Question

Will Cadmium hydroxide precipitate from 0.01 M solution of `"CdCl"_2` at pH 9. `K_(sp)` of `"Cd"("OH")_2 = 2.5 xx 10^(-14)"M"^3`? Please solve,thanks!

Answer 1

Yes, a precipitate will form.

Explanation:

Your strategy here will be to calculate the concentrations of the cadmium(II) cations, `"Cd"^(2+)`, and of the hydroxide anions, `"OH"^(-)`, and figure out if these concentrations are high enough to ensure that a precipitate is formed.

So, you know that the pH of the solution is equal to `9`. As you know, an aqueous solution at room temperature has

`color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " POH" = 14)color(white)(a/a)|)))`

This means that the pOH of the solution, which can be used to find the concentration of hydroxide anions, will be equal to

`"pOH" = 14 - 9 = 5`

This means that you have

`["OH"^(-)] = 10^(-"pOH") = 10^(-5)"M"`

Cadmium chloride, `"CdCl"_2`, is a soluble ionic compound that dissociates completely in aqueous solution to form cadium(II) cations and chloride anions

`"CdCl"_text(2(aq]) -> "Cd"_text((aq])^(2+) |+ 2"Cl"_text((aq])^(-)`

Notice that one mole of cadmium chloride produces one mole of cadmium cations in solution. This means that you have

`["Cd"^(2+)] = ["CdCl"_2] = 10^(-2)"M"`

The partial dissociation of cadmium hydroxide in aqueous solution looks like this

`"Cd"("OH")_text(2(s]) rightleftharpoons "Cd"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)`

By definition, the solubility product constant, `K_(sp)`, for cadmium hydroxide will be equal to

`K_(sp) = ["Cd"^(2+)] * ["OH"^(-)]^color(red)(2)`

Now, in order to determine if a precipitate if formed or not, you need to calculate the reaction quotient, `Q_(sp)`.

The difference between `K_(sp)` and `Q_(sp)` lies in the fact that the former is calculated using equilibrium concentrations, while the latter is calculated using the concentrations of the ions at a given time.

In other words, `Q_(sp)` can be used to determine if a precipitate will form. In order for a precipitate to form, you need to have

`Q_(sp) > K_(sp)`

At this point, a precipitate will form until you reach saturation, i.e until `Q_(sp) = K_(sp)`.

If `Q_(sp) < K_(sp)`, the solution is unsaturated and a precipitate will not form.

Plug in the concentrations of the two ions to get

`Q_(sp) = 10^(-2)"M" * (10^(-5))^color(red)(2)"M"^color(red)(2) = 10^(-12)"M"^3`

Since you have

`10^(-12)"M"^3 > 2.5 * 10^(-14)"M"^3`

cadmium hydroxide will indeed precipitate out of solution.