# Will a precipitate form?

## What happens when 10.0 mL of 0.2 M Sr(OH)2 is added to 10.0 mL of 0.2 M CuSO4? I don't have a ksp for Cu(OH)2 (in my booklet, I can find it online though), so how do I compare the trial ksp with the ksp?

Mar 1, 2017

I think that two precipitates will form.

#### Explanation:

More specifically, this reaction will produce copper(II) hydroxide, "Cu"("OH")_2, and strontium sulfate, ${\text{SrSO}}_{4}$, which will precipitate out of solution.

"Sr"("OH")_ (2(aq)) + "CuSO"_ (4(aq)) -> "SrSO"_ (4(s)) darr + "Cu"("OH") _(2(s)) darr

Keep in mind that strontium hydroxide is only slightly soluble in water, but its dissociation will be aided by the fact that the hydroxide anions will form a precipitate with the copper(II) cations.

You can calculate the concentration of the ions involved in the reaction after the two solutions are mixed. The total volume of the solution will be

${V}_{\text{sol" = "10.0 mL" + "10.0 mL" = "20.0 mL}}$

Since the volume of the solution doubles, the concentrations of the ions will be halved. You can thus say that you will have

["Sr"^(2+)] = "0.2 M"/2 = "0.1 M"

["OH"^(-)] = (2 xx "0.2 M")/2 = "0.2 M"

["Cu"^(2+)] = "0.2 M"/2 = "0.1 M"

["SO"_4^(2-)] = "0.2 M"/2 = "0.1 M"

For strontium sulfate, the ${K}_{s p}$ will look like this

${K}_{s p} = \left[{\text{Sr"^(2+)] * ["SO}}_{4}^{2 -}\right] = 3.44 \cdot {10}^{- 7}$

The solubility product quotient, ${Q}_{s p}$, is calculated using the concentrations of the ions in the resulting solution. You should know that

• ${Q}_{s p} > {K}_{s p} \to$ a precipitate will form until ${Q}_{s p} = {K}_{s p}$
• ${Q}_{s p} < {K}_{s p} \to$ a precipitate will not form

For strontium sulfate, you have

${Q}_{s p} = \text{0.1 M" * "0.1 M} = 1.0 \cdot {10}^{- 2}$

Since

${Q}_{s p} > {K}_{s p}$

so a precipitate of strontium sulfate will form.

For copper(II) hydroxide, you have

${K}_{s p} = {\left[{\text{Cu"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}} = 2.20 \cdot {10}^{- 20}$

In this case, the solubility product quotient will be equal to

Q_(sp) = "0.1 M" * ("0.2 M")^color(red)(2) = 4.0 * 10^(-3)

Since it's obvious that

${Q}_{s p} > {K}_{s p}$

a precipitate of copper(II) hydroxide will form.

Therefore, you can say that two precipitates will form when you mix the two solutions.