Question

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^-5`
Ksp Ag2SO4= `1.5*10^-5`

Answer 1

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

Explanation:

Step 1. Write the equations for the equilibria

`"CaSO"_4"(s)" ⇌ "Ca"^"2+""(aq)" + "SO"_4^"2-""(aq)"`

`K_text(sp) = ["Ca"^"2+"]["SO"_4^"2-"] = 2.4 × 10^"-5"`

`"Ag"_2"SO"_4"(s)" ⇌ "2Ag"^"+""(aq)" + "SO"_4^"2-""(aq)"`

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

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

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

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

`["Ca"^"2+"]["SO"_4^"2-"] = 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"`

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

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"`

`Q_text(sp) > K_text(sp)`

Thus, some silver ion will precipitate.

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