Fractional — Precipitation Pogil Answer Key

To find when a specific ion will start to precipitate, you set . If you are adding a carbonate ( cap C cap O sub 3 raised to the 2 minus power ) to a solution of cap Z n raised to the 2 plus power , you use the formula:

Using the known initial concentrations of your anions, solve for the exact concentration of Silver (

[ [Cl^-] \text to ppt Ag^+ = \fracK_sp(AgCl)[Ag^+] = \frac1.8\times 10^-100.10 = 1.8\times 10^-9 \text M ] [ [Cl^-] \text to ppt Pb^2+ = \sqrt\fracK_sp(PbCl_2)[Pb^2+] = \sqrt\frac1.7\times 10^-50.10 = \sqrt1.7\times 10^-4 \approx 0.013 \text M ] Since (1.8\times 10^-9 \text M < 0.013 \text M), AgCl precipitates first . fractional precipitation pogil answer key

Fractional Precipitation: A Guide to the Process and POGIL Concepts

To find the exact concentration needed to start precipitation, set and solve for For : To find when a specific ion will start

Find the concentration of the added reagent needed to start the precipitation. Plug that value back into the cap K sub s p end-sub expression of the substance.

AgI(s)⇌Ag+(aq)+I−(aq)AgI open paren s close paren is in equilibrium with Ag raised to the positive power open paren a q close paren plus I raised to the negative power open paren a q close paren Look at the provided table or reference data: POGIL Answer Key Logic: Because the Kspcap K sub s p end-sub is significantly smaller than the Kspcap K sub s p end-sub is much less soluble. Therefore, will precipitate first . Plug that value back into the cap K

Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq)Ksp=[Ag+]2[CrO42−]Ag sub 2 CrO sub 4 open paren s close paren is in equilibrium with 2 Ag raised to the positive power open paren a q close paren plus CrO sub 4 raised to the 2 minus power open paren a q close paren space cap K sub s p end-sub equals open bracket Ag raised to the positive power close bracket squared open bracket CrO sub 4 raised to the 2 minus power close bracket

At the moment AgCl just begins to precipitate, what is the concentration of I⁻ remaining in the solution? What fraction of the original I⁻ remains?