For AgI: (K_sp = [Ag^+][I^-] \Rightarrow [Ag^+] = \fracK_sp[I^-] = \frac8.5 \times 10^-170.010 = 8.5 \times 10^-15 , M)
AgI requires a much lower [Ag⁺] ((8.5 \times 10^-15 M)) to precipitate than AgCl ((1.8 \times 10^-8 M)). Therefore, AgI precipitates first . fractional precipitation pogil answer key best
AgCl begins to precipitate when [Ag⁺] reaches (1.8 \times 10^-8 M). At this [Ag⁺], the remaining [I⁻] is found from the (K_sp) of AgI: For AgI: (K_sp = [Ag^+][I^-] \Rightarrow [Ag^+] =
[ [I^-] = \fracK_sp(\textAgI)[Ag^+] = \frac8.5 \times 10^-171.8 \times 10^-8 = 4.7 \times 10^-9 , M ] At this [Ag⁺], the remaining [I⁻] is found
Use the detailed explanations above to check your POGIL answers, but more importantly, practice the calculations repeatedly. Cover the answers, re-derive the [Ag⁺] thresholds, and test yourself on the “what if” scenarios. That’s the pathway from rote answers to genuine mastery.
Now, go separate those ions with confidence.
The salt with the smaller (K_sp) requires a lower concentration of the common ion to reach saturation. This is the cardinal rule of fractional precipitation. Learning Objective 2: Calculating Ion Concentration at the Second Precipitation Point Question: As you continue adding AgNO₃, AgI continues to precipitate. At the moment just before AgCl begins to precipitate, what is the concentration of I⁻ remaining in solution?