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)#

Answer 1

Here's how you can do that.

What you need to do here is set up an ICE table based on the equilibrium reaction that describes the way barium sulfate, #"BaSO"_4#, dissolves in aqueous solution.

Since most of the compound will remain undissociated as a solid and the salt is thought to be insoluble in water, you can tell from the outset that the solution will contain very little dissolved ions.

Thus, very little amounts of barium sulfate will dissolve 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, #"Ba"^(2+)#, and one mole of sulfate anions, #"SO"_4^(2-)#.

If you take #s# to be the molar solubility of the salt, i.e. the number of moles of barium sulfate that dissolve to release ions per liter of solution, you can use an ICE table to write

#"BaSO"_ (4(aq)) rightleftharpoons " ""Ba"_ ((aq))^(2+) " "+" " "SO"_ (4(aq))^(2-)#

#color(purple)("I")color(white)(aaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaacolor(black)(0)# #color(purple)("C")color(white)(aaaacolor(black)(-)aaaaaaaaacolor(black)((+s))aaaaaacolor(black)((+s))# #color(purple)("E")color(white)(aaaacolor(black)(-)aaaaaaaaaaacolor(black)(s)aaaaaaaaaacolor(black)(s)#

By definition, the solubility product constant, #K_(sp)#, for this solubility equilibrium will be equal to

#K_(sp) = ["Ba"^(2+)] * ["SO"_4^(2-)]#

Since the expression of #K_(sp)# uses equilibrium concentrations, you can say that

#K_(sp) = s * s = s^2#

As for you, you've

#1.07 * 10^(-10) = s^2#

Solve for #s# to find

#s = sqrt(1.07 * 10^(-10)) = 1.03 * 10^(-5)#

Since #s# represents molar solubility, you can say that you have

#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 #1.03 * 10^(-5)# moles for every liter of solution.

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Answer 2

Jeremiah Adams

To calculate the molar solubility of a compound using the $K_{sp}$ value, you need to set up an equilibrium expression using the solubility of the compound (let's call it $x$ ). Then, you can substitute this value into the equilibrium expression and solve for $x$ , which represents the molar solubility of the compound in pure water.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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)#

#"BaSO"_4#

#K_"sp" = 1.07xx10^(-10)#