What is the minimum molar concentration of hydroxide ions needed to selectively precipitate chromium(III) hydroxide from a mixture of #"0.025 M"# #"Ca"^(2+)# and #"0.0050 M"# #"Cr"^(3+)#?

At least #["OH"^(-)] = 6.8 xx 10^(-10) "M"# is required to precipitate #"Cr"("OH")_3(s)#.

For comparison, to precipitate #"Ca"("OH")_2#, you would need #["OH"^(-)] = "0.015 M"#, much larger than what you need to precipitate #"Cr"("OH")_3#.

We begin with

#["Ca"^(2+)] = "0.025 M" -= s#
#["Cr"^(3+)] = "0.0050 M" -= s#.

where #s# is the solubility of the substance, given that it is #1:1# with the specified ions above.

We say that a solid precipitates out of solution when the reaction quotient #Q# exceeds the #K_(sp)#. Therefore, if #Q = K_(sp)#, that is the minimum value of #Q# that forces precipitation.

With the given metal ion concentration, we then need an accompanying hydroxide concentration to cause precipitation.

The calcium reaction is:

#"Ca"("OH")_2(s) rightleftharpoons "Ca"^(2+)(aq) + 2"OH"^(-)(aq)#,

with #K_(sp)# #">>>>"# #1.6 xx 10^(-30)# and

#K_(sp) = overbrace(["Ca"^(2+)])^(s)overbrace(["OH"^(-)]^2)^((2s)^2)#

#= s(2s)^2 = 4s^3#.

Usually, #"Ca"("OH")_2# is considered a strong base... I would consider it somewhat strong, although it does dissociate quite easily. Actually, #K_(sp) ~~ 5.5 xx 10^(-6)#.

The chromium reaction is:

#"Cr"("OH")_3(s) rightleftharpoons "Cr"^(3+)(aq) + 3"OH"^(-)(aq)#

with #K_(sp) = 1.6 xx 10^(-30) = overbrace(["Cr"^(3+)])^(s)overbrace(["OH"^(-)]^3)^((3s)^3)#

#= s(3s)^3 = 27s^4#.

At equilibrium, the concentrations of calcium and chromium ions are as given above.

For chromium (III) hydroxide, if we want the minimum hydroxide concentration to precipitate #"Cr"^(3+)# first, we require:

#Q = K_(sp) = "0.0050 M" cdot ["OH"^(-)]^3 = 1.6 xx 10^(-30)#

#= s(3s)^3#

Thus,

#color(blue)(["OH"^(-)]) = (K_(sp)/s)^(1//3)#

#= ((1.6 xx 10^(-30))/("0.0050 M"))^(1//3)#

#= color(blue)(6.8 xx 10^(-10) "M")#

If we get any higher, we disturb the system by overadding ion products, and by Le Chatelier's principle, it would shift the equilibrium to the left, and make #"Cr"("OH")_3(s)# precipitate.