[ [I^-] = \fracK_sp(\textAgI)[Ag^+] = \frac8.5 \times 10^-171.8 \times 10^-8 = 4.7 \times 10^-9 , M ]
For AgCl: ([Ag^+] = \frac1.8 \times 10^-100.010 = 1.8 \times 10^-8 , 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
A common mistake is to assume the ion with the smaller (K_sp) always precipitates first regardless of concentration. Is that true? Explain.
| Salt | (K_sp) | |------|------------| | AgCl | (1.8 \times 10^-10) | | AgI | (8.5 \times 10^-17) | [ [I^-] = \fracK_sp(\textAgI)[Ag^+] = \frac8
The 1:2 stoichiometry dramatically changes the required cation concentration. Conclusion: From Answer Key to Mastery Searching for the "fractional precipitation pogil answer key best" is a smart move—but the best key is the one that teaches you to think like a chemist. It doesn’t just confirm that AgI precipitates first; it shows you why the difference in (K_sp) values by seven orders of magnitude guarantees a clean separation. It warns you about concentration reversals and stoichiometry traps. And it prepares you for lab applications and exams alike.
Now, go separate those ions with confidence. Is that true
By the time AgCl starts to precipitate, the [I⁻] has dropped from 0.010 M to (4.7 \times 10^-9 M). That’s a decrease by a factor of over 2 million. The separation is essentially complete.