By working through these guided steps, you build a foundation in chemical equilibrium that is essential for more advanced topics like qualitative analysis and complex ion formation.
Always calculate the required precipitant concentration . For (Ag_2S) (very small (K_sp)) vs. (CuS), the sulfide ion needed might be different due to stoichiometry.
Precipitation begins when the concentration of ions in the solution is high enough that the reaction quotient ( ) exceeds the cap K sub s p end-sub of the salt. Chemistry LibreTexts : The solution is unsaturated; no precipitate forms. : The solution is saturated; it is at equilibrium. : The solution is supersaturated; a precipitate will form. Chemistry LibreTexts Step 1: Identifying the Salts and cap K sub s p end-sub
Find ([Cl^-]) when ([Ag^+] = 1.0\times 10^-5) M (complete precipitation): [ [Cl^-] = \fracK_sp(AgCl)[Ag^+] \textfinal = \frac1.8\times 10^-101.0\times 10^-5 = 1.8\times 10^-5 \text M ] At this ([Cl^-]), check if (PbCl_2) has started: (Q = [Pb^2+][Cl^-]^2 = (0.10)(1.8\times 10^-5)^2 = 3.24\times 10^-11) Compare to (K sp(PbCl_2) = 1.7\times 10^-5). (Q \ll K_sp), so (Pb^2+) is still in solution. Separation is possible.
Since (1.8 \times 10^-8 \text M) is much less than (0.041 \text M), (AgCl) reaches its (K_sp) first and precipitates.
