Fractional Precipitation Pogil Answer Key Best File

For AgI: (K_sp = [Ag^+][I^-] \Rightarrow [Ag^+] = \fracK_sp[I^-] = \frac8.5 \times 10^-170.010 = 8.5 \times 10^-15 , M)

A common mistake is to assume the ion with the smaller (K_sp) always precipitates first regardless of concentration. Is that true? Explain. fractional precipitation pogil answer key best

Second precipitate (PbBr₂) begins at [Pb²⁺] = (2.64 \times 10^-3 M). At that [Pb²⁺], [CrO₄²⁻] remaining is: [ [CrO_4^2-] = \frac2.8 \times 10^-132.64 \times 10^-3 = 1.06 \times 10^-10 M ] For AgI: (K_sp = [Ag^+][I^-] \Rightarrow [Ag^+] =

PbCrO₄ precipitates first (much lower [Pb²⁺]). fractional precipitation pogil answer key best

The [Br⁻] is still essentially 0.050 M (negligible precipitation of PbBr₂ has occurred yet).

For AgCl: ([Ag^+] = \frac1.8 \times 10^-100.010 = 1.8 \times 10^-8 , M)