Will a solid form when two solutions are mixed? Compare the reaction quotient of the ions with Ksp. The one step people skip is dilution: after mixing, every ion is in a larger volume.
Key idea
Find each ion's concentration after mixing, $c = \dfrac{c_0V_0}{V_\text{total}}$. Compute $Q$ the same way as $K_{sp}$. If $Q > K_{sp}$, a precipitate forms; if $Q < K_{sp}$, it does not; if $Q = K_{sp}$, the solution is just saturated.
Why it works
$Q$ above $K_{sp}$ means more ions are in solution than the equilibrium allows, so some combine into solid until the product drops back to $K_{sp}$. Below $K_{sp}$, the solution is unsaturated and more solid could even dissolve.
Worked examples
Worked example: A precipitate forms
$100\ \text{mL}$ of $0.0010\ \text{M}$ $\mathrm{AgNO_3}$ is mixed with $100\ \text{mL}$ of $0.0010\ \text{M}$ NaCl. Does AgCl ($K_{sp} = 1.8 \times 10^{-10}$) precipitate?
- The total volume is $200\ \text{mL}$, so each concentration halves: $[\mathrm{Ag^+}] = [\mathrm{Cl^-}] = 5.0 \times 10^{-4}\ \text{M}$.
- $Q = (5.0 \times 10^{-4})^2 = 2.5 \times 10^{-7}$.
- $Q > K_{sp}$ by more than a thousandfold: AgCl precipitates.
Answer: Yes.
Worked example: No precipitate
$50.0\ \text{mL}$ of $2.0 \times 10^{-4}\ \text{M}$ $\mathrm{Pb(NO_3)_2}$ is mixed with $50.0\ \text{mL}$ of $2.0 \times 10^{-3}\ \text{M}$ KI. Does $\mathrm{PbI_2}$ ($K_{sp} = 9.8 \times 10^{-9}$) form?
- After mixing (100 mL): $[\mathrm{Pb^{2+}}] = 1.0 \times 10^{-4}\ \text{M}$, $[\mathrm{I^-}] = 1.0 \times 10^{-3}\ \text{M}$.
- $$Q = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2 = (1.0 \times 10^{-4})(1.0 \times 10^{-3})^2 = 1.0 \times 10^{-10}$$
- $Q < K_{sp}$: no precipitate.
Answer: No.
Worked example: The concentration that starts precipitation
What $[\mathrm{SO_4^{2-}}]$ is needed to begin precipitating $\mathrm{BaSO_4}$ ($K_{sp} = 1.1 \times 10^{-10}$) from $0.0010\ \text{M}$ $\mathrm{Ba^{2+}}$?
- Precipitation starts when $Q = K_{sp}$: $(0.0010)[\mathrm{SO_4^{2-}}] = 1.1 \times 10^{-10}$.
- $[\mathrm{SO_4^{2-}}] = 1.1 \times 10^{-7}\ \text{M}$.
Answer: $1.1 \times 10^{-7}\ \text{M}$.
Worked example: Which precipitates first?
A solution is $0.010\ \text{M}$ in both $\mathrm{Cl^-}$ and $\mathrm{CrO_4^{2-}}$. Silver nitrate is added slowly. Which forms first, AgCl ($K_{sp} = 1.8 \times 10^{-10}$) or $\mathrm{Ag_2CrO_4}$ ($1.1 \times 10^{-12}$)?
- AgCl starts when $[\mathrm{Ag^+}] = \dfrac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8}\ \text{M}$.
- $\mathrm{Ag_2CrO_4}$ starts when $[\mathrm{Ag^+}]^2(0.010) = 1.1 \times 10^{-12}$: $[\mathrm{Ag^+}] = 1.0 \times 10^{-5}\ \text{M}$.
- AgCl needs far less silver, so it forms first. By the time red $\mathrm{Ag_2CrO_4}$ appears, $[\mathrm{Cl^-}]$ is down to $1.7 \times 10^{-5}\ \text{M}$: over 99.8% of the chloride has precipitated.
Answer: AgCl.
Common mistake
Forgetting that mixing two solutions dilutes both. Using the original concentrations here would make $Q$ four times too big for a 1:1 salt (and eight times for a 1:2 salt with equal volumes). Check: after mixing equal volumes, every concentration is half what it was.