(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
Here's what I got.
Explanation:
So, you know that you're dealing with a
You also know that this solution contains
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
Likewise, the
So, using this mole ratio, you can say that the molarity of the iodate anions will be
#["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) = ["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.