If #"K"_"sp"# of #"Mg(OH)"_2# is more than that of #"MgCO"_3# which gets precipitated first?

NCERT XI Chemistry

It is because of high solubility product of #"Mg(OH)"_2# as compared to that of #"MgCO"_3#, that #"Mg(OH)"_2# is precipated.

Is this statement from my textbook correct?
How a more soluble compound gets precipated easily than a less soluble compound?

1 Answer
anor277
Mar 24, 2018

#K_"sp"# is an equilibrium quantity....and I think your text book has a typo....

Explanation:

The section of the text tries to explain that BICARBONATES, i.e. salts of #HCO_3^(-)# are generally soluble, yet carbonates, salts of #CO_3^(2-)# are generally insoluble. This is often observed for insoluble sulfates versus soluble bisulfates, salts of #HSO_4^-#.

To look at the data, (and as physical scientists we MUST look at data...)...this site quotes #K_"sp"=5.61xx10^-12# for #Mg(OH)_2#.....solubilities in the range of a few #"ppm"#...

and this site quotes #K_"sp"=10^-7# for #MgCO_3#, i.e. #0.139*g*L^-1#, or #139*"ppm"#..

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