Suppose you are given a question that asks whether a precipitate of a particular compound will form?

This is actually a common question to ask.  Many solutions have a complex mix of metal cations and anions.  It is quite likely that some of these combinations have small Ksp values, and so are sparingly soluble.  In this case, we might be interested to understand ahead of time whether it is likely that a precipitate will form in the solution.  Another common example is that many metals form insoluble hydroxide complexes.  We therefore may want to know whether a change in pH is going to cause a dissolved metal ion to precipitate out of solution.

The thing to keep in mind is that the solubility product can never exceed the value of Ksp.  For example, suppose you were to consider the species silver carbonate (Ag2CO3).  The solubility reaction and Ksp expression is shown below.

$\ce{Ag2CO3 \leftrightarrow 2Ag+ + CO3^2-}$

$\mathrm{K_{sp} = [Ag^+]^2[CO_3^{2-}] = 7.7\times10^{-12}}$

Suppose we had a process that would lead to a solution with silver and carbonate ions in it.  Suppose that we were also able to calculate the starting value of each ion that we expected in the solution.

If we expected a concentration of silver of 5×10-4 M and a concentration of carbonate of 1×10-3 M, would a precipitate form?  What we need to do is take these values and put them into the form of the Ksp expression.  Since these are not likely to be equilibrium concentrations, instead of calling this expression K, we use the notation Q.

$\mathrm{Q = [Ag^+]^2[CO_3^{2-}] }$

$\mathrm{Q = [5×10^{-4}]^2[1×10^{-3}] = 2.5×10^{-10}}$

What we now need to do is compare the magnitude of Q (2.5×10-10) to the magnitude of Ksp (7.7×10-12).  If Q is greater than Ksp, a precipitate will form since the solubility product term can never exceed Ksp.  If Q is less than Ksp, no precipitate will form (this is not yet a saturated solution).

Since 2.5×10-10 > 7.7×10-12, a precipitate will form in this case.  Not all of the silver and carbonate will precipitate out of solution.  Instead, the concentrations will be lowered so that the concentrations exactly satisfy the Ksp expression.

Another common question is whether it is possible to quantitatively precipitate (99.9%) of one metal cation in the presence of another.

If we assume that the concentrations of the metal ions in the solution are known, we can calculate the concentration of the precipitating anion that is the highest possible value that will not cause any precipitation.  We can also calculate the concentration of the precipitating anion that is needed to precipitate 99.9% of the metal ion.

For example, suppose we had a solution that was 1×10-3 M in Pb2+, and we wanted to try to precipitate 99.9% of the lead as its bromide salt.  The relevant reaction and equilibrium expression is shown below.

$\ce{PbBr2 \leftrightarrow Pb^2+ + 2Br-}$

$\mathrm{K_{sp} = [Pb^{2+}][Br^-]^2 = 6.2\times 10^{-6}}$

We could calculate the concentration of bromide ion that is the highest one at which none of the lead ion will precipitate.  This will be the value where the solubility product exactly equals the value of Ksp.

$\mathrm{K_{sp} = [Pb^{2+}][Br^-]^2 = 6.2\times 10^{-6} = (1\times 10^{-3})[Br^-]^2}$

$\mathrm{[Br^-]^2 = 6.2\times 10^{-3} \hspace{60px} [Br^-] = 7.87\times 10^{-2}}$

Any concentration of bromide higher than 7.87×10-2 M will cause some of the lead to precipitate as lead bromide.  Suppose we had another metal ion in solution besides lead, and this other ion formed a bromide complex that was much less soluble than lead bromide.  We could calculate the concentration of bromide needed to precipitate 99.9% of this other ion, and then compare that value to 7.87×10-2 M.  If the value was less than 7.87×10-2 M, it is theoretically possible to precipitate this other ion in the presence of lead.  If the value is greater than 7.87×10-2 M, lead bromide will start to precipitate and interfere with the separation.

If we want to precipitate 99.9% of the lead, that means that 0.1% remains.  Since the lead concentration was initially 1×10-3 M, the final concentration of Pb2+after 99.9% precipitates will be 1×10-6 M.  We can plug this into the Ksp expression to solve for the concentration of bromide that is needed to precipitate 99.9% of the lead.

$\mathrm{K_{sp} = [Pb^{2+}][Br^-]^2 = 6.2\times 10^{-6} = (1\times 10^{-6})[Br^-]^2}$

$\mathrm{[Br^-]^2 = 6.2 \hspace{60px} [Br^-] = 2.49}$

So a bromide concentration of 2.49 M would be needed to precipitate 99.9% of the lead ion as lead bromide in this solution.  This is a reasonably high concentration of bromide ion.  We could presumably get that high a level with a solution of hydrobromic acid.  The solubility of sodium bromide might be as high as this, but it is getting to be a bit of a high concentration of bromide to precipitate out the lead.