#CrO_4^(2-)# is added to a solution where the original concentration of #Sr^(2+)# is #1.0 * 10^-3#. Assuming the concentration of #Sr^(2+)# stays constant, will a precipitate of #SrCrO_4# (Ksp = #3.6 * 10^-5#) form when #[CrO_4^(2-)]# = #3.0 * 10^-5# M?

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Answer 1 · 2018-03-23T16:17:59

And so we consider the solubility equilibrium....

#Sr^(2+) + CrO_4^(2-)rarrSrCrO_4(s)darr#

For which #K_"sp"=[Sr^(2+)][CrO_4^(2-)]=3.6xx10^-5#...

And we are given that i. #[CrO_4^(2-)]-=3.0xx10^-5*mol*L^-1#..and that (ii) #[Sr^(2+)]-=1.0xx10^-3*mol*L^-1#....

And so all we have to do is take the ion product...

#Q=[Sr^(2+)][CrO_4^(2-)]=(1.0xx10^-3)(3.0xx10^-5)=3.0xx10^-8#...

And since #Q"<<"K_"sp"#...strontium chromate should remain in solution...