(1)425ml of a saturated solution of lanthanum iodate, #La(IO_3)_3#, has #2.93 xx 10^-4# mole of #La^(3+)# (a)what is the concentration of #IO_3^-#? (b)calculate the solubility product of lantham iodate?

1 Answer
Feb 9, 2016

Answer:

Here's what I got.

Explanation:

So, you know that you're dealing with a #"425-mL"# sample of a saturated solution of lanthanum(III) iodate, #"La"("IO"_3)_3#.

You also know that this solution contains #2.93 * 10^(-4)# moles of lanthanum(III) cations, #"La"^(3+)#.

Use this information to find the molarity of the lanthanum(III) cations - do not forget that molarity uses liters of solution!

#color(blue)(c = n/V)#

#["La"^(3+)] = (2.93 * 10^(-4)"moles")/(425 * 10^(-3)"L") = 6.89 * 10^(-4)"M"#

Now, the equilibrium reaction that describes the partial dissociation of lanthanum iodate in aqueous solution looks like this

#"La"("IO"_3)_text(3(s]) rightleftharpoons "La"_text((aq])^(3+) + color(red)(3)"IO"_text(3(aq])^(-)#

Notice that you have a #1:1# mole ratio between lanthanum(III) iodate and the lanthanum(III) cations. This tells you that every mole of the solid that dissolves in solution will produce #1# mole of lanthanum(III) cations.

Likewise, the #1:color(red)(3)# mole ratio that exists between the solid and the iodate anions, #"IO"_3^(-)#, tells you that every mole of the solid that dissolves in solution will produce #color(red)(3)# moles of iodate anions.

So, using this mole ratio, you can say that the molarity of the iodate anions will be #color(red)(3)# times higher than that of the lanthanum(III) cations.

#["IO"_3^(-)] = color(red)(3) xx ["La"^(3+)]#

#["IO"_3^(-)] = color(red)(3) xx 6.89 * 10^(-4)"M" = color(green)(2.07 * 10^(-3)"M"#

The expression of the solubility product constant, #K_(sp)#, for lanthanum(III) iodate will look like this

#K_(sp) = ["La"^(3+)] * ["IO"_3^(-)]^color(red)(3)#

Plug in your values to get

#K_(sp) = 6.89 * 10^(-4) * (2.07 * 10^(-3))^color(red)(3)#

#K_(sp) = color(green)(6.12 * 10^(-12))#

The answers are rounded to three sig figs.