathrm{(aq)} + 2OH-_\mathrm{(aq)}->Mg(OH)2_\mathrm{(s)}}$$ $\endgroup$ – Adnan AL-Amleh Aug 30 at 8:37
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    $\begingroup$ That's a good possibility. One approach would be to 1) analyze the hard water for Mg and Ca, in % or ppm. Then, 2) do an experiment by heating some of the temp hard water, and then, 3) (most important!) analyze the ppt. All we really know is that something ppts out. If we limit ourselves to Mg(OH)2 and MgCO3 (probably reasoning that Mg is like Ca), we may miss those mixed ppts, which may not always be well catalogued. I think the Ca system is simpler than the Mg system. But Mg is very interesting! $\endgroup$ – James Gaidis Aug 30 at 13:32

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