A solution contains 0.025 M Ca2+ and 0.015 M Ag+. Determine whether it is possible to precipitate 99% of Ca2+ by sulphate without precipitating Ag+?

Ksp CaSO4= 2.4*10^-52.4105
Ksp Ag2SO4= 1.5*10^-51.5105

1 Answer
Dec 12, 2017

It is not possible to precipitate 99 % of the "Ca"^"2+"Ca2+ without also precipitating some of the "Ag"^"+"Ag+.

Explanation:

Step 1. Write the equations for the equilibria

"CaSO"_4"(s)" ⇌ "Ca"^"2+""(aq)" + "SO"_4^"2-""(aq)"CaSO4(s)Ca2+(aq)+SO2-4(aq)

K_text(sp) = ["Ca"^"2+"]["SO"_4^"2-"] = 2.4 × 10^"-5"Ksp=[Ca2+][SO2-4]=2.4×10-5

"Ag"_2"SO"_4"(s)" ⇌ "2Ag"^"+""(aq)" + "SO"_4^"2-""(aq)"Ag2SO4(s)2Ag+(aq)+SO2-4(aq)

K_text(sp) = ["Ag"^"+"]^2["SO"_4^"2-"] = 1.5 × 10^"-5"Ksp=[Ag+]2[SO2-4]=1.5×10-5

Step 2. Calculate the sulfate concentration needed to precipitate the calcium

If we precipitate 99 % of the "Ca"^"2+"Ca2+, we have 1 % of the original calcium remaining.

0.01 × "0.025 mol/L" = 2.5 × 10^"-4" color(white)(l)"mol/L"0.01×0.025 mol/L=2.5×10-4lmol/L

["Ca"^"2+"]["SO"_4^"2-"] = 2.4 × 10^"-5"[Ca2+][SO2-4]=2.4×10-5

["SO"_4^"2-"] = (2.4 × 10^"-5")/(["Ca"^"2+"]) = (2.4 × 10^"-5")/(2.5 ×10^"-4") = 9.6 × 10^"-2"[SO2-4]=2.4×10-5[Ca2+]=2.4×10-52.5×10-4=9.6×10-2

So, ["SO"_4^"2-"][SO2-4] must equal 9.6 × 10^"-2"color(white)(l)"mol/L"9.6×10-2lmol/L tp precipitate 99 % of the "Ca"^"2+"Ca2+.

Step 3. Will this precipitate any of the silver?

Q_text(sp) = ["Ag"^"+"]^2["SO"_4^"2-"] = (0.015)^2 × 9.6 × 10^"-2" = 2.2 × 10^"-5"Qsp=[Ag+]2[SO2-4]=(0.015)2×9.6×10-2=2.2×10-5

Q_text(sp) > K_text(sp)Qsp>Ksp

Thus, some silver ion will precipitate.

It is not possible to precipitate 99 % of the "Ca"^"2+"Ca2+ without also precipitating some of the "Ag"^"+"Ag+.