Calcium carbonate, #CaCO_3# has a Ksp value of #1.4 x 10^-8#, what is the solubility of #CaCO_3#?

1 Answer
anor277
Aug 24, 2016

#CaCO_3(s) rightleftharpoons Ca^(2+) + CO_3^(2-)#

Explanation:

#K_(sp)=[Ca^(2+)][CO_3^(2-)]# #=# #1.4xx10^-8#.

If we call the solubility #S#, then #S=[Ca^(2+)]=[CO_3^(2-)]#, and,

#K_(sp)=[Ca^(2+)][CO_3^(2-)]# #=# #1.4xx10^-8# #=# #S^2#

So #Ca^(2+)# #=# #sqrt{K_"sp"}# #=# #sqrt(1.4xx10^-8)# #=# #1.18xx10^(-4)# #mol*L^-1#.

#"Solubility"# #=# #1.18xx10^-4xx100.09*g*mol^-1# #=# #??g*L^-1#

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