# Can anyone please explain the statement on solubility and precipitation [closed]

"It is because of high solubility product of Mg(OH)2 as compared to that of MgCO3 that Mg(OH)2 is precipitated." Can anyone please explain the statement.

As it is written, it is nonsense, as the higher sulubility product generally means higher solubility.

But the direct comparison of solubility, based on solubility products, is possible just for precipitations of the same type ( binary as $$\ce{BaSO4}$$ , ternary as $$\ce{PbI2}$$) and for solutions not containing the common ion, ideally just water.

$$\ce{Mg(OH)2}$$ precipitates if $$[\ce{Mg^2+}] \gt \frac {K_\mathrm{sp, \ce{Mg(OH)2}}}{{[\ce{OH-}]}^2}$$

$$\ce{MgCO3}$$ precipitates if $$[\ce{Mg^2+}] \gt \frac {K_\mathrm{sp, \ce{MgCO3}}}{[\ce{CO3^2-}]}$$

The priority of precipitation depends on which of magnesium concentration thresholds above is lower.

For kinetic reasons, both precipitates may form, if both thresholds are crossed. But in long term, the precipitate with the lower threshold takes precedence for thermodynamic reasons.

$$K_\mathrm{sp, \ce{Mg(OH)2}}=5.61\cdot 10^{-12}$$

$$\mathrm{p}K_\mathrm{sp, \ce{Mg(OH)2}}=11.25$$

$$\mathrm{pMg}=11.25 - 2 \cdot \mathrm{pOH} =-16.75 + 2 \cdot \mathrm{pH}$$

It means at $$\mathrm{pH}=10$$ the threshold magnesium precipitation concentration is at level of typical magnesium concentration in drink water.

$$\mathrm{p}K_\mathrm{sp, \ce{MgCO3}}=7.8$$

• Should be $\ce{[CO3^2-]}$ without a square in the denominator. – Aniruddha Deb Dec 2 at 14:58
• @Aniruddha Deb Yes, sure, it was a copy/paste/should_edit error. Fixed now. thanks. – Poutnik Dec 2 at 15:00
• what does threshold mean? – PRITIPRIYA DASBEHERA Dec 2 at 16:01
• Generally, it is a limit value, when something happens, or a different decision is made, if the value is crossed. In this case - precipitation. – Poutnik Dec 2 at 16:23