What Sulfate Concentration Initiates Precipitation in a Mixed Nitrate Solution?

[jpd5184]

Homework Statement

consider a solution that initially contains 0.15M Sr(NO3)2 and 0.25M Ba(NO3)2. The sulfate ion concentration is gradually increased by the slow addition of solid Na2SO4.
a. What is the sulfate ion concentration when barium sulfate just begins to precipitate
b. what is the sulfate concentration when strontium sulfate just begins to precipitate

Homework Equations

a. Ba(NO3)2 ---- Ba^2+ + 2NO3-
Ksp=[NO3-][Ba^2+]
Qsp=[NO3-][Ba^2+] (initial concentrations)
b. Sr(NO3)2 ---- Sr^2+ + NO3-
ksp=[NO3-][Sr^2+]
Qsp=[NO3-][Sr^2+] (initial concentrations)

The Attempt at a Solution

knowing the ksp for each i can solve the equations for the unknown

 

[Borek]

Something wrong either with the wording or with your understanding of the problem. Nitrates don't precipitate, sulfates do.

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[jpd5184]

i changed the problem. this is what it should be but doesn't make sense since their is no sulfate in the problem

 

[chemisttree]

Ignore the nitrate ions. They are spectators in this process, only there to provide a countercharge to the Sr and Ba. Given the known [Sr] and [Ba], how much sulfate (read that as [SO4-2]) do you need to add to reach Ksp for SrSO4 and BaSO4?

 

[DeeNos]

concentration at which precipitation begins for each compound. This is known as the solubility product, and it represents the maximum concentration of ions that can exist in a solution before precipitation occurs.

a. For barium sulfate, the Ksp is 1.1x10^-10. Using the equation Ksp=[NO3-][Ba^2+], we can solve for the concentration of sulfate ions when barium sulfate begins to precipitate.
1.1x10^-10 = (x)(0.25)
x = 4.4x10^-10 M

Therefore, when the sulfate ion concentration reaches 4.4x10^-10 M, barium sulfate will begin to precipitate.

b. For strontium sulfate, the Ksp is 3.9x10^-7. Using the equation Ksp=[NO3-][Sr^2+], we can solve for the concentration of sulfate ions when strontium sulfate begins to precipitate.
3.9x10^-7 = (x)(0.15)
x = 2.6x10^-6 M

Therefore, when the sulfate ion concentration reaches 2.6x10^-6 M, strontium sulfate will begin to precipitate.

In conclusion, by knowing the solubility product for each compound and using the appropriate equations, we can determine the concentration of sulfate ions at which precipitation will occur for both barium sulfate and strontium sulfate in the given solution. This information can be useful in controlling and predicting precipitation in various chemical reactions and processes.