How do you use the #K_"sp"# value to calculate the molar solubility of the following compound in pure water?
#"BaSO"_4#
#K_"sp" = 1.07xx10^(-10)#
1 Answer
Here's how you can do that.
Explanation:
What you need to do here is set up an ICE table based on the equilibrium reaction that describes the way barium sulfate,
The salt is considered insoluble in water, so right from the start you know that the solution will contain very little amounts of dissolved ions, since most of the compound will remain undissociated as a solid.
So, barium sulfate will dissolve in very small quantities to produce
#"BaSO"_ (4(aq)) rightleftharpoons "Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-)#
Notice that for every mole of barium sulfate that dissolves in solution, you get one mole of barium cations,
If you take
#"BaSO"_ (4(aq)) rightleftharpoons " ""Ba"_ ((aq))^(2+) " "+" " "SO"_ (4(aq))^(2-)#
By definition, the solubility product constant,
#K_(sp) = ["Ba"^(2+)] * ["SO"_4^(2-)]#
Since the expression of
#K_(sp) = s * s = s^2#
In your case, you have
#1.07 * 10^(-10) = s^2#
Solve for
#s = sqrt(1.07 * 10^(-10)) = 1.03 * 10^(-5)#
Since
#s = color(green)(|bar(ul(color(white)(a/a)color(black)(1.03 * 10^(-5)"M")color(white)(a/a)|)))#
This means that when you add barium sulfate to water, you can only hope to dissolve