A solution is 0.10 M in each of Ca^2+ and Mg^2+. The carbonate ion concentration is raised slowly (no volume change). Take K_sp(CaCO3) = 3.0x10^-9 and K_sp(MgCO3) = 6.0x10^-6. What percentage of the original Ca^2+ is still in solution when MgCO3 is just about to precipitate?

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc145

A solution is 0.10 M in each of $Ca X^{2 +}$ and $Mg X^{2 +}$ . The carbonate ion concentration is raised slowly (no volume change). Take $K_{s p} (CaCO X_{3}) = 3.0 \times 1 0^{- 9}$ and $K_{s p} (MgCO X_{3}) = 6.0 \times 1 0^{- 6}$ . What percentage of the original $Ca X^{2 +}$ is still in solution when $MgCO X_{3}$ is just about to precipitate?

MgCO X_{3}

starts to precipitate at

[CO X_{3} X^{2 -}] = 6.0 \times 1 0^{- 6} /0.10 = 6.0 \times 1 0^{- 5}

M. At this point

[Ca X^{2 +}] = 3.0 \times 1 0^{- 9} /6.0 \times 1 0^{- 5} = 5.0 \times 1 0^{- 5}

M. The fraction left is 5.00e-05/0.10 = 5.0e-04, i.e. 0.050 %, so calcium is essentially removed before magnesium comes down.