but using the current concentrations of ions. Precipitation begins when
Equation: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ Expression: $K_sp = [Ag^+][Cl^-]$ Calculation: $$[Ag^+] = \fracK_sp[Cl^-] = \frac1.8 \times 10^-100.010 = \mathbf1.8 \times 10^-8\ M$$ fractional precipitation pogil answer key best
The first precipitate will continue to form as more $Ag^+$ is added. Eventually, the $[Ag^+]$ rises high enough that the second anion begins to precipitate. This is the critical moment for separation. but using the current concentrations of ions
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 ] This is the critical moment for separation
Suppose [I⁻] = (1.0 \times 10^-10 M) and [Cl⁻] = 0.10 M. Then:
A "best-in-class" answer key for this topic should clearly explain the following steps:
"Right," Leo said. "But the constants in the textbook—the $K_sp$ for Silver Chromate—is listed as $1.1 \times 10^-12$. But the constants on the sheet you're projecting... they use $1.2 \times 10^-12$."